Integration is an essential part of basic calculus. Algebra plays a very important part to become proficient in this topic.
I have compiled some of the questions that I have encountered during my Math tutoring classes. Do take your time to try the questions and learn from the solutions I have provided below. Cheers ! =) .
\({\small 1.\enspace}\)
9709/32/F/M/17 – Paper 32 Feb March 2017 Pure Maths 3 No 10 \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \( \ {\small y \ = \ {(\ln x)}^{2} }\). The
x-coordinate of the point
P is equal to e, and the normal to the curve at
P meets the
x-axis at
Q.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the
x-coordinate of
Q.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Show that \({\small \displaystyle \int \ln x \ \mathrm{d}x \ = \ x \ln x \ – \ x \ + \ c }\), where
c is a constant.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the
x-axis and the normal
PQ.
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CMYjESYl61N/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 2.\enspace}\)
9709/32/F/M/19 – Paper 32 Feb March 2019 Pure Maths 3 No 10 \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \( \ {\small y \ = \ {\sin}^{3} x \sqrt{(\cos x)} \ }\) for \( \ {\small 0 \leq x \leq \large{ \frac{1}{2}} \pi } \), and its maximum point
M.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Using the substitution \({\small \ u \ = \ \cos x }\), find by integration the exact area of the shaded region bounded by the curve and the
x-axis.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Showing all your working, find the
x-coordinate of
M, giving your answer correct to 3 decimal places.
Check out my solution here:
\(\\[1pt]\)
Part (i):
https://www.instagram.com/p/CK2puaJliok/ Part (ii):
https://www.instagram.com/p/CK5_vibF5Co/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 3.\enspace}\)
9709/32/M/J/20 – Paper 32 May June 2020 Pure Maths 3 No 9 \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curves \( \ {\small \ y \ = \ \cos x \ }\) and \( \ {\small \ y \ = \ \large{ \frac{k}{1 \ + \ x} } }\), where
k is a constant, for \( \ {\small \ 0 \leq x \leq \large{ \frac{1}{2}} \pi } \). The curves touch at the point where
x =
p.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Show that
p satisfies the equation \( {\small \ \tan p \ = \ \large{ \frac{1}{1 \ + \ p} } }\).
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CLIxp2eFy7P/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 4.\enspace} \displaystyle \int_{1}^{a} \ln 2x \ \mathrm{d}x = 1.\) Find \({\small a} \).
We will use integration by parts to solve this question. The form can be written as follows:
\(\\[1pt]\)
\(\displaystyle \int {u\frac{{\mathrm{d}v}}{{\mathrm{d}x}}} \ \mathrm{d}x = uv \ – \int {\frac{{\mathrm{d}u}}{{\mathrm{d}x}}} v \ \mathrm{d}x \)
\(\\[1pt]\)
Let \(u = \ln 2x\) and \(v = x\)
\(\\[1pt]\)
Then,
\(\\[1pt]\)
\( \displaystyle \int {\ln 2x\frac{{\mathrm{d}x}}{{\mathrm{d}x}}} \ \mathrm{d}x = x \ln 2x \ – \int {\frac{{\mathrm{d} \ln 2x}}{{\mathrm{d}x}}} x \ \mathrm{d}x \)
\(\\[1pt]\)
\( \displaystyle \int {\ln 2x\frac{{\mathrm{d}x}}{{\mathrm{d}x}}} \ \mathrm{d}x = x \ln 2x \ – \int {\frac{{\large\frac{1}{2x}} . 2 . \mathrm{d}x} {{\mathrm{d}x}}} x \ \mathrm{d}x \)
\(\\[1pt]\)
Cancel out the like terms,
\(\\[1pt]\)
\( \require{cancel} \displaystyle \int {\ln 2x\frac{{\cancel{\mathrm{d}x}}}{\cancel{\mathrm{d}x}}} \ \mathrm{d}x = x \ln 2x \ – \int {\frac{{\large\frac{1}{\cancel{2}\cancel{x}}} . \cancel{2} . \cancel{\mathrm{d}x}} {\cancel{\mathrm{d}x}}} \cancel{x} \ \mathrm{d}x \)
\(\\[1pt]\)
\( \displaystyle \int \ln 2x \ \mathrm{d}x = x \ln 2x \ – \int \mathrm{d}x \)
\(\\[1pt]\)
\( \displaystyle \int \ln 2x \ \mathrm{d}x = x \ln 2x \ – \ x \)
\(\\[1pt]\)
After the simplification above, we will just have to substitute the upper and lower bounds of the integration and use algebra to get the solution for a.
\(\\[1pt]\)
\(\hspace{1.2em} \displaystyle \int_{1}^{a} \ln 2x \ \mathrm{d}x = 1\)
\(\\[1pt]\)
\(\hspace{1.2em} \Big[ x \ln 2x \ – \ x \Big]^{1}_{a} = 1 \)
\(\\[1pt]\)
\(\hspace{1.2em} \Big[ (a \ln 2a \ – \ a) \ – \ (\ln 2 \ – \ 1) \Big] = 1 \)
\(\\[1pt]\)
\(\hspace{1.2em} a \ln 2a \ – a \ – \ \ln 2 \ + \ 1 = 1 \)
\(\\[1pt]\)
\(\hspace{1.2em} a \ln 2a = a \ + \ \ln 2 \)
\(\\[1pt]\)
Divide both side of the equation by a,
\(\\[1pt]\)
\(\hspace{1.2em} \ln 2a = 1 \ + \ \frac{\ln 2}{a} \)
\(\\[1pt]\)
\(\hspace{1.2em} 2a = \exp(1 \ + \ \frac{\ln 2}{a}) \)
\(\\[1pt]\)
\(\hspace{1.2em} a = {\large\frac{1}{2}} \exp(1 \ + \ \frac{\ln 2}{a}) \)
\(\\[1pt]\)
\({\small 5.\enspace}\) Use the substitution \(u = \sin 4x\) to find the exact value of \(\displaystyle \int_{0}^{{\Large\frac{\pi}{24}}} \cos^{3} 4x \ \mathrm{d}x.\)
In this question, we will use the direct substitution method and some “tweaking” of the cosine function.
\(\\[1pt]\)
\(\\[7pt]\) Let \(u = \sin 4x\)
\(\\[10pt]\hspace{1.1em} \frac{\mathrm{d}u}{\mathrm{d}x} = 4\cos 4x \)
\(\hspace{1.1em} \mathrm{d}u = 4\cos 4x \ \mathrm{d}x \)
\(\\[1pt]\)
Then we tweak the cube of cosine function to make use the direct substitution with u and du.
\(\\[1pt]\)
\(\\[12pt]\) Since,
\(\\[15pt] \cos^{3} 4x = \cos^{2} 4x \times \cos 4x \)
\(\\[12pt]\) then,
\(\displaystyle \int \cos^{3} 4x \ \mathrm{d}x = \int (\cos^{2} 4x )( \cos 4x) \ \mathrm{d}x \)
\(\\[1pt]\)
\(\hspace{2.1em} = \displaystyle \int (1 \ – \ \sin^{2} 4x )( \cos 4x) \ \mathrm{d}x \)
\(\\[1pt]\)
\(\hspace{2.1em} = \displaystyle \int (1 \ – \ u^2 ) \times \frac{1}{4} \ \mathrm{d}u \)
\(\\[1pt]\)
\(\hspace{2.1em} = \frac{1}{4} \Big( u \ – \ {\large\frac{u^3}{3}} \Big) \ + \ C \)
\(\\[1pt]\)
We can then substitute u back to sin 4x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\hspace{1.2em} = \frac{1}{4} \Big[ \sin 4x \ – \ {\large\frac{{\sin}^{3} 4x}{3}} \Big]_{0}^{{\Large\frac{\pi}{24}}} \)
\(\\[1pt]\)
\(\hspace{1.2em} = \frac{1}{4} \Big[\Big( \sin \frac{\pi}{6} \ – \ {\large\frac{{\sin}^{3} \frac{\pi}{6}}{3}} \Big) \ – \ \Big( \sin 0 \ – \ {\large\frac{{\sin}^{3} 0}{3}} \Big) \Big] \)
\(\\[1pt]\)
\(\hspace{1.2em} = \frac{1}{4} \Big[\Big( 0.5 \ – \ {\large\frac{{0.5}^{3}}{3}} \Big) \ – \ \Big( 0 \ – \ 0 \Big) \Big] \)
\(\\[1pt]\)
\(\hspace{1.2em} = {\large \frac{11}{96}}\)
\(\\[1pt]\)
\({\small 6. \hspace{0.8em}(i).\hspace{0.8em}}\) Use the trapezium rule with 3 intervals to estimate the value of: \(\displaystyle \int_{{\Large\frac{\pi}{9}}}^{{\Large\frac{2\pi}{3}}} \csc x \ \mathrm{d}x\) giving your answer correct to 2 decimal places.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Using a sketch of the graph of \(y = \csc x\), explain whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).
\({\small \hspace{1.2em}(i). \enspace }\) We use trapezium rule when solving the integral analytically is not possible. The formula for trapezium rule with 3 intervals is shown as follows:
\(\\[1pt]\)
\(\hspace{1.2em} \displaystyle \int_{x_0}^{x_3} f(x) \ \mathrm{d}x \)
\(\\[1pt]\)
\(\hspace{1.8em} \approx \frac{1}{2}h \Big[ f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3) \Big]\)
\(\\[1pt]\)
\(\hspace{1.2em}\) with \(h = {\large \frac{ {x_3} \ – \ {x_0} }{3}}\)
\(\\[1pt]\)
So let’s us plug in each of the corresponding variables,
\(\\[1pt]\)
\(\\[12pt] \hspace{1.2em} x_0 = {\large\frac{\pi}{9}}\)
\(\\[12pt] \hspace{1.2em} x_3 = {\large\frac{2\pi}{3}}\)
\(\\[10pt] \hspace{1.2em} f(x) = \csc x \)
\(\\[12pt] \hspace{1.2em} h = {\large \frac{ {{\large\frac{2\pi}{3}}} \ – \ {{\large\frac{\pi}{9}}} }{3}}\)
\(\\[12pt] \hspace{2em} = \ {\large\frac{5\pi}{27}} \)
\(\\[12pt] \hspace{1.2em} f(x_0) = \csc {\large\frac{\pi}{9}} \)
\(\\[12pt] \hspace{3.8em} \approx \ 2.9238 \)
\(\\[12pt] \hspace{1.2em} f(x_1) = \csc {\large\frac{8\pi}{27}} \)
\(\\[12pt] \hspace{3.8em} \approx \ 1.2467 \)
\(\\[12pt] \hspace{1.2em} f(x_2) = \csc {\large\frac{13\pi}{27}} \)
\(\\[12pt] \hspace{3.8em} \approx \ 1.0017 \)
\(\\[12pt] \hspace{1.2em} f(x_3) = \csc {\large\frac{2\pi}{3}} \)
\(\\[12pt] \hspace{3.8em} \approx \ 1.1547 \)
\(\\[12pt] \displaystyle \int_{x_0}^{x_3} f(x) \ \mathrm{d}x \)
\(\\[10pt] \approx \ \frac{1}{2} \Big(\frac{5\pi}{27}\Big) {\small \big[ 2.9238 + 2(1.2467) + 2(1.0017) + 1.1547\big] }\)
\(\\[10pt] \approx \ 2.49 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}(ii). \enspace }\) The sketch of \(y = \csc x\) with the corresponding trapezium rule drawing is shown below.
\(\\[1pt]\)
\(\\[1pt]\)
\(\hspace{1.2em}\) It can be seen that the trapezium rule gives an overestimate of the true value of the integral. The green region shows the extra or “the error” from the approximation of the trapezium rule.
\(\\[1pt]\)
\({\small 7.\enspace}\) Solve these integrations.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{-\infty}^{\infty} \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}} \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(e\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(f\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x\)
Each of these integrals can be solved using trigonometric substitution. There are 3 generic forms to be used in conjunction with the trigonometric substitution. They are:
\(\\[1pt]\)
\({\small\hspace{1.2em} 1.\hspace{0.8em}}\) If the integral contains: \( (a^2 \ + \ x^2)\) then use \(x = a \tan \theta\)
\(\\[1pt]\)
\({\small\hspace{1.2em} 2.\hspace{0.8em}}\) If the integral contains: \( (a^2 \ – \ x^2)\) then use \(x = a \sin \theta\)
\(\\[1pt]\)
\({\small\hspace{1.2em} 3.\hspace{0.8em}}\) If the integral contains: \( (x^2 \ – \ a^2)\) then use \(x = a \sec \theta\)
\(\\[1pt]\)
Using one of the above forms, we can solve these integrals,
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[7pt] \hspace{1.2em}\) Let \({\small x = 2 \ \tan \theta}\)
\(\\[17pt] \hspace{2.1em}\) \({\small dx = 2 \ {\sec}^{2} \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x \)
\(\\[17pt]{\small\hspace{1.2em}} = \ \displaystyle \int \frac{2 \ {\sec}^{2} \theta}{ 4 \ {\tan}^{2} \theta \ + \ 4} \ \mathrm{d}\theta \)
\(\\[17pt]{\small\hspace{1.2em}} = \ \displaystyle \int \frac{2 \ {\sec}^{2} \theta}{ 4 \ ({\tan}^{2} \theta \ + \ 1)} \ \mathrm{d}\theta \)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\Large\frac{1}{2}} \ \require{cancel} \displaystyle \int \frac{\cancel{{\sec}^{2} \theta}}{ \cancel{{\sec}^{2} \theta }} \ \mathrm{d}\theta \)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\large\frac{1}{2}} \theta \ + \ C \)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\large\frac{1}{2}} \Big[ {\tan}^{-1} \big( \frac{x}{2} \big) \Big]_{0}^{\infty} \)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\large\frac{1}{2}} \Big[ \frac{\pi}{2} \ – \ 0 \Big] \)
\({\small\hspace{1.2em}} = \ {\large \frac{\pi}{4} }\)
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[7pt] \hspace{1.2em}\) Let \({\small x = 3 \ \sin \theta}\)
\(\\[17pt] \hspace{2.1em}\) \({\small dx = 3 \ \cos \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{3 \ \cos \theta}{\sqrt{9 \ – \ 9 \ {\sin}^{2}\theta}} \ \mathrm{d}\theta }\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{3 \ \cos \theta}{\sqrt{9} \ \sqrt{(1 \ – \ {\sin}^{2}\theta)}} \ \mathrm{d}\theta }\)
\(\\[25pt]{\small\hspace{1.2em} = \ \require{cancel} \displaystyle \int \frac{\cancel{3} \ \cancel{\cos \theta}}{\cancel{\sqrt{9}}\cancel{\sqrt{(1 \ – \ {\sin}^{2}\theta)}}} \ \mathrm{d}\theta }\)
Note that: \({\small (1 \ – \ {\sin}^{2}\theta) \ = \ {\cos}^{2}\theta}\)
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em} = \ \displaystyle \int \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \ \theta \ + \ C }\)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} = \ \Big[ {\sin}^{-1} \big( \frac{x}{3} \big) \Big]_{0}^{3} \)
\(\\[17pt]{\small\hspace{1.2em}} = \ \Big[ \frac{\pi}{2} \ – \ 0 \Big] \)
\({\small\hspace{1.2em}} = \ {\large \frac{\pi}{2} }\)
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{-\infty}^{\infty} \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[15pt] \hspace{1.2em}\) Let \({\small x = {\large\frac{2}{3}} \ \tan \theta}\)
\(\\[20pt] \hspace{2.1em}\) \({\small dx = {\large\frac{2}{3}} \ {\sec}^{2} \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[25pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{{\large\frac{2}{3}} \ {\sec}^{2} \theta}{9 \ \big({\large\frac{2}{3}} \ \tan \theta \big)^{2} \ + \ 4 } \ \mathrm{d}\theta }\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{{\large\frac{2}{3}} \ {\sec}^{2} \theta}{4 \ {\tan}^{2} \theta \ + \ 4 } \ \mathrm{d}\theta }\)
\(\\[25pt]{\small\hspace{1.2em} = \ \require{cancel} \displaystyle \ \int \frac{{\large\frac{2}{3}} {\cancel{\sec}^{2} \theta}}{4 \ (\cancel{{\tan}^{2} \theta \ + \ 1}) } \ \mathrm{d}\theta }\)
Note that: \({\small (1 \ + \ {\tan}^{2}\theta) \ = \ {\sec}^{2}\theta}\)
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em} = \ \displaystyle \frac{1}{6} \ \int \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \ {\large\frac{1}{6}} \theta \ + \ C }\)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\large\frac{1}{6}} \Big[ {\tan}^{-1} \big( \frac{3x}{2} \big) \Big]_{-\infty}^{\infty} \)
\(\\[17pt]{\small\hspace{1.2em}} = \ {\large\frac{1}{6}} \Big[ \frac{\pi}{2} \ + \ \frac{\pi}{2} \Big] \)
\({\small\hspace{1.2em}} = \ {\large \frac{\pi}{6} }\)
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}} \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[7pt] \hspace{1.2em}\) Let \({\small \sqrt{x} = \sin \theta}\)
\(\\[7pt] \hspace{3.6em} {\small x = {\sin}^{2} \theta}\)
\(\\[17pt] \hspace{2.9em}\) \({\small dx = 2 \ \sin \theta \ \cos \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{2 \ \sin \theta \ \cos \theta}{\sin \theta \ \sqrt{1 \ – \ {\sin}^{2}\theta}} \ \mathrm{d}\theta }\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{2 \ \sin \theta \ \cos \theta}{\sin \theta \ \sqrt{{\cos}^{2}\theta}} \ \mathrm{d}\theta }\)
Note that: \({\small (1 \ – \ {\sin}^{2}\theta) \ = \ {\cos}^{2}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} = \ \require{cancel} \displaystyle 2 \ \int \frac{\cancel{\sin \theta \ \cos \theta}}{ \cancel{\sin \theta \ \cos \theta}} \ \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \ 2 \ \displaystyle \int \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \ 2 \ \theta \ + \ C }\)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} = \ 2 \ \Big[ {\sin}^{-1} ( \sqrt{x} ) \Big]_{0}^{1} \)
\(\\[17pt]{\small\hspace{1.2em}} = \ 2 \ \Big[ \frac{\pi}{2} \ – \ 0 \Big] \)
\({\small\hspace{1.2em}} = \ \pi \)
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(e\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[7pt] \hspace{1.2em}\) Let \({\small x = \tan \theta}\)
\(\\[17pt] \hspace{2.1em}\) \({\small dx = {\sec}^{2} \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[25pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x\)
\(\\[25pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{{\sec}^{2} \theta}{{(1 \ + \ {\tan}^{2} \theta)}^{{\large\frac{3}{2}}}} \ \mathrm{d}\theta }\)
\(\\[25pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{{\sec}^{2} \theta}{{\sec}^{3} \theta} \ \mathrm{d}\theta }\)
Note that: \({\small (1 \ + \ {\tan}^{2}\theta) \ = \ {\sec}^{2}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} = \ \displaystyle \int \frac{1}{\sec \theta } \ \mathrm{d}\theta \)
\(\\[20pt]{\small\hspace{1.2em}} = \ \displaystyle \int \cos \theta \ \mathrm{d}\theta \)
\(\\[20pt]{\small\hspace{1.2em}} = \ \sin \theta \ + \ C \)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} = \ \Big[ {\large\frac{x}{\sqrt{{x}^{2} \ + \ 1}}} \Big]_{1}^{\infty} \)
\(\\[20pt]{\small\hspace{1.2em}} = \ \Big[ 1 \ – \ {\large\frac{1}{\sqrt{2}} } \Big] \)
\({\small\hspace{1.2em}} = \ 1 \ – \ {\large\frac{1}{2}}\sqrt{2}\)
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(f\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x\)
\(\\[1pt]\)
\(\\[7pt] \hspace{1.2em}\) Let \({\small x = \sec \theta}\)
\(\\[17pt] \hspace{2.1em}\) \({\small dx = \sec \theta \ \tan \theta \ \mathrm{d}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em}} \displaystyle \int \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{\sec \theta \ \tan \theta}{\sec \theta \ \sqrt{ {\sec}^{2}\theta \ – \ 1} } \ \mathrm{d}\theta }\)
\(\\[20pt]{\small\hspace{1.2em} = \ \displaystyle \int \frac{\sec \theta \ \tan \theta}{\sec \theta \ \tan \theta } \ \mathrm{d}\theta }\)
Note that: \({\small ({\sec}^{2}\theta \ – \ 1) \ = \ {\tan}^{2}\theta}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} = \ \require{cancel} \displaystyle \int \frac{\cancel{\sec \theta \ \tan \theta}}{\cancel{\sec \theta \ \tan \theta} } \ \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \ \int \mathrm{d}\theta }\)
\(\\[17pt]{\small\hspace{1.2em} = \theta \ + \ C }\)
Then we revert the variable \(\theta\) to
x and calculate the upper and lower bound of the integration.
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em}} = \Big[ {\cos}^{-1} \Big( {\large\frac{1}{x}} \Big) \Big]_{1}^{\infty} \)
\(\\[17pt]{\small\hspace{1.2em}} = \Big[ \frac{\pi}{2} \ – \ 0 \Big] \)
\({\small\hspace{1.2em}} = \ \frac{\pi}{2} \)
\(\\[1pt]\)
\({\small 8.\enspace}\) The diagram shows the curve \({\small y = {e}^{{\large – \frac{1}{2}x}} \ \sqrt{(1 \ + \ 2x)}}\) and its maximum point
M. The shaded region between the curve and the axes is denoted by
R.
\(\\[1pt]\)
\(\\[1pt]\)
\({\small \hspace{1.2em}(i). \enspace }\) Find the
x-coordinate of
M.
\(\\[1pt]\)
\({\small \hspace{1.2em}(ii). \enspace }\) Find by integration the volume of the solid obtained when
R is rotated completely about the
x-axis. Give your answer in terms of \({\small \pi}\) and e.
\({\small \hspace{1.2em}(i). \enspace }\) To find the x-coordinate of the maximum point M, we set \({\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 0}\).
\(\\[1pt]\)
To solve for \({\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}}}\), we could use the Product Rule of Differentiation.
\(\\[1pt]\)
\(\boxed{ \frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)}\)
\(\\[1pt]\)
\(\\[15pt]\) Let \(y = {f\left( x \right)g\left( x \right)}\)
\(\\[15pt] \hspace{2.1em} f(x) = \ {e}^{- \frac{1}{2}x} \)
\(\\[25pt] \hspace{2em} \frac{\mathrm{d}f(x)}{\mathrm{d}x} = \ – \frac{1}{2} \ {e}^{- \frac{1}{2}x} \)
\(\\[15pt] \hspace{2.1em} g(x) = \ \sqrt{(1 \ + \ 2x)} \)
\(\\[15pt] \hspace{4.3em} = \ (1 \ + \ 2x)^{{\large\frac{1}{2}}} \)
\(\\[20pt] \hspace{2.1em} \frac{\mathrm{d}g(x)}{\mathrm{d}x} = \ ({\large\frac{1}{2}}) \ (2) \ (1 \ + \ 2x)^{{- \large\frac{1}{2}}} \)
\(\\[15pt] \hspace{4.3em} = \ (1 \ + \ 2x)^{{- \large\frac{1}{2}}} \)
Then,
\(\\[1pt]\)
\(\\[15pt]\hspace{1.2em} {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 0}\)
\(\\[25pt] ({e}^{- \frac{1}{2}x})\Big({\large\frac{1}{\sqrt{1 \ + \ 2x}}}\Big) \ + \ (- \frac{1}{2} \ {e}^{- \frac{1}{2}x})(\sqrt{1+2x}) \ = \ 0\)
\(\\[10pt] ({e}^{- \frac{1}{2}x})\Big({\large\frac{1}{\sqrt{1 \ + \ 2x}}}\Big) \ = \ (\frac{1}{2} \ {e}^{- \frac{1}{2}x})(\sqrt{1+2x}) \)
\(\\[1pt]\)
Cancel out the like term \( ({e}^{- \frac{1}{2}x}) \),
\(\\[1pt]\)
\(\\[25pt] \require{cancel} (\cancel {{e}^{- \frac{1}{2}x}})\Big({\large\frac{1}{\sqrt{1 \ + \ 2x}}}\Big) \ = \ (\frac{1}{2}) \ (\cancel {{e}^{- \frac{1}{2}x}})(\sqrt{1+2x}) \)
\(\\[25pt] \hspace{3em} \Big({\large\frac{1}{\sqrt{1 \ + \ 2x}}}\Big) \ = \ (\frac{1}{2})(\sqrt{1+2x}) \)
\(\\[15pt] \hspace{4.6em} 1 \ + \ 2x \ = \ 2 \)
\(\\[25pt] \hspace{7.3em} x \ = \ {\large \frac{1}{2}} \)
\({\small \hspace{1.2em}(ii). \enspace }\) The volume of revolution of solid rotated about the x-axis is
\(\\[1pt]\)
\(\\[25pt] \hspace{2.5em} \displaystyle V \ = \ \pi \int_{a}^{b} {r}^{2} \ \mathrm{d}x \) – The disk method
The lower limit of the integration, a is the x value when \({\small y = 0}\).
\(\\[1pt]\)
\(\\[15pt] \hspace{3.5em} y \ = \ 0 \)
\(\\[20pt] {\small {e}^{{\large – \frac{1}{2}x}} \ \sqrt{(1 \ + \ 2x)} \ = \ 0}\)
\(\\[20pt] {\small {e}^{{\large – \frac{1}{2}{x}_{1}}} \ = \ 0} \quad \) or \(\quad {\small \sqrt{(1 \ + \ 2{x}_{2})} \ = \ 0}\)
\(\\[20pt] \hspace{1.6em} {\small {x}_{1} \ = \ \infty}\) (rejected)
\(\\[20pt] {\small (1 \ + \ 2{x}_{2}) \ = \ 0}\)
\(\\[20pt] \hspace{3em} {\small {x}_{2} \ = \ {\large -\frac{1}{2}} }\)
Therefore, the lower limit, \({\small a \ = \ {\large -\frac{1}{2}} }\).
\(\\[1pt]\)
The upper limit of the integration, \({\small b \ = \ 0 }\).
\(\\[1pt]\)
\(\\[15pt] \hspace{3.7em} {r}^{2} \ = \ {y}^{2} \)
\(\\[20pt] \hspace{5em} = \ \big[ {\small {e}^{{\large – \frac{1}{2}x}} \ \sqrt{(1 \ + \ 2x)}} \ \big]^{2} \)
\(\\[18pt] \hspace{5em} = \ {\small {e}^{-x} \ (1 \ + \ 2x) }\)
Hence, the volume is
\(\\[1pt]\)
\(\\[25pt] \hspace{2.5em} \displaystyle V \ = \ \pi \int_{{\large -\frac{1}{2}}}^{0} {e}^{-x} \ (1 \ + \ 2x) \ \mathrm{d}x \)
To solve \({\small \displaystyle \int {e}^{-x} \ (1 \ + \ 2x) \ \mathrm{d}x }\), we’ll use integration by parts,
\(\\[1pt]\)
\(\displaystyle \int {u\frac{{\mathrm{d}v}}{{\mathrm{d}x}}} \ \mathrm{d}x = uv \ – \int {\frac{{\mathrm{d}u}}{{\mathrm{d}x}}} v \ \mathrm{d}x \)
\(\\[1pt]\)
Let \(\quad {\small u \ = \ 1 \ + \ 2x }\)
\(\\[1pt]\)
\({\small \hspace{2em} \mathrm{d}u \ = \ 2 \ \mathrm{d}x }\)
\(\\[1pt]\)
\({\small \hspace{2.5em} v \ = \ -{e}^{-x} }\)
\(\\[1pt]\)
\({\small \hspace{2em} \mathrm{d}v \ = \ {e}^{-x} \ \mathrm{d}x }\)
\(\\[1pt]\)
\(\\[20pt] {\small \displaystyle \int (1 \ + \ 2x) \ ({e}^{-x}) \ \mathrm{d}x \ = \ (-{e}^{-x}) \ (1 \ + \ 2x) \ + \ 2 \int {e}^{-x} \ \mathrm{d}x }\)
\(\\[20pt] \hspace{2em} = \ {\small (-{e}^{-x}) \ (1 \ + \ 2x) \ – \ 2 {e}^{-x} \ + \ C }\)
\(\\[20pt] \hspace{2em} = \ {\small -{e}^{-x} \ – \ 2x {e}^{-x} \ – \ 2 {e}^{-x} \ + \ C }\)
\(\\[20pt] \hspace{2em} = \ {\small -2x {e}^{-x} \ – \ 3 {e}^{-x} \ + \ C }\)
Then, apply the upper and lower limit of the integration,
\(\\[1pt]\)
\(\\[25pt]{\small \displaystyle \int_{{\large -\frac{1}{2}}}^{0} \ {e}^{-x} \ (1 \ + \ 2x) \ \mathrm{d}x }\)
\(\\[20pt] \hspace{2em} = \ -2 \ \Big[ x {e}^{-x}\Big]_{ -\frac{1}{2}}^{0} \ – \ 3 \Big[ {e}^{-x}\Big]_{ -\frac{1}{2}}^{0} \)
\(\\[20pt] \hspace{2em} = \ -2 \ \Big[ 0 \ + \ \frac{1}{2}{e}^{\frac{1}{2}}\Big] \ – \ 3 \Big[1 \ – \ {e}^{\frac{1}{2}} \Big] \)
\(\\[20pt] \hspace{2em} = \ 2{e}^{\frac{1}{2}} \ – \ 3 \)
Finally, the volume is
\(\\[1pt]\)
\(\\[25pt] \hspace{2.5em} \displaystyle V \ = \ \pi \int_{{\large -\frac{1}{2}}}^{0} {e}^{-x} \ (1 \ + \ 2x) \ \mathrm{d}x \)
\(\hspace{3.8em} = \ ( 2 \pi \ \sqrt{e} \ – \ 3 \pi )\) cubic units
\(\\[1pt]\)
\({\small 9.\enspace}\)
9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 2 \(\\[1pt]\)
Find the exact value of \(\displaystyle \int_{0}^{1} (2 \ – \ x) \mathrm{e}^{-2x} \ \mathrm{d}x \).
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\(\\[1pt]\)
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\({\small 10.\enspace}\)
9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 7(a), (b), (c) \(\\[1pt]\)
Let \({\small f(x) \ = \ {\large \frac{ 2 }{(2x \ – \ 1)( 2x \ + \ 1 ) }} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using your answer to part (a), show that
\(\\[1pt]\)
\({\scriptsize {\Big( f(x) \Big)}^{2} \ = \ {\large \frac{ 1 }{ {(2x \ – \ 1)}^{2} }} \ – \ {\large \frac{ 1 }{ (2x \ – \ 1) }} }\)
\(\\[1pt]\)
\({\hspace{3em} \scriptsize \ + \ {\large \frac{ 1 }{ (2x \ + \ 1)}} \ + \ {\large \frac{ 1 }{ {(2x \ + \ 1)}^{2} }} . }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Hence show that \(\displaystyle \int_{1}^{2} {\Big( f(x) \Big)}^{2} \ \mathrm{d}x = \frac{2}{5} + \frac{1}{2} \ln \frac{5}{9}\).
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\({\small 11.\enspace}\)
9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 3 \(\\[1pt]\)
Find the exact value of \(\displaystyle \int_{1}^{4} x^{\frac{3}{2}} \ln x \ \mathrm{d}x \).
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\(\\[1pt]\)
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\({\small 12.\enspace}\)
9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 4 \(\\[1pt]\)
A curve has equation \( y = \cos x \sin 2x \).
\(\\[1pt]\)
Find the \(x\)-coordinate of the stationary point in the interval \(0 \lt x \lt \frac{1}{2}\pi\), giving your answer correct to 3 significant figures.
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\(\\[1pt]\)
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\({\small 13.\enspace}\)
9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 6(a), (b) \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \(y = {\large\frac{x}{1+{3x}^{4}} } \), for \(x \geq 0\), and its maximum point \(M\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using the substitution \(u = \sqrt{3}{x}^{2}\), find by integration the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
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\(\\[1pt]\)
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\({\small 14.\enspace}\)
9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 9(a) \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curves \( y = \cos x \) and \( y = \frac{k}{1 \ + \ x} \), where \(k\) is a constant for \(0 \leq x \leq \frac{1}{2\pi}\). The curves touch at the point where \(x = p\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \(p\) satisfies the equation \( \tan p = \frac{1}{1+p} \).
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\({\small 15.\enspace}\)
9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 4(a), (b) \(\\[1pt]\)
The curve with equation \(y = { \mathrm{e} }^{2x} (\sin x + 3 \cos x) \) has a stationary point in the interval \(0 \leq x \leq \pi \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of this point, giving your answer correct to 2 decimal places.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Determine whether the stationary point is a maximum or a minimum.
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\({\small 16.\enspace}\)
9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 5(a), (b) \(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the quotient and remainder when \(2x^3 − x^2 + 6x + 3\) is divided by \(x^2 + 3\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using your answer to part (a), find the exact value of \(\displaystyle \int_{1}^{3} \frac{2x^3 \ – \ x^2 \ + \ 6x \ + \ 3}{x^2 \ + \ 3} \mathrm{d}x \).
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\(\\[1pt]\)
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\({\small 17.\enspace}\)
9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 7(a), (b) \(\\[1pt]\)
Let \( f(x) = { \large \frac{\cos x}{1 \ + \ \sin x} } \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \(f'(x) \lt 0 \) for all \(x\) in the interval \(-\frac{1}{2} \pi \lt x \lt \frac{3}{2} \pi\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find \(\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2} } f(x) \ \mathrm{d}x \). Give your answer in a simplified exact form.
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\(\\[1pt]\)
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\({\small 18.\enspace}\)
9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 6(a), (b) \(\\[1pt]\)
Let \({\small f(x) \ = \ {\large \frac{ 5a }{(2x \ – \ a)( 3a \ – \ x ) }} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \(\displaystyle \int_{a}^{ 2a } f(x) \ \mathrm{d}x = \ln 6\).
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\({\small 19.\enspace}\)
9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 9(c) \(\\[1pt]\)
Let \({\small f(x) \ = \ {\large \frac{ { \mathrm{e} }^{2x} \ + \ 1 }{{ \mathrm{e} }^{2x} \ – \ 1 }} }\), for \(x \gt 0\).
\(\\[1pt]\)
Find \(f'(x)\). Hence find the exact value of x for which \( f'(x) = -8 \).
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\({\small 20.\enspace}\)
9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 10(a), (b) \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \( y = \sin 2x \ {\cos}^{2} x \) for \(0 \leq x \leq \frac{1}{2} \pi \), and its maximum point \(M\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Using the substitution \(u = \sin x\), find the exact area of the region bounded by the curve and the \(x\)-axis.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the exact \(x\)-coordinate of \(M\).
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\(\\[1pt]\)
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\({\small 21.\enspace}\)
9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 4(a),(b) \(\\[1pt]\)
Let \({\small f(x) \ = \ {\large \frac{ 15 \ – \ 6x }{(1 \ + \ 2x)( 4 \ – \ x ) }} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence find \(\displaystyle \int_{1}^{ 2 } f(x) \ \mathrm{d}x \), giving your answer in the form \( \ln ( \frac{a}{b} ) \), where \(a\) and \(b\) are integers.
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\({\small 22.\enspace}\)
9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 8(a),(b) \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \( y = {\large \frac{ \ln x }{ x^4 } } \) and its maximum point \(M\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the exact \(x\)-coordinate of \(M\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) By using integration by parts, show that for all \( a \gt 1, \displaystyle \int_{1}^{ a } \frac{ \ln x }{ x^4 } \ \mathrm{d}x \lt \frac{1}{9} \).
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\({\small 23.\enspace}\)
9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 4 \(\\[1pt]\)
Using integration by parts, find the exact value of \( \displaystyle \int_{0}^{ 2 } {\tan}^{-1} \big( \frac{1}{2} x \big) \ \mathrm{d}x \).
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\(\\[1pt]\)
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\({\small 24.\enspace}\)
9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 6(a), (b) \(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Prove that \( \mathrm{cosec} 2\theta − \cot 2\theta \equiv tan \theta \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \( \displaystyle \int_{\frac{1}{4}\pi}^{ \frac{1}{3}\pi } (\mathrm{cosec} 2\theta − \cot 2\theta) \ \mathrm{d}\theta = \frac{1}{2} \ln 2 \).
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\({\small 25.\enspace}\)
9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 8 \(\\[1pt]\)
The equation of a curve is \( y = { \mathrm{e} }^{-5x} \ {\tan}^{2} x \) for \( -\frac{1}{2}\pi \lt x \lt \frac{1}{2}\pi \).
\(\\[1pt]\)
Find the \(x\)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.
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\({\small 26.\enspace}\)
9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 7(a) \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows the curve \( y = {\large \frac{ {\tan}^{-1} x }{ \sqrt{x} } } \) and its maximum point \(M\) where \(x = a\).
\(\\[1pt]\)
Show that a satisfies the equation \( a = \tan \Big( {\large \frac{2a}{1 \ + \ a^2} }\Big) \).
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\({\small 27.\enspace}\)
9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 9(a), (b) \(\\[1pt]\)
The equation of a curve is \( y = { x^{-\frac{2}{3}}} \ \ln x \ \) for \( x \gt 0 \). The curve has one stationary point.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the exact coordinates of the stationary point.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Show that \( \displaystyle \int_{1}^{ 8 } y \ \mathrm{d}x = 18 \ln 2 \ – \ 9 \).
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\({\small 28.\enspace}\)
9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 8(a), (b) \(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the quotient and remainder when \({\small 8x^3 + 4x^2 + 2x + 7}\) is divided by \({\small 4x^2 + 1}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence find the exact value of \( \displaystyle \int_{0}^{ \frac{1}{2} } \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} \ \mathrm{d}x \).
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\({\small 29.\enspace}\)
9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 11 \(\\[1pt]\)
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The diagram shows the curve \( \ y \ = \ \sin x \ \cos 2x \ \) for \({\small \ 0 \le x \le {\large\frac{1}{2}}\pi}\), and its maximum point \(M\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 significant figures.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using the substitution \( \ u = \cos x \), find the area of the shaded region enclosed by the curve and the \(x\)-axis in the first quadrant, giving your answer in a simplified exact form.
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\({\small 30.\enspace}\)
9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 8 \(\\[1pt]\)
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The diagram shows the curves with equations \( \ y \ = \ {x}^{ { \large – \frac{1}{2} } } \ \) and \( \ y \ = \ \frac{5}{2} \ – \ {x}^{ { \large \frac{1}{2} } } \). The curves intersect at the points \( \ A( {\large\frac{1}{4}},2) \ \) and \( \ B( 4 , {\large\frac{1}{2}}) \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the area of the region between the two curves.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The normal to the curve \( \ y \ = \ {x}^{ { \large – \frac{1}{2} } } \ \) at the point \( (1, 1) \) intersects the \(y\)-axis at the point \( (0, p) \). Find the value of \(p\).
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\({\small 31.\enspace}\)
9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 10 \(\\[1pt]\)
A curve has equation \( \ y = \mathrm{f}(x) \ \) and it is given that
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\( \hspace{2em} \mathrm{f}^{\prime}(x) = { \big( \frac{1}{2}x \ + \ k \big) }^{-2} \ – \ { ( 1 \ + \ k ) }^{-2} \),
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where \({\small \ k \ }\) is a constant. The curve has a minimum point at \({\small \ x \ = \ 2 }\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}^{\prime\prime}(x) \) in terms of \(k\) and \(x\), and hence find the set of possible values of \(k\).
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It is now given that \( {\small \ k \ = \ −3 \ }\) and the minimum point is at \({\small \ (2, \ 3\frac{1}{2}) }\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}(x) \).
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\({\small 32.\enspace}\)
9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 9 \(\\[1pt]\)
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The diagram shows part of the curve with equation \( \ {\small y^2 \ = \ x \ − \ 2 } \ \) and the lines \( \ {\small x \ = \ 5 } \ \) and \( \ {\small y \ = \ 1 } \). The shaded region enclosed by the curve and the lines is rotated through \( \ {\small 360^{\circ} \ }\) about the \(x\)-axis.
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Find the volume obtained.
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\({\small 33.\enspace}\)
9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 11 \(\\[1pt]\)
The gradient of a curve is given by
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\( \hspace{2em} {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 6{(3x \ – \ 5)}^{3} \ – \ k{x}^{2}}\),
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where \({\small \ k \ }\) is a constant. The curve has a stationary point at \( \ {\small (2, -3.5) }\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of \( \ k\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the equation of the curve.
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find \( {\small {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
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\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Determine the nature of the stationary point at \( \ {\small (2, -3.5) }\).
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\({\small 34.\enspace}\)
9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 8 \(\\[1pt]\)
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The diagram shows part of the curve with equation \( \ {\small y \ = \ {\large\frac{6}{x}} } \). The points \( \ {\small (1, 6) \ }\) and \( \ {\small (3, 2) \ }\) lie on the curve. The shaded region is bounded by the curve and the lines \( \ {\small y \ = \ 2 } \ \) and \( \ {\small x \ = \ 1 } \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the volume generated when the shaded region is rotated through \( \ {\small 360^{\circ} \ }\) about the \(y\)-axis.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The tangent to the curve at a point \(X\) is parallel to the line \( \ {\small y \ + \ 2x \ = \ 0 } \). Show that \(X\) lies on the line \( \ {\small y \ = \ 2x } \).
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\({\small 35.\enspace}\)
9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 10 \(\\[1pt]\)
The equation of a curve is
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\( \hspace{2em} y \ = 54x \ – \ {(2x \ – \ 7)}^{3} \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \( {\small \ {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the coordinates of each of the stationary points on the curve.
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Determine the nature of each of the stationary points.
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\({\small 36.\enspace}\)
9709/12/O/N/19 – Paper 12 June 2019 Pure Maths 1 No 10 \(\\[1pt]\)
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The diagram shows part of the curve \( \ {\small y \ = \ 1 \ – \ {\large\frac{4}{ {(2x \ + \ 1)}^{2} }} } \). The curve intersects the \(x\)-axis at \(A\). The normal to the curve at A intersects the \(y\)-axis at \(B\).
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\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Obtain expressions for \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \(\displaystyle \int y \ \mathrm{d}x\).
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\({\small\hspace{1.2em}\left(ii\right).\hspace{0.7em}}\) Find the coordinates of \(B\).
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\({\small\hspace{1.2em}\left(iii\right).\hspace{0.6em}}\) Find, showing all necessary working, the area of the shaded region.
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Follow my instagram for daily updates on many more solutions on other topics too!\({\small 1.\enspace}\) Find \(\displaystyle \int \frac{1}{x^2\sqrt{x^2 \ – \ 4}} \ \mathrm{d}x\) using the substitution \({\small x \ = \ 2 \sec \theta }\).
\({\small 2. \enspace}\) Find the exact value of \(\displaystyle \int_{1}^{e} x^4 \ \ln \ x \ \mathrm{d}x \).
\({\small 3. \enspace}\) Find the exact value of \(\displaystyle \int_{4}^{10} \frac{2x \ + \ 1}{(x \ – \ 3)^2} \ \mathrm{d}x \), giving your answer in the form of \({\small a \ + \ b \ \ln \ c}\), where a, b and c are integers.
\({\small 4. \enspace}\) Find the exact value of \(\displaystyle \int_{1}^{4} \frac{\ln \ x}{\sqrt{x}} \ \mathrm{d}x \).
\({\small 5. \enspace}\) Find the exact value of
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} {e}^{1 \ – \ 2x} \ \mathrm{d}x\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{-1}^{0} \big(
2 \ + \ \frac{1}{x \ – \ 1} \big) \ \mathrm{d}x\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \cot x \ \mathrm{d}x\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Using your result in (c), find also the exact value of \(\displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \csc 2x \ \mathrm{d}x\) by using the identity \(\cot x \ – \ \cot 2x \ \equiv \ \csc 2x\).
\({\small 6. \enspace}\) The diagram shows the part of the curve \({\small y \ = \ f(x)}\), where \({\small f(x) \ = \ p \ – \ {e}^{x} }\) and p is a constant. The curve crosses the y-axis at (0, 2).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the coordinates of the point where the curve crosses the x-axis.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) What is the area of the shaded region R?
\({\small 7. \enspace}\) Integrate the following:
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int \frac{x^2}{1 \ + \ {x}^{3}} \ \mathrm{d}x\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int x^4 \ \sin (x^5 \ + \ 2) \ \mathrm{d}x\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int e^{x} \ \sin x \ \mathrm{d}x\)
\({\small 8. \enspace}\) Let \(I \ = \ \displaystyle \int_{0}^{1} {\large \frac{\sqrt{x}}{2 \ – \ \sqrt{x}}} \ \mathrm{d}x\).
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Using the substitution \({\small u = \ 2 \ – \ \sqrt{x}}\), show that \(I \ = \ \displaystyle \int_{1}^{2} {\large \frac{2 {(2 \ – \ u)}^{2}}{u}} \ \mathrm{d}u\).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \(I \ = \ 8 \ \ln 2 \ – \ 5 \).
\({\small 9. \enspace}\) The constant a is such that
\({\small\hspace{3em}} \displaystyle \int_{0}^{a} x{e}^{{\large \frac{1}{2}x}} \mathrm{d}x \ = \ 6 \).
\({\small\hspace{3em}} a \ = \ 2 \ + \ {e}^{{\large -\frac{1}{2}a}}\).
\({\small 10. \enspace}\) Use the substitution \({\small u \ = \ 1 \ + \ 3 \ \tan x }\) to find the exact value of
\({\small\hspace{3em}} \ \displaystyle \int_{0}^{{\large\frac{\pi}{4}}} {\large \frac{\sqrt{1 \ + \ 3 \ \tan x}}{{\cos}^{2}x}} \ \mathrm{d}x\).
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
