9709/32/F/M/19 – Paper 32 Feb March 2019 No 10

Integration and Differentiation

Integration and Differentiation

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 ! =) .

More Integration Exercises can be found here.


EXAMPLE:

\({\small 1.\enspace}\) 9709/32/F/M/17 – Paper 32 Feb March 2017 Pure Maths 3 No 10
\(\\[1pt]\)
9709/32/F/M/17 – Paper 32 Feb March 2017 No 10
\(\\[1pt]\)
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.

\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/32/F/M/19 – Paper 32 Feb March 2019 Pure Maths 3 No 10
\(\\[1pt]\)
9709/32/F/M/19 – Paper 32 Feb March 2019 No 10
\(\\[1pt]\)
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.

\(\\[1pt]\)
\({\small 3.\enspace}\) 9709/32/M/J/20 – Paper 32 May June 2020 Pure Maths 3 No 9
\(\\[1pt]\)
9709/32/M/J/20 – Paper 32 May June 2020 No 9
\(\\[1pt]\)
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 \( {\small \ \tan p \ = \ \large{ \frac{1}{1 \ + \ p} } }\).

\(\\[1pt]\)
\({\small 4.\enspace} \displaystyle \int_{1}^{a} \ln 2x \ \mathrm{d}x = 1.\) Find \({\small a} \).

\(\\[1pt]\)
\({\small 5.\enspace}\) Use the substitution \(u = \sin 4x\) to find the exact value of \(\displaystyle \int_{0}^{{\Large\frac{\pi}{24}}} \cos^{3} 4x \ \mathrm{d}x.\)

\(\\[1pt]\)
\({\small 6. \hspace{0.8em}(i).\hspace{0.8em}}\) 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).

\(\\[1pt]\)
\({\small 7.\enspace}\) Solve these integrations.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{-\infty}^{\infty} \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}} \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(e\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(f\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x\)

\(\\[1pt]\)
\({\small 8.\enspace}\) 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.
\(\\[1pt]\)
Integration Example 5
\(\\[1pt]\)
\({\small \hspace{1.2em}(i). \enspace }\) 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.

\(\\[1pt]\)
\({\small 9.\enspace}\) 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 2
\(\\[1pt]\)
Find the exact value of \(\displaystyle \int_{0}^{1} (2 \ – \ x) \mathrm{e}^{-2x} \ \mathrm{d}x \).

\(\\[1pt]\)
\({\small 10.\enspace}\) 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 \(\displaystyle \int_{1}^{2} {\Big( f(x) \Big)}^{2} \ \mathrm{d}x = \frac{2}{5} + \frac{1}{2} \ln \frac{5}{9}\).

\(\\[1pt]\)
\({\small 11.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 3
\(\\[1pt]\)
Find the exact value of \(\displaystyle \int_{1}^{4} x^{\frac{3}{2}} \ln x \ \mathrm{d}x \).

\(\\[1pt]\)
\({\small 12.\enspace}\) 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.

\(\\[1pt]\)
\({\small 13.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 6(a), (b)
\(\\[1pt]\)
 Question no 6 Paper 32 June 2020
\(\\[1pt]\)
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 \(x = 1\).

\(\\[1pt]\)
\({\small 14.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 9(a)
\(\\[1pt]\)
Question 9 (a) Paper 32 June 2020
\(\\[1pt]\)
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 \( \tan p = \frac{1}{1+p} \).

\(\\[1pt]\)
\({\small 15.\enspace}\) 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.

\(\\[1pt]\)
\({\small 16.\enspace}\) 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 \(\displaystyle \int_{1}^{3} \frac{2x^3 \ – \ x^2 \ + \ 6x \ + \ 3}{x^2 \ + \ 3} \mathrm{d}x \).

\(\\[1pt]\)
\({\small 17.\enspace}\) 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.

\(\\[1pt]\)
\({\small 18.\enspace}\) 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 \(\displaystyle \int_{a}^{ 2a } f(x) \ \mathrm{d}x = \ln 6\).

\(\\[1pt]\)
\({\small 19.\enspace}\) 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 \( f'(x) = -8 \).

\(\\[1pt]\)
\({\small 20.\enspace}\) 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 10(a), (b)
\(\\[1pt]\)
Question 10 Paper 32 March 2021
\(\\[1pt]\)
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 \(M\).

\(\\[1pt]\)
\({\small 21.\enspace}\) 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.

\(\\[1pt]\)
\({\small 22.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 8(a),(b)
\(\\[1pt]\)
Question 8 Paper 33 June 2021
\(\\[1pt]\)
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 \( a \gt 1, \displaystyle \int_{1}^{ a } \frac{ \ln x }{ x^4 } \ \mathrm{d}x \lt \frac{1}{9} \).

\(\\[1pt]\)
\({\small 23.\enspace}\) 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 4
\(\\[1pt]\)
Using integration by parts, find the exact value of \( \displaystyle \int_{0}^{ 2 } {\tan}^{-1} \big( \frac{1}{2} x \big) \ \mathrm{d}x \).

\(\\[1pt]\)
\({\small 24.\enspace}\) 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 \( \displaystyle \int_{\frac{1}{4}\pi}^{ \frac{1}{3}\pi } (\mathrm{cosec} 2\theta − \cot 2\theta) \ \mathrm{d}\theta = \frac{1}{2} \ln 2 \).

\(\\[1pt]\)
\({\small 25.\enspace}\) 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.

\(\\[1pt]\)
\({\small 26.\enspace}\) 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 7(a)
\(\\[1pt]\)
Question 7 Paper 31 June 2021
\(\\[1pt]\)
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 \( a = \tan \Big( {\large \frac{2a}{1 \ + \ a^2} }\Big) \).

\(\\[1pt]\)
\({\small 27.\enspace}\) 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 \( \displaystyle \int_{1}^{ 8 } y \ \mathrm{d}x = 18 \ln 2 \ – \ 9 \).

\(\\[1pt]\)
\({\small 28.\enspace}\) 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 \( \displaystyle \int_{0}^{ \frac{1}{2} } \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} \ \mathrm{d}x \).

\(\\[1pt]\)
\({\small 29.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 11
\(\\[1pt]\)
Integration - Direct substitution & Differentiation - Maximum point - Paper 32 March 2022
\(\\[1pt]\)
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 \(M\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 significant figures.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) 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.

\(\\[1pt]\)
\({\small 30.\enspace}\) 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 8
\(\\[1pt]\)
Integration - Finding Area Between Two Curves - Paper 13 Nov 2021 No 8
\(\\[1pt]\)
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 \( \ B( 4 , {\large\frac{1}{2}}) \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the area of the region between the two curves.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) 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 \(p\).

\(\\[1pt]\)
\({\small 31.\enspace}\) 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
\(\\[1pt]\)
\( \hspace{2em} \mathrm{f}^{\prime}(x) = { \big( \frac{1}{2}x \ + \ k \big) }^{-2} \ – \ { ( 1 \ + \ k ) }^{-2} \),
\(\\[1pt]\)
where \({\small \ k \ }\) is a constant. The curve has a minimum point at \({\small \ x \ = \ 2 }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}^{\prime\prime}(x) \) in terms of \(k\) and \(x\), and hence find the set of possible values of \(k\).
\(\\[1pt]\)
It is now given that \( {\small \ k \ = \ −3 \ }\) and the minimum point is at \({\small \ (2, \ 3\frac{1}{2}) }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}(x) \).

\(\\[1pt]\)
\({\small 32.\enspace}\) 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 9
\(\\[1pt]\)
Volume Revolution Integration - Paper 12 June 2021 No 9
\(\\[1pt]\)
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.
\(\\[1pt]\)
Find the volume obtained.

\(\\[1pt]\)
\({\small 33.\enspace}\) 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 11
\(\\[1pt]\)
The gradient of a curve is given by
\(\\[1pt]\)
\( \hspace{2em} {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 6{(3x \ – \ 5)}^{3} \ – \ k{x}^{2}}\),
\(\\[1pt]\)
where \({\small \ k \ }\) is a constant. The curve has a stationary point at \( \ {\small (2, -3.5) }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of \( \ k\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the equation of the curve.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find \( {\small {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Determine the nature of the stationary point at \( \ {\small (2, -3.5) }\).

\(\\[1pt]\)
\({\small 34.\enspace}\) 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 8
\(\\[1pt]\)
Volume Revolution - Paper 12 June 2020 No 8
\(\\[1pt]\)
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 \( \ {\small x \ = \ 1 } \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the volume generated when the shaded region is rotated through \( \ {\small 360^{\circ} \ }\) about the \(y\)-axis.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) 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 \( \ {\small y \ = \ 2x } \).

\(\\[1pt]\)
\({\small 35.\enspace}\) 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 10
\(\\[1pt]\)
The equation of a curve is
\(\\[1pt]\)
\( \hspace{2em} y \ = 54x \ – \ {(2x \ – \ 7)}^{3} \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \( {\small \ {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the coordinates of each of the stationary points on the curve.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Determine the nature of each of the stationary points.

\(\\[1pt]\)
\({\small 36.\enspace}\) 9709/12/O/N/19 – Paper 12 June 2019 Pure Maths 1 No 10
\(\\[1pt]\)
Integration - Paper 12 Nov 2019 No 10
\(\\[1pt]\)
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 \(B\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Obtain expressions for \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \(\displaystyle \int y \ \mathrm{d}x\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.7em}}\) Find the coordinates of \(B\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iii\right).\hspace{0.6em}}\) Find, showing all necessary working, the area of the shaded region.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) 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).

Integration Practice 6

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of p.

\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) 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

\({\small\hspace{3em}} \displaystyle \int_{0}^{a} x{e}^{{\large \frac{1}{2}x}} \mathrm{d}x \ = \ 6 \).

Show that a satisfies the equation

\({\small\hspace{3em}} a \ = \ 2 \ + \ {e}^{{\large -\frac{1}{2}a}}\).

\({\small 10. \enspace}\) 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 =) .