Question 8 - A-Level Maths

OCR C4 — Question 8 13 marks

Exam Board	OCR
Module	C4 (Core Mathematics 4)
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Parts
Type	Independent multi-part (different techniques)
Difficulty	Standard +0.3 Part (i) requires two applications of integration by parts with an exponential, which is standard C4 technique but involves careful bookkeeping. Part (ii) requires recognizing that sin(2t) = 2sin(t)cos(t), then applying the substitution to reduce it to a polynomial integral. Both parts are routine applications of taught methods with no novel insight required, making this slightly easier than average.
Spec	1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates 1.08h Integration by substitution

8. (i) Find $$\int x ^ { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ (ii) Using the substitution $u = \sin t$, evaluate $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$

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(i)

Answer	Marks
$u = x^2$, $u' = 2x$, $v' = e^{\frac{1}{4}x}$, $v = 2e^{\frac{1}{4}x}$	M1
$I = 2x^2e^{\frac{1}{4}x} - \int 4xe^{\frac{1}{4}x}dx$	A2
$u = 4x$, $u' = 4$, $v' = e^{\frac{1}{4}x}$, $v = 2e^{\frac{1}{4}x}$	M1
$I = 2x^2e^{\frac{1}{4}x} - [8xe^{\frac{1}{4}x} - \int 8e^{\frac{1}{4}x}dx]$	A1
$= 2x^2e^{\frac{1}{4}x} - 8xe^{\frac{1}{4}x} + 16e^{\frac{1}{4}x} + c$ or $2e^{\frac{1}{4}x}(x^2 - 4x + 8) + c$	A1

(ii)

Answer	Marks
$u = \sin t \Rightarrow \frac{du}{dt} = \cos t$	M1
$t = 0 \Rightarrow u = 0$, $t = \frac{\pi}{2} \Rightarrow u = 1$	B1
$\sin^2 2t = 4\sin^2 t \cos^2 t = 4\sin^2 t(1-\sin^2 t)$	M1
$I = \int_0^1 4u^2(1-u^2)du$	M1
$= 4[\frac{1}{3}u^3 - \frac{1}{5}u^5]_0^1$	M1
$= 4(\frac{1}{3} - \frac{1}{5}) - (0)) = \frac{8}{15}$	M1 A1
(13)

Total (72)

**(i)**
$u = x^2$, $u' = 2x$, $v' = e^{\frac{1}{4}x}$, $v = 2e^{\frac{1}{4}x}$ | M1 |
$I = 2x^2e^{\frac{1}{4}x} - \int 4xe^{\frac{1}{4}x}dx$ | A2 |
$u = 4x$, $u' = 4$, $v' = e^{\frac{1}{4}x}$, $v = 2e^{\frac{1}{4}x}$ | M1 |
$I = 2x^2e^{\frac{1}{4}x} - [8xe^{\frac{1}{4}x} - \int 8e^{\frac{1}{4}x}dx]$ | A1 |
$= 2x^2e^{\frac{1}{4}x} - 8xe^{\frac{1}{4}x} + 16e^{\frac{1}{4}x} + c$ or $2e^{\frac{1}{4}x}(x^2 - 4x + 8) + c$ | A1 |

**(ii)**
$u = \sin t \Rightarrow \frac{du}{dt} = \cos t$ | M1 |
$t = 0 \Rightarrow u = 0$, $t = \frac{\pi}{2} \Rightarrow u = 1$ | B1 |
$\sin^2 2t = 4\sin^2 t \cos^2 t = 4\sin^2 t(1-\sin^2 t)$ | M1 |
$I = \int_0^1 4u^2(1-u^2)du$ | M1 |
$= 4[\frac{1}{3}u^3 - \frac{1}{5}u^5]_0^1$ | M1 |
$= 4(\frac{1}{3} - \frac{1}{5}) - (0)) = \frac{8}{15}$ | M1 A1 |
| | **(13)** |

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**Total (72)**

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8. (i) Find

$$\int x ^ { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$

(ii) Using the substitution $u = \sin t$, evaluate

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$

\hfill \mbox{\textit{OCR C4  Q8 [13]}}

This paper (8 questions)

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Q1 5 Q2 6 Q3 7 Q4 8 Q5 9 Q6 12 Q7 12 Q8 13

Answer	Marks
\(u = x^2\), \(u' = 2x\), \(v' = e^{\frac{1}{4}x}\), \(v = 2e^{\frac{1}{4}x}\)	M1
\(I = 2x^2e^{\frac{1}{4}x} - \int 4xe^{\frac{1}{4}x}dx\)	A2
\(u = 4x\), \(u' = 4\), \(v' = e^{\frac{1}{4}x}\), \(v = 2e^{\frac{1}{4}x}\)	M1
\(I = 2x^2e^{\frac{1}{4}x} - [8xe^{\frac{1}{4}x} - \int 8e^{\frac{1}{4}x}dx]\)	A1
\(= 2x^2e^{\frac{1}{4}x} - 8xe^{\frac{1}{4}x} + 16e^{\frac{1}{4}x} + c\) or \(2e^{\frac{1}{4}x}(x^2 - 4x + 8) + c\)	A1

Answer	Marks
\(u = \sin t \Rightarrow \frac{du}{dt} = \cos t\)	M1
\(t = 0 \Rightarrow u = 0\), \(t = \frac{\pi}{2} \Rightarrow u = 1\)	B1
\(\sin^2 2t = 4\sin^2 t \cos^2 t = 4\sin^2 t(1-\sin^2 t)\)	M1
\(I = \int_0^1 4u^2(1-u^2)du\)	M1
\(= 4[\frac{1}{3}u^3 - \frac{1}{5}u^5]_0^1\)	M1
\(= 4(\frac{1}{3} - \frac{1}{5}) - (0)) = \frac{8}{15}\)	M1 A1
(13)