Question 7 - A-Level Maths

OCR C4 — Question 7 11 marks

Exam Board	OCR
Module	C4 (Core Mathematics 4)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Parts
Type	Independent multi-part (different techniques)
Difficulty	Standard +0.3 Part (i) is a standard integration by parts question requiring two applications (routine C4 technique). Part (ii) is a straightforward substitution with clear limits. Both are textbook-style exercises with no novel insight required, making this slightly easier than average for C4 material.
Spec	1.08h Integration by substitution 1.08i Integration by parts

(i) Find

$$\int x ^ { 2 } \sin x \mathrm {~d} x$$ (ii) Use the substitution $u = 1 + \sin x$ to find the value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos x ( 1 + \sin x ) ^ { 3 } d x$$

Show mark scheme Show mark scheme source

Question 7:

Part (i):

Answer	Marks
$u = x^2,\ u' = 2x,\ v' = \sin x,\ v = -\cos x$	M1
$I = -x^2\cos x - \int -2x\cos x\ dx = -x^2\cos x + \int 2x\cos x\ dx$	A1
$u = 2x,\ u' = 2,\ v' = \cos x,\ v = \sin x$	M1
$I = -x^2\cos x + 2x\sin x - \int 2\sin x\ dx$	A1
$= -x^2\cos x + 2x\sin x + 2\cos x + c$	A1

Part (ii):

Answer	Marks	Guidance
$u = 1 + \sin x \Rightarrow \frac{du}{dx} = \cos x$	M1
$x=0 \Rightarrow u=1,\ x=\frac{\pi}{2} \Rightarrow u=2$	B1
$I = \int_1^2 u^3\ du$	M1 A1
$= \left[\frac{1}{4}u^4\right]_1^2$	M1
$= 4 - \frac{1}{4} = \frac{15}{4}$	A1	(11)

# Question 7:

## Part (i):
| $u = x^2,\ u' = 2x,\ v' = \sin x,\ v = -\cos x$ | M1 | |
|---|---|---|
| $I = -x^2\cos x - \int -2x\cos x\ dx = -x^2\cos x + \int 2x\cos x\ dx$ | A1 | |
| $u = 2x,\ u' = 2,\ v' = \cos x,\ v = \sin x$ | M1 | |
| $I = -x^2\cos x + 2x\sin x - \int 2\sin x\ dx$ | A1 | |
| $= -x^2\cos x + 2x\sin x + 2\cos x + c$ | A1 | |

## Part (ii):
| $u = 1 + \sin x \Rightarrow \frac{du}{dx} = \cos x$ | M1 | |
|---|---|---|
| $x=0 \Rightarrow u=1,\ x=\frac{\pi}{2} \Rightarrow u=2$ | B1 | |
| $I = \int_1^2 u^3\ du$ | M1 A1 | |
| $= \left[\frac{1}{4}u^4\right]_1^2$ | M1 | |
| $= 4 - \frac{1}{4} = \frac{15}{4}$ | A1 | **(11)** |

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Show LaTeX source

\begin{enumerate}
  \item (i) Find
\end{enumerate}

$$\int x ^ { 2 } \sin x \mathrm {~d} x$$

(ii) Use the substitution $u = 1 + \sin x$ to find the value of

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos x ( 1 + \sin x ) ^ { 3 } d x$$

\hfill \mbox{\textit{OCR C4  Q7 [11]}}

This paper (8 questions)

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Q1 4 Q2 5 Q3 7 Q4 9 Q5 9 Q6 11 Q7 11 Q8 16

Answer	Marks
\(u = x^2,\ u' = 2x,\ v' = \sin x,\ v = -\cos x\)	M1
\(I = -x^2\cos x - \int -2x\cos x\ dx = -x^2\cos x + \int 2x\cos x\ dx\)	A1
\(u = 2x,\ u' = 2,\ v' = \cos x,\ v = \sin x\)	M1
\(I = -x^2\cos x + 2x\sin x - \int 2\sin x\ dx\)	A1
\(= -x^2\cos x + 2x\sin x + 2\cos x + c\)	A1

Answer	Marks	Guidance
\(u = 1 + \sin x \Rightarrow \frac{du}{dx} = \cos x\)	M1
\(x=0 \Rightarrow u=1,\ x=\frac{\pi}{2} \Rightarrow u=2\)	B1
\(I = \int_1^2 u^3\ du\)	M1 A1
\(= \left[\frac{1}{4}u^4\right]_1^2\)	M1
\(= 4 - \frac{1}{4} = \frac{15}{4}\)	A1	(11)