Edexcel C4 2010 June — Question 2 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow definite integral equals specific value (algebraic/exponential substitution)
DifficultyModerate -0.3 This is a straightforward substitution question where the substitution is given explicitly. Students need to find du/dx = -sin x, change limits (x=0 gives u=2, x=π/2 gives u=1), rewrite the integral as -∫₂¹ eᵘ du, evaluate to get [eᵘ]₁² = e - e², then recognize this equals e(e-1) after factoring. While it requires careful execution of multiple steps (substitution, limit changes, integration, simplification), each step is standard technique with no novel insight required, making it slightly easier than average.
Spec1.08h Integration by substitution
2. Using the substitution \(u = \cos x + 1\), or otherwise, show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { \cos x + 1 } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$ (6)
AnswerMarks Guidance
\(\frac{du}{dx} = -\sin x\)B1
\(\int \sin x e^{\cos x} dx = -\int e^u du = -e^u = -e^{\cos x + 1}\)M1 A1 ft sign error, A1ft
\(\left[-e^{\cos x+1}\right]_0^{\pi/2} = -e^1 - (-e^2) = e(e-1)\) ⭐M1 A1 or equivalent with \(u\), cso 
| $\frac{du}{dx} = -\sin x$ | B1 | |
| $\int \sin x e^{\cos x} dx = -\int e^u du = -e^u = -e^{\cos x + 1}$ | M1 A1 | ft sign error, A1ft |
| $\left[-e^{\cos x+1}\right]_0^{\pi/2} = -e^1 - (-e^2) = e(e-1)$ ⭐ | M1 A1 | or equivalent with $u$, cso |

---
2. Using the substitution $u = \cos x + 1$, or otherwise, show that

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { \cos x + 1 } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$

(6)\\

\hfill \mbox{\textit{Edexcel C4 2010 Q2 [6]}}
This paper (8 questions)
View full paper
Q1 8 Q2 6 Q3 7 Q4 10 Q5 11 Q6 10 Q7 12 Q8 11