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Integration by partsAQA A-Level Maths: Subtopic test

10 questions, 27 marks

AQA A-Level Maths

Integration by parts

Total 27 marks

Name

Class

Date

  1. 1
    Let I=∫x e2x dxI=\int x\,e^{2x}\,dx.
    (a)
    Which choice of uu and dvdx\frac{dv}{dx} is best for integration by parts?
    [1 mark]
    • Au=e2x, dvdx=xu=e^{2x},\ \frac{dv}{dx}=x
    • Bu=xe2x, dvdx=1u=xe^{2x},\ \frac{dv}{dx}=1
    • Cu=2x, dvdx=exu=2x,\ \frac{dv}{dx}=e^{x}
    • Du=x, dvdx=e2xu=x,\ \frac{dv}{dx}=e^{2x}
    (b)
    Find II.
    [1 mark]
    • Ae2x4(2x−1)+c\frac{e^{2x}}{4}(2x-1)+c
    • Be2x2(x−1)+c\frac{e^{2x}}{2}(x-1)+c
    • Ce2x2(x+1)+c\frac{e^{2x}}{2}(x+1)+c
    • Dxe2x−e2x2+cxe^{2x}-\frac{e^{2x}}{2}+c
    (c)
    Hence find the exact value of ∫01x e2x dx\int_0^1x\,e^{2x}\,dx.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A student tries to find ∫xcos⁡x dx\int x\cos x\,dx by integration by parts, choosing u=cos⁡xu=\cos x and dvdx=x\frac{dv}{dx}=x.
    (a)
    What is wrong with the student's choice?
    [1 mark]
    • AIntegration by parts cannot be used when both functions are of xx.
    • BIt leads to ∫x22sin⁡x dx\int\frac{x^2}{2}\sin x\,dx, which is harder than the original integral.
    • CNothing: this choice always gives a simpler integral than u=xu=x.
    • Dcos⁡x\cos x cannot be differentiated.
    (b)
    Find ∫xcos⁡x dx\int x\cos x\,dx.
    [1 mark]
    • Axsin⁡x−cos⁡x+cx\sin x-\cos x+c
    • Bx22sin⁡x+c\frac{x^2}{2}\sin x+c
    • Cxsin⁡x+cos⁡x+cx\sin x+\cos x+c
    • D−xsin⁡x+cos⁡x+c-x\sin x+\cos x+c
    (c)
    Hence find the exact value of ∫0π/2xcos⁡x dx\int_0^{\pi/2}x\cos x\,dx.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The function y=ln⁡xy=\ln x is defined for x>0x>0.
    (a)
    Use integration by parts with u=ln⁡xu=\ln x to show that ∫ln⁡x dx=xln⁡x−x+c\int\ln x\,dx=x\ln x-x+c.
    [3 marks]
    (b)
    Hence find the exact area of the region under the curve y=ln⁡xy=\ln x between x=1x=1 and x=e2x=e^2.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    Give exact answers in each part.
    (a)
    Use integration by parts twice to find ∫01x2ex dx\int_0^1x^2e^x\,dx.
    [6 marks]
    (b)
    Find ∫1ex2ln⁡x dx\int_1^ex^2\ln x\,dx.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).