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Sheet 3 · IntegrationAI-marked, provisional
2(b)

Find ∫xe2x dx\displaystyle\int x e^{2x}\,dx.

3/5

Let u=xu = x and dv=e2x dxdv = e^{2x}\,dx, so

1

v=2e2xv = 2e^{2x}

.


2

∫xe2x dx=2xe2x+∫2e2x dx\int x e^{2x}\,dx = 2xe^{2x} + \int 2e^{2x}\,dx


=2xe2x+e2x+C= 2xe^{2x} + e^{2x} + C

  1. 1

    Careless slip · Algebra or sign slip

    v=2e2xv = 2e^{2x}

    Differentiate your vv. Do you get back e2xe^{2x}?

  2. 2

    Structural mistake · Method

    ∫xe2x dx=2xe2x+∫2e2x dx\int x e^{2x}\,dx = 2xe^{2x} + \int 2e^{2x}\,dx

    Write out the integration by parts formula first. Which sign sits in front of ∫v du\int v\,du?

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