Ivan Shishkin, Rye (1878)

Problems/Riemann integrationExerciseUnreviewed

Quick integrations by part (level 11)

by Sequoia·
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Compute the following integrals using an integration by part.

  1. ∫0π/2x cos⁡(x) dx.\displaystyle\int_0^{\pi/2}x\,\cos(x)\,\mathrm{d}x.

  2. ∫1euln⁡(u) du.\displaystyle\int_1^{e}u\ln(u)\,\mathrm{d}u.

  3. ∫01(2t+1)et dt.\displaystyle\int_0^{1}(2t+1)e^{t}\,\mathrm{d}t.

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  1. Posons f(x)=xf(x)=x et g′(x)=cos⁡xg'(x)=\cos x, d’où f′(x)=1f'(x)=1 et g(x)=sin⁡xg(x)=\sin x. Ces deux fonctions étant de classe C1\mathcal{C}^1 sur [0,π/2][0,\pi/2] :
    ∫0π/2xcos⁡x dx=[xsin⁡x]0π/2−∫0π/2sin⁡x dx=π2−[−cos⁡x]0π/2=π2−1.\int_0^{\pi/2} x\cos x\,\mathrm{d}x=\Bigl[x\sin x\Bigr]_0^{\pi/2}-\int_0^{\pi/2}\sin x\,\mathrm{d}x=\frac{\pi}{2}-\Bigl[-\cos x\Bigr]_0^{\pi/2}=\frac{\pi}{2}-1 .
  2. Ici le logarithme doit être dérivé, ce qui impose f(u)=ln⁡uf(u)=\ln u et g′(u)=ug'(u)=u, donc f′(u)=1/uf'(u)=1/u et g(u)=u2/2g(u)=u^2/2 :
    ∫1euln⁡u du=[u22ln⁡u]1e−∫1eu2 du=e22−[u24]1e=e22−e24+14=e2+14.\int_1^e u\ln u\,\mathrm{d}u=\left[\frac{u^2}{2}\ln u\right]_1^e-\int_1^e \frac{u}{2}\,\mathrm{d}u=\frac{e^2}{2}-\left[\frac{u^2}{4}\right]_1^e=\frac{e^2}{2}-\frac{e^2}{4}+\frac14=\frac{e^2+1}{4}.
  3. Posons f(t)=2t+1f(t)=2t+1 et g′(t)=etg'(t)=e^t, d’où f′(t)=2f'(t)=2 et g(t)=etg(t)=e^t :
    ∫01(2t+1)et dt=[(2t+1)et]01−2∫01et dt=(3e−1)−2(e−1)=e+1.\int_0^1 (2t+1)e^t\,\mathrm{d}t=\Bigl[(2t+1)e^t\Bigr]_0^1-2\int_0^1 e^t\,\mathrm{d}t=(3e-1)-2\bigl(e-1\bigr)=e+1 .
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