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Integration by Substitution

AP Calculus AB· difficulty 4/5

∫01x1−x2 dx=\displaystyle \int_0^1 x \sqrt{1-x^2}\, dx =

  • A

    11

  • B

    12\frac{1}{2}

  • C

    13\frac{1}{3}

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  • D

    23\frac{2}{3}

Explanation

Let u=1−x2u=1-x^2, du=−2x dxdu=-2x\,dx. Limits: x=0→u=1x=0\to u=1, x=1→u=0x=1\to u=0. Then ∫10u⋅(−du/2)=12∫01u1/2 du=13\int_1^0 \sqrt{u}\cdot(-du/2) = \frac{1}{2}\int_0^1 u^{1/2}\,du = \frac{1}{3}.

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