Corrigé du 77 P. 384

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a. \begin{align*} \int_0^1 \left(x^4 - 4\right)\mathrm dx &=\left[\frac{x^5}5 - 4x\right]_0^1& \\ &=\left(\frac{1^5}5 - 4\times 1\right) - \left(\frac 0 5 - 4\times 0\right)& \\ &=\frac 1 5 - 4& \\ &=-\frac{19}5.& \end{align*}

b. \begin{align*} \int_2^3 \frac 1 {x^4}\:\mathrm dx &=\int_2^3 x^{-4}\:\mathrm dx& \\ &=\left[\frac{x^{-3}}{-3}\right]_2^3& \\ &=\left[-\frac 1 {3x^3}\right]_2^3& \\ &=-\frac 1 {3\times 3^3} - \left(-\frac 1 {3\times 2^3}\right)& \\ &=-\frac 1 {81} + \frac 1 {24}& \\ &=\frac{-8+27}{648}& \\ &=\frac{19}{648}.& \end{align*}

c. \[\begin{aligned} \int_0^{\frac{\pi}2}\cos(2x)\mathrm dx &=\left[\frac 1 2\sin(2x)\right]_0^{\frac{\pi}2}& \\ &=\frac 1 2\sin(\pi) - \frac 1 2 \sin(0)& \\ &=0 - 0& \\ &=0.& \end{aligned}\]

d. \[ \int_1^2 \mathrm e^{2t+1}\mathrm dt =\left[\frac 1 2 \mathrm e^{2t+1}\right]_1^2 =\frac 1 2\mathrm e^5 - \frac 1 2 \mathrm e^3. \]

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