integral from 1 to t^2 of sqrt(s+1)

解答

\int_{1}^{t^{2}} s + 1 d s

\frac{2 ( 1 + t ^{2} ) ^{\frac{3}{2}} - 4 2}{3}

求解步骤

\int_{1}^{t^{2}} s + 1 d s

使用换元积分法

= \int_{2}^{1 + t^{2}} u d u

使用幂法则

= [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{1 + t^{2}}

带入上下限计算后得:

\frac{2}{3} (1 + t^{2})^{\frac{3}{2}} - \frac{4 2}{3}

= \frac{2}{3} (1 + t^{2})^{\frac{3}{2}} - \frac{4 2}{3}

化简

= \frac{2 ( 1 + t ^{2} ) ^{\frac{3}{2}} - 4 2}{3}

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