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受欢迎的微积分 >

integral from 1 to 3 of sqrt(x^2+3)

解答

\int_{1}^{3} x^{2} + 3 d x

解答

33 - 1 + \frac{3}{2} ln (3 + 2) - \frac{3}{4} ln (3)

+1

十进制

5.34763 \dots

求解步骤

\int_{1}^{3} x^{2} + 3 d x

使用三角换元法

= \int_{\frac{π}{6}}^{\frac{π}{3}} 3 sec^{3} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec^{3} (u) d u

使用降次积分法

= 3 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec (u) d u)

\int_{\frac{π}{6}}^{\frac{π}{3}} sec (u) d u = ln (3 + 2) - \frac{1}{2} ln (3)

= 3 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

化简

3 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3))) : 3 ([\frac{1}{2} sec (u) tan (u)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

= 3 ([\frac{1}{2} sec (u) tan (u)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

带入上下限计算后得:

3 - \frac{1}{3}

= 3 (3 - \frac{1}{3} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

化简

= 33 - 1 + \frac{3}{2} ln (3 + 2) - \frac{3}{4} ln (3)

作图

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