integral from 0 to 3 of sqrt((3t+4))

解答

\int_{0}^{3} (3 t + 4) d t

\frac{26 13 - 16}{9}

十进制

8.63825 \dots

求解步骤

\int_{0}^{3} 3 t + 4 d t

使用换元积分法

= \int_{4}^{13} \frac{u}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int_{4}^{13} u d u

使用幂法则

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

带入上下限计算后得:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{26 13}{3} - \frac{16}{3})

化简

= \frac{26 13 - 16}{9}

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12 y^{^{''}} - 12 y^{^{'}} + 78 y = 0

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