integral from 0 to 2 of xsqrt(3x^2+4)

解答

\int_{0}^{2} x 3 x^{2} + 4 d x

\frac{56}{9}

十进制

6.22222 \dots

求解步骤

\int_{0}^{2} x 3 x^{2} + 4 d x

使用换元积分法

= \int_{4}^{16} \frac{u}{6} d u

提出常数:

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

= \frac{1}{6} \cdot \int_{4}^{16} u d u

使用幂法则

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

带入上下限计算后得:

\frac{112}{3}

= \frac{1}{6} \cdot \frac{112}{3}

化简

= \frac{56}{9}

\int_{0}^{15} \frac{x}{36 + 3 x} d x

\int_{1}^{4} \frac{1}{x ^{2} - 4} d x

\int_{2.7}^{8.3} 12 x^{2} - 6 x + 4 d x

\int_{0}^{2 π} - 16 d t

\int_{- π}^{π} cos^{2} (m x) d x