integral from 1 to sqrt(2 of)xe^{4x^2}

解答

\int_{1}^{2} x e^{4 x^{2}} d x

\frac{e ^{8} - e ^{4}}{8}

十进制

365.79497 \dots

求解步骤

\int_{1}^{2} x e^{4 x^{2}} d x

使用换元积分法

= \int_{4}^{8} \frac{e ^{u}}{8} d u

提出常数:

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

= \frac{1}{8} \cdot \int_{4}^{8} e^{u} d u

使用积分常见定则:

\int e^{u} d u = e^{u}

= \frac{1}{8} [e^{u}]_{4}^{8}

带入上下限计算后得:

e^{8} - e^{4}

= \frac{1}{8} (e^{8} - e^{4})

化简

= \frac{e ^{8} - e ^{4}}{8}

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

\int_{0}^{1} sin^{4} (x) cos (x) d x

\int_{0}^{1} (2 x + x) d x

\int_{π}^{\frac{3 π}{2}} - sin (x) d x

\int_{- 1}^{2} 8 x - x^{2} - 7 d x