integral from 0 to 1 of te^{2t}

解答

\int_{0}^{1} t e^{2 t} d t

\frac{e ^{2} + 1}{4}

十进制

2.09726 \dots

求解步骤

\int_{0}^{1} t e^{2 t} d t

使用换元积分法

= \int_{0}^{2} \frac{e ^{u} u}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{0}^{2} e^{u} u d u

使用分布积分法

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{2}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{2}

带入上下限计算后得:

e^{2} + 1

= \frac{1}{4} (e^{2} + 1)

化简

= \frac{e ^{2} + 1}{4}

t an g e n t tan (x), at x = - \frac{π}{3}

\int_{0}^{14} 10 e^{x} d x

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x \to - \frac{3 π}{2} - lim (tan (x))

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