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受欢迎的微积分 >

integral from 0 to 1 of e^{2sqrt(x)}

解答

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

解答

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

+1

十进制

4.19452 \dots

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

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

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

带入上下限计算后得:

e^{2} + 1

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

化简

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

作图

流行的例子

integral of 1/(x(sqrt(x-1)))

\int \frac{1}{x ( x - 1 )} d x

integral from-1 to 1 of xsqrt(1-x^2)

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

derivative y=4arcsin(5)x^5

d er i v a t i v e y = 4 arcsin (5) x^{5}

泰勒 (1+x)^{-1}

t a y l or (1 + x)^{- 1}

tangent y=-3x^3-2x^2-x+3,(-1,5)

t an g e n t y = - 3 x^{3} - 2 x^{2} - x + 3, (- 1, 5)