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受欢迎的微积分 >

integral from 0 to 1/2 of 1/(64x^2+16)

解答

\int_{0}^{\frac{1}{2}} \frac{1}{64 x ^{2} + 16} d x

解答

\frac{π}{128}

+1

十进制

0.02454 \dots

求解步骤

\int_{0}^{\frac{1}{2}} \frac{1}{64 x ^{2} + 16} d x

使用换元积分法

= \int_{0}^{1} \frac{1}{2 ( 16 u ^{2} + 16 )} 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}^{1} \frac{1}{16 u ^{2} + 16} d u

\frac{1}{16 u ^{2} + 16} = \frac{1}{16} \frac{1}{u ^{2} + 1}

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{16} \cdot \frac{1}{u ^{2} + 1} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{16} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

使用积分常见定则:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} \cdot \frac{1}{16} [arctan (u)]_{0}^{1}

化简

\frac{1}{2} \cdot \frac{1}{16} [arctan (u)]_{0}^{1} : \frac{1}{32} [arctan (u)]_{0}^{1}

= \frac{1}{32} [arctan (u)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{32} \cdot \frac{π}{4}

化简

= \frac{π}{128}

作图

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