integral from 0 to 2 of 1/(4+x^2)

解答

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

\frac{π}{8}

十进制

0.39269 \dots

求解步骤

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

使用换元积分法

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

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

化简

= \frac{π}{8}

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