integral from 0 to 1 of 9/(1+x^2)

解答

\int_{0}^{1} \frac{9}{1 + x ^{2}} d x

\frac{9 π}{4}

十进制

7.06858 \dots

求解步骤

\int_{0}^{1} \frac{9}{1 + x ^{2}} d x

提出常数:

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

= 9 \cdot \int_{0}^{1} \frac{1}{1 + x ^{2}} d x

使用积分常见定则:

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

= 9 [arctan (x)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= 9 \cdot \frac{π}{4}

化简

= \frac{9 π}{4}

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