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受欢迎的微积分 >

integral from 4 to 6 of 1/(2x^2-16x+40)

解答

\int_{4}^{6} \frac{1}{2 x ^{2} - 16 x + 40} d x

解答

\frac{π}{16}

+1

十进制

0.19634 \dots

求解步骤

\int_{4}^{6} \frac{1}{2 x ^{2} - 16 x + 40} d x

对

2 x^{2} - 16 x + 40

配方:

2 (x - 4)^{2} + 8

= \int_{4}^{6} \frac{1}{2 ( x - 4 ) ^{2} + 8} d x

使用换元积分法

= \int_{0}^{2} \frac{1}{2 u ^{2} + 8} d u

使用换元积分法

= \int_{0}^{1} \frac{1}{4 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

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

= \frac{1}{4} [arctan (v)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

化简

= \frac{π}{16}

作图

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