integral from 0 to pi of (9-sin(4x))/4

解答

\int_{0}^{π} \frac{9 - sin ( 4 x )}{4} d x

\frac{9 π}{4}

十进制

7.06858 \dots

求解步骤

\int_{0}^{π} \frac{9 - sin ( 4 x )}{4} d x

提出常数:

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

= \frac{1}{4} \cdot \int_{0}^{π} 9 - sin (4 x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int_{0}^{π} 9 d x - \int_{0}^{π} sin (4 x) d x)

\int_{0}^{π} 9 d x = 9 π

\int_{0}^{π} sin (4 x) d x = 0

= \frac{1}{4} (9 π - 0)

化简

\frac{1}{4} (9 π - 0) : \frac{9 π}{4}

= \frac{9 π}{4}