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受欢迎的微积分 >

integral from 0 to 2 of 15sqrt(4x+1)

解答

\int_{0}^{2} 15 4 x + 1 d x

解答

65

求解步骤

\int_{0}^{2} 15 4 x + 1 d x

提出常数:

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

= 15 \cdot \int_{0}^{2} 4 x + 1 d x

使用换元积分法

= 15 \cdot \int_{1}^{9} \frac{u}{4} d u

提出常数:

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

= 15 \cdot \frac{1}{4} \cdot \int_{1}^{9} u d u

使用幂法则

= 15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

化简

15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

带入上下限计算后得:

\frac{52}{3}

= \frac{15}{4} \cdot \frac{52}{3}

化简

= 65

作图

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