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受欢迎的微积分 >

integral from 0 to 1 of (15sqrt(5x+4))/2

解答

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

解答

19

求解步骤

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

提出常数:

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

= \frac{15}{2} \cdot \int_{0}^{1} 5 x + 4 d x

使用换元积分法

= \frac{15}{2} \cdot \int_{4}^{9} \frac{u}{5} d u

提出常数:

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

= \frac{15}{2} \cdot \frac{1}{5} \cdot \int_{4}^{9} u d u

使用幂法则

= \frac{15}{2} \cdot \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

化简

\frac{15}{2} \cdot \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9} : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

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

带入上下限计算后得:

\frac{38}{3}

= \frac{3}{2} \cdot \frac{38}{3}

化简

= 19

作图

流行的例子

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y^{^{'}} = \frac{11 x ^{2} - 1}{11 + 4 y}

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d er i v a t i v e h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 5)^{3}

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