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受欢迎的微积分 >

integral from 1 to 4 of x/(1+sqrt(x))

解答

\int_{1}^{4} \frac{x}{1 + x} d x

解答

2 (- ln (3) + ln (2) + \frac{11}{6})

+1

十进制

2.85573 \dots

求解步骤

\int_{1}^{4} \frac{x}{1 + x} d x

使用换元积分法

= \int_{1}^{2} \frac{2 u ^{3}}{1 + u} d u

提出常数:

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

= 2 \cdot \int_{1}^{2} \frac{u ^{3}}{1 + u} d u

使用换元积分法

= 2 \cdot \int_{2}^{3} \frac{( v - 1 ) ^{3}}{v} d v

乘开

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

= 2 \cdot \int_{2}^{3} v^{2} - 3 v + 3 - \frac{1}{v} d v

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

= 2 (\int_{2}^{3} v^{2} d v - \int_{2}^{3} 3 v d v + \int_{2}^{3} 3 d v - \int_{2}^{3} \frac{1}{v} d v)

\int_{2}^{3} v^{2} d v = \frac{19}{3}

\int_{2}^{3} 3 v d v = \frac{15}{2}

\int_{2}^{3} 3 d v = 3

\int_{2}^{3} \frac{1}{v} d v = ln (3) - ln (2)

= 2 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2)))

化简

2 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2))) : 2 (- ln (3) + ln (2) + \frac{11}{6})

= 2 (- ln (3) + ln (2) + \frac{11}{6})

作图

流行的例子

integral from 1 to 16 of x^{-3/4}

\int_{1}^{16} x^{- \frac{3}{4}} d x

integral from-10 to 10 of (100-x^2)

\int_{- 10}^{10} (100 - x^{2}) d x

integral from-7 to 0 of sqrt(49-x^2)

\int_{- 7}^{0} 49 - x^{2} d x

integral from 0 to 1/2 of-2x+1

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

integral from 0 to pi/2 of sqrt(cos(x))

\int_{0}^{\frac{π}{2}} cos (x) d x