integral from 0 to 1 of 3(1+sqrt(x))^8

解答

\int_{0}^{1} 3 (1 + x)^{8} d x

\frac{4097}{15}

十进制

273.13333 \dots

求解步骤

\int_{0}^{1} 3 (1 + x)^{8} d x

提出常数:

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

= 3 \cdot \int_{0}^{1} (1 + x)^{8} d x

乘开

(1 + x)^{8} : 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4}

= 3 \cdot \int_{0}^{1} 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4} 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

= 3 (\int_{0}^{1} 1 d x + \int_{0}^{1} 8 x d x + \int_{0}^{1} 28 x d x + \int_{0}^{1} 56 x^{\frac{3}{2}} d x + \int_{0}^{1} 70 x^{2} d x + \int_{0}^{1} 56 x^{\frac{5}{2}} d x + \int_{0}^{1} 28 x^{3} d x + \int_{0}^{1} 8 x^{\frac{7}{2}} d x + \int_{0}^{1} x^{4} d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 8 x d x = \frac{16}{3}

\int_{0}^{1} 28 x d x = 14

\int_{0}^{1} 56 x^{\frac{3}{2}} d x = \frac{112}{5}

\int_{0}^{1} 70 x^{2} d x = \frac{70}{3}

\int_{0}^{1} 56 x^{\frac{5}{2}} d x = 16

\int_{0}^{1} 28 x^{3} d x = 7

\int_{0}^{1} 8 x^{\frac{7}{2}} d x = \frac{16}{9}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= 3 (1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5})

化简

3 (1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5}) : \frac{4097}{15}

= \frac{4097}{15}

\int csc^{4} (5 x) cot^{4} (5 x) d x

\frac{d}{d x} (2 x e^{- x^{2}})

(y^{16}) d y = \frac{( 1 + x )}{x} d x, y (1) = 5

in v erse l a pl a ce \frac{( s ^{2} + s )}{2 s ^{2} + 2 s + 1}

d er i v a t i v e f (x) = \frac{1}{( 3 x + 8 ) ^{2}}