integral from 1 to-1 of (9r+8)^2

解答

\int_{1}^{- 1} (9 r + 8)^{2} d r

- 182

求解步骤

\int_{1}^{- 1} (9 r + 8)^{2} d r

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} (9 r + 8)^{2} d r

乘开

(9 r + 8)^{2} : 81 r^{2} + 144 r + 64

= - \int_{- 1}^{1} 81 r^{2} + 144 r + 64 d r

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

= - (\int_{- 1}^{1} 81 r^{2} d r + \int_{- 1}^{1} 144 r d r + \int_{- 1}^{1} 64 d r)

\int_{- 1}^{1} 81 r^{2} d r = 54

\int_{- 1}^{1} 144 r d r = 0

\int_{- 1}^{1} 64 d r = 128

= - (54 + 0 + 128)

化简

= - 182

\int_{1}^{4} \frac{1}{( x + 1 ) ^{2}} d x

\int_{0}^{4} ln (x) d x

\int_{0}^{1 - x} 2 d y

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

\int_{0}^{8} \frac{1}{x ^{2} + 64} d x