integral from 0 to 1 of 7e^x-7xe^{x^2}

解答

\int_{0}^{1} 7 e^{x} - 7 x e^{x^{2}} d x

7 (e - 1) - \frac{7 ( e - 1 )}{2}

十进制

6.01398 \dots

求解步骤

\int_{0}^{1} 7 e^{x} - 7 x e^{x^{2}} 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

= \int_{0}^{1} 7 e^{x} d x - \int_{0}^{1} 7 x e^{x^{2}} d x

\int_{0}^{1} 7 e^{x} d x = 7 (e - 1)

\int_{0}^{1} 7 x e^{x^{2}} d x = \frac{7 ( e - 1 )}{2}

= 7 (e - 1) - \frac{7 ( e - 1 )}{2}

\frac{d}{d x} (sin (\frac{π}{2} x))

\int_{0}^{6} \frac{7 t}{( t - 7 ) ^{2}} d t

\int cos (4 x) cos (8 x) d x

y^{^{''}} - 4 y^{^{'}} - 12 y = 8 e^{- 2 x}

\frac{d}{d x} (x e^{x} cos (x))