integral from 0 to 2 of ax^2+bx

解答

\int_{0}^{2} a x^{2} + b x d x

\frac{8 a}{3} + b \cdot 2

求解步骤

\int_{0}^{2} a x^{2} + b x 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}^{2} a x^{2} d x + \int_{0}^{2} b x d x

\int_{0}^{2} a x^{2} d x = \frac{8 a}{3}

\int_{0}^{2} b x d x = b \cdot 2

= \frac{8 a}{3} + b \cdot 2

\int_{- 1}^{3} (x^{3} - 3 x^{2}) d x

\int_{0}^{x^{2}} e^{- 2 t^{2}} d t

\int_{0}^{\frac{π}{3}} y ln (z) tan (x) d x

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

\int_{- 2}^{0} x^{4} - 2 x^{2} d x