integral from 0 to 2 of x^2+x+1

解答

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

\frac{20}{3}

十进制

6.66666 \dots

求解步骤

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

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

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

\int_{0}^{2} 1 d x = 2

= \frac{8}{3} + 2 + 2

化简

\frac{8}{3} + 2 + 2 : \frac{20}{3}

= \frac{20}{3}

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

\int_{0}^{1} 3 x (x^{2} + 1)^{3} d x

\int_{0}^{x} (x - t) f (t) d t

\int_{- π}^{π} x^{4} cos (n x) d x

\int_{0}^{0.5} e^{- x} d x