integral from-1 to 3 of x^2+1

解答

\int_{- 1}^{3} x^{2} + 1 d x

\frac{40}{3}

十进制

13.33333 \dots

求解步骤

\int_{- 1}^{3} x^{2} + 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_{- 1}^{3} x^{2} d x + \int_{- 1}^{3} 1 d x

\int_{- 1}^{3} x^{2} d x = \frac{28}{3}

\int_{- 1}^{3} 1 d x = 4

= \frac{28}{3} + 4

化简

\frac{28}{3} + 4 : \frac{40}{3}

= \frac{40}{3}

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