integral from 0 to 1 of (x^2+y^2)

解

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

y^{2} + \frac{1}{3}

解答ステップ

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= y^{2} + \frac{1}{3}