integral from 1 to 3 of (xy^2+x^2+y+4)

解

\int_{1}^{3} (x y^{2} + x^{2} + y + 4) d x

8 + 2 y + \frac{26}{3} + y^{2} \cdot 4

解答ステップ

\int_{1}^{3} x y^{2} + x^{2} + y + 4 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_{1}^{3} 4 d x + \int_{1}^{3} y d x + \int_{1}^{3} x^{2} d x + \int_{1}^{3} y^{2} x d x

\int_{1}^{3} 4 d x = 8

\int_{1}^{3} y d x = 2 y

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} y^{2} x d x = y^{2} \cdot 4

= 8 + 2 y + \frac{26}{3} + y^{2} \cdot 4