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人気のある微分積分 >

integral from 0 to 1 of (x^4y+y^2)

解

\int_{0}^{1} (x^{4} y + y^{2}) d y

解

\frac{1}{3} + \frac{x ^{4}}{2}

解答ステップ

\int_{0}^{1} x^{4} y + y^{2} d y

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 y + \int_{0}^{1} x^{4} y d y

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

\int_{0}^{1} x^{4} y d y = \frac{x ^{4}}{2}

= \frac{1}{3} + \frac{x ^{4}}{2}

人気の例

integral from-0.25 to 0.25 of (cos(x))^2

\int_{- 0.25}^{0.25} (cos (x))^{2} d x

integral from-9 to-2 of 1/x

\int_{- 9}^{- 2} \frac{1}{x} d x

integral from x+2 to 4-x^2 of 1

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

integral from 0 to 2 of (3t^3+5t)

\int_{0}^{2} (3 t^{3} + 5 t) d t

integral from 1 to 9 of sqrt(x)-1

\int_{1}^{9} x - 1 d x