What is the integral from 0 to 1 of (x^4y+y^2) ?

The integral from 0 to 1 of (x^4y+y^2) is 1/3+(x^4)/2

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

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

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

Solution steps

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

Apply the Sum Rule:

\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}

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What is the integral from 0 to 1 of (x^4y+y^2) ?
The integral from 0 to 1 of (x^4y+y^2) is 1/3+(x^4)/2