What is the integral from 0 to 4 of (sqrt(y))^2 ?

The integral from 0 to 4 of (sqrt(y))^2 is 8

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integral from 0 to 4 of (sqrt(y))^2

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

8

Solution steps

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

Simplify

(y)^{2} : y

= \int_{0}^{4} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{4}

= [\frac{y ^{2}}{2}]_{0}^{4}

Compute the boundaries:

8

= 8

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

\int_{0}^{4} π y d y

What is the integral from 0 to 4 of (sqrt(y))^2 ?
The integral from 0 to 4 of (sqrt(y))^2 is 8