What is the integral from 0 to 1-y of y^2 ?

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

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integral from 0 to 1-y of y^2

\int_{0}^{1 - y} y^{2} d y

\frac{( - y + 1 ) ^{3}}{3}

Solution steps

\int_{0}^{1 - y} y^{2} d y

Apply the Power Rule

: [\frac{y ^{3}}{3}]_{0}^{1 - y}

= [\frac{y ^{3}}{3}]_{0}^{1 - y}

Compute the boundaries:

\frac{1}{3} (- y + 1)^{3}

= \frac{1}{3} (- y + 1)^{3}

Simplify

= \frac{( - y + 1 ) ^{3}}{3}

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