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

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

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integral from 0 to 1-x of y

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

\frac{( - x + 1 ) ^{2}}{2}

Solution steps

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

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{1 - x}

= [\frac{y ^{2}}{2}]_{0}^{1 - x}

Compute the boundaries:

\frac{( - x + 1 ) ^{2}}{2}

= \frac{( - x + 1 ) ^{2}}{2}

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

\int_{0}^{b} x d x

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