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

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

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integral from 0 to 1-x of xy

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

\int_{- 1}^{3} (x) d x

\int_{0}^{π^{7}} 1

\int_{0}^{1} x^{2} - 2 x + 5 d x

\int_{a}^{\infty} x^{2} d x

\int_{0}^{1} x^{2} - 2 x + 6 d x

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