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

The integral from 2-y to-y^2+2 of 1/2 y is (y^4-5y^2+4y)/4

Popular >

integral from 2-y to-y^2+2 of 1/2 y

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

\frac{y ^{4} - 5 y ^{2} + 4 y}{4}

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2 - y}^{- y^{2} + 2} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{2 - y}^{- y^{2} + 2}

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

Compute the boundaries:

\frac{( - y ^{2} + 2 ) ^{2}}{2} - \frac{( 2 - y ) ^{2}}{2}

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

Simplify

= \frac{y ^{4} - 5 y ^{2} + 4 y}{4}

\int_{1}^{4} 25 - 5 x d x

\int_{0}^{6 - 2 x} (6 - 2 x - y) d y

\int_{- 0}^{+ 5} \frac{x}{4 + x ^{2}} d x

\int_{- 1}^{1} 4 x^{2} + 1 d x

\int_{- 3}^{3} \frac{1}{( x + 2 ) ^{3}} d x

What is the integral from 2-y to-y^2+2 of 1/2 y ?
The integral from 2-y to-y^2+2 of 1/2 y is (y^4-5y^2+4y)/4