What is the integral from-1 to 0 of y(2y^2-3)^5 ?

The integral from-1 to 0 of y(2y^2-3)^5 is 91/3

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integral from-1 to 0 of y(2y^2-3)^5

\int_{- 1}^{0} y (2 y^{2} - 3)^{5} d y

\frac{91}{3}

Decimal

30.33333 \dots

Solution steps

\int_{- 1}^{0} y (2 y^{2} - 3)^{5} d y

Apply u-substitution

: \int_{- 1}^{- 3} \frac{u ^{5}}{4} d u

= \int_{- 1}^{- 3} \frac{u ^{5}}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 3}^{- 1} \frac{u ^{5}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{- 3}^{- 1} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 3}^{- 1}

= - \frac{1}{4} [\frac{u ^{6}}{6}]_{- 3}^{- 1}

Compute the boundaries:

- \frac{364}{3}

= - \frac{1}{4} (- \frac{364}{3})

Simplify

= \frac{91}{3}

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What is the integral from-1 to 0 of y(2y^2-3)^5 ?
The integral from-1 to 0 of y(2y^2-3)^5 is 91/3