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

The integral from 0 to 1 of (x^2+2x)^2(x+1) is 9/2

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integral from 0 to 1 of (x^2+2x)^2(x+1)

\int_{0}^{1} (x^{2} + 2 x)^{2} (x + 1) d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{0}^{1} (x^{2} + 2 x)^{2} (x + 1) d x

Apply u-substitution

: \int_{0}^{3} \frac{u ^{2}}{2} d u

= \int_{0}^{3} \frac{u ^{2}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{3} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{3}

= \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{3}

Compute the boundaries:

9

= \frac{1}{2} \cdot 9

Simplify

= \frac{9}{2}

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What is the integral from 0 to 1 of (x^2+2x)^2(x+1) ?
The integral from 0 to 1 of (x^2+2x)^2(x+1) is 9/2