What is the integral from-1 to 1 of x^2(x^3+1)^4 ?

The integral from-1 to 1 of x^2(x^3+1)^4 is 32/15

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integral from-1 to 1 of x^2(x^3+1)^4

\int_{- 1}^{1} x^{2} (x^{3} + 1)^{4} d x

\frac{32}{15}

Decimal

2.13333 \dots

Solution steps

\int_{- 1}^{1} x^{2} (x^{3} + 1)^{4} d x

Apply u-substitution

: \int_{0}^{2} \frac{u ^{4}}{3} d u

= \int_{0}^{2} \frac{u ^{4}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{2} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{2}

= \frac{1}{3} [\frac{u ^{5}}{5}]_{0}^{2}

Compute the boundaries:

\frac{32}{5}

= \frac{1}{3} \cdot \frac{32}{5}

Simplify

= \frac{32}{15}

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What is the integral from-1 to 1 of x^2(x^3+1)^4 ?
The integral from-1 to 1 of x^2(x^3+1)^4 is 32/15