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

The integral from-1 to 1 of x^2(x^3+2)^4 is 242/15

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

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

\frac{242}{15}

Decimal

16.13333 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{3} \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_{1}^{3} u^{4} d u

Apply the Power Rule

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

Compute the boundaries:

\frac{242}{5}

= \frac{1}{3} \cdot \frac{242}{5}

Simplify

= \frac{242}{15}

What is the integral from-1 to 1 of x^2(x^3+2)^4 ?
The integral from-1 to 1 of x^2(x^3+2)^4 is 242/15