What is the integral from 0 to 2 of (x^2-2)^4x ?

The integral from 0 to 2 of (x^2-2)^4x is 32/5

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integral from 0 to 2 of (x^2-2)^4x

\int_{0}^{2} (x^{2} - 2)^{4} x d x

\frac{32}{5}

Decimal

6.4

Solution steps

\int_{0}^{2} (x^{2} - 2)^{4} x d x

Apply u-substitution

= \int_{- 2}^{2} \frac{u ^{4}}{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_{- 2}^{2} u^{4} d u

Apply the Power Rule

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

Compute the boundaries:

\frac{64}{5}

= \frac{1}{2} \cdot \frac{64}{5}

Simplify

= \frac{32}{5}

What is the integral from 0 to 2 of (x^2-2)^4x ?
The integral from 0 to 2 of (x^2-2)^4x is 32/5