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

Popular >

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

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

\frac{32}{5}

Decimal Notation

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

= \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{u ^{5}}{5}]_{- 2}^{2}

= \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}

y = 4 x, y = \frac{5}{4} x, y = 21 - x^{2}

x \to - 1 - lim (\frac{x ^{3}}{3 x + 3})

3 x^{4}

\frac{\partial}{\partial x} (\frac{x ^{2} + y ^{2}}{x + y})

y = \frac{590}{x}

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