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

The integral from 2 to 3 of x(x^2-4)^3 is 625/8

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

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

\frac{625}{8}

Decimal

78.125

Solution steps

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

Apply u-substitution

: \int_{0}^{5} \frac{u ^{3}}{2} d u

= \int_{0}^{5} \frac{u ^{3}}{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}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{5}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{0}^{5}

Compute the boundaries:

\frac{625}{4}

= \frac{1}{2} \cdot \frac{625}{4}

Simplify

= \frac{625}{8}

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What is the integral from 2 to 3 of x(x^2-4)^3 ?
The integral from 2 to 3 of x(x^2-4)^3 is 625/8