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

The integral from-2 to 4 of x^2(x^3+8)^2 is 41472

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integral from-2 to 4 of x^2(x^3+8)^2

\int_{- 2}^{4} x^{2} (x^{3} + 8)^{2} d x

41472

Solution steps

\int_{- 2}^{4} x^{2} (x^{3} + 8)^{2} d x

Apply u-substitution

: \int_{0}^{72} \frac{u ^{2}}{3} d u

= \int_{0}^{72} \frac{u ^{2}}{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}^{72} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{72}

= \frac{1}{3} [\frac{u ^{3}}{3}]_{0}^{72}

Compute the boundaries:

124416

= \frac{1}{3} \cdot 124416

Simplify

= 41472

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x \to 0 lim (\frac{ln ( 1 + 3 x )}{x})

\frac{d}{d y} (- \frac{y}{t ^{2} + y ^{2}})

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What is the integral from-2 to 4 of x^2(x^3+8)^2 ?
The integral from-2 to 4 of x^2(x^3+8)^2 is 41472