What is the integral from-2 to 0 of 6x^2(2x^3+4)^2 ?

The integral from-2 to 0 of 6x^2(2x^3+4)^2 is 1792/3

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integral from-2 to 0 of 6x^2(2x^3+4)^2

\int_{- 2}^{0} 6 x^{2} (2 x^{3} + 4)^{2} d x

\frac{1792}{3}

Decimal

597.33333 \dots

Solution steps

\int_{- 2}^{0} 6 x^{2} (2 x^{3} + 4)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{- 2}^{0} x^{2} (2 x^{3} + 4)^{2} d x

Apply u-substitution

: \int_{- 12}^{4} \frac{u ^{2}}{6} d u

= 6 \cdot \int_{- 12}^{4} \frac{u ^{2}}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{6} \cdot \int_{- 12}^{4} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 12}^{4}

= 6 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{- 12}^{4}

Simplify

6 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{- 12}^{4} : [\frac{u ^{3}}{3}]_{- 12}^{4}

= [\frac{u ^{3}}{3}]_{- 12}^{4}

Compute the boundaries:

\frac{1792}{3}

= \frac{1792}{3}

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What is the integral from-2 to 0 of 6x^2(2x^3+4)^2 ?
The integral from-2 to 0 of 6x^2(2x^3+4)^2 is 1792/3