What is the integral from-1 to 0 of 18x^2(3x^3+3)^2 ?

The integral from-1 to 0 of 18x^2(3x^3+3)^2 is 18

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integral from-1 to 0 of 18x^2(3x^3+3)^2

\int_{- 1}^{0} 18 x^{2} (3 x^{3} + 3)^{2} d x

18

Solution steps

\int_{- 1}^{0} 18 x^{2} (3 x^{3} + 3)^{2} d x

Take the constant out:

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

= 18 \cdot \int_{- 1}^{0} x^{2} (3 x^{3} + 3)^{2} d x

Apply u-substitution

: \int_{0}^{3} \frac{u ^{2}}{9} d u

= 18 \cdot \int_{0}^{3} \frac{u ^{2}}{9} d u

Take the constant out:

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

= 18 \cdot \frac{1}{9} \cdot \int_{0}^{3} u^{2} d u

Apply the Power Rule

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

= 18 \cdot \frac{1}{9} [\frac{u ^{3}}{3}]_{0}^{3}

Simplify

18 \cdot \frac{1}{9} [\frac{u ^{3}}{3}]_{0}^{3} : 2 [\frac{u ^{3}}{3}]_{0}^{3}

= 2 [\frac{u ^{3}}{3}]_{0}^{3}

Compute the boundaries:

9

= 2 \cdot 9

Simplify

= 18

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What is the integral from-1 to 0 of 18x^2(3x^3+3)^2 ?
The integral from-1 to 0 of 18x^2(3x^3+3)^2 is 18