What is the integral from-1 to 0 of x^2(4x^3+5)^3 ?

The integral from-1 to 0 of x^2(4x^3+5)^3 is 13

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integral from-1 to 0 of x^2(4x^3+5)^3

\int_{- 1}^{0} x^{2} (4 x^{3} + 5)^{3} d x

13

Solution steps

\int_{- 1}^{0} x^{2} (4 x^{3} + 5)^{3} d x

Apply u-substitution

= \int_{1}^{5} \frac{u ^{3}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{1}^{5} u^{3} d u

Apply the Power Rule

= \frac{1}{12} [\frac{u ^{4}}{4}]_{1}^{5}

Compute the boundaries:

156

= \frac{1}{12} \cdot 156

Simplify

= 13

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What is the integral from-1 to 0 of x^2(4x^3+5)^3 ?
The integral from-1 to 0 of x^2(4x^3+5)^3 is 13