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

The integral from 0 to 1 of x^2(3x^3-1)^4 is 11/15

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

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

\frac{11}{15}

Decimal

0.73333 \dots

Solution steps

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

Apply u-substitution

= \int_{- 1}^{2} \frac{u ^{4}}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{- 1}^{2} u^{4} d u

Apply the Power Rule

= \frac{1}{9} [\frac{u ^{5}}{5}]_{- 1}^{2}

Compute the boundaries:

\frac{33}{5}

= \frac{1}{9} \cdot \frac{33}{5}

Simplify

= \frac{11}{15}

What is the integral from 0 to 1 of x^2(3x^3-1)^4 ?
The integral from 0 to 1 of x^2(3x^3-1)^4 is 11/15