What is the integral from 0 to 1 of 2x^2(x^3-1)^8 ?

The integral from 0 to 1 of 2x^2(x^3-1)^8 is 2/27

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integral from 0 to 1 of 2x^2(x^3-1)^8

\int_{0}^{1} 2 x^{2} (x^{3} - 1)^{8} d x

\frac{2}{27}

Decimal

0.07407 \dots

Solution steps

\int_{0}^{1} 2 x^{2} (x^{3} - 1)^{8} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x^{2} (x^{3} - 1)^{8} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{u ^{8}}{3} d u

= 2 \cdot \int_{- 1}^{0} \frac{u ^{8}}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{- 1}^{0} u^{8} d u

Apply the Power Rule

: [\frac{u ^{9}}{9}]_{- 1}^{0}

= 2 \cdot \frac{1}{3} [\frac{u ^{9}}{9}]_{- 1}^{0}

Simplify

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

= \frac{2}{3} [\frac{u ^{9}}{9}]_{- 1}^{0}

Compute the boundaries:

\frac{1}{9}

= \frac{2}{3} \cdot \frac{1}{9}

Simplify

= \frac{2}{27}

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What is the integral from 0 to 1 of 2x^2(x^3-1)^8 ?
The integral from 0 to 1 of 2x^2(x^3-1)^8 is 2/27