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

The integral from 0 to 1 of 3x(4x^2-1)^5 is 91/2

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

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

\frac{91}{2}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 1}^{3} \frac{u ^{5}}{8} d u

= 3 \cdot \int_{- 1}^{3} \frac{u ^{5}}{8} d u

Take the constant out:

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

= 3 \cdot \frac{1}{8} \cdot \int_{- 1}^{3} u^{5} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{8} [\frac{u ^{6}}{6}]_{- 1}^{3}

Simplify

3 \cdot \frac{1}{8} [\frac{u ^{6}}{6}]_{- 1}^{3} : \frac{3}{8} [\frac{u ^{6}}{6}]_{- 1}^{3}

= \frac{3}{8} [\frac{u ^{6}}{6}]_{- 1}^{3}

Compute the boundaries:

\frac{364}{3}

= \frac{3}{8} \cdot \frac{364}{3}

Simplify

= \frac{91}{2}

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What is the integral from 0 to 1 of 3x(4x^2-1)^5 ?
The integral from 0 to 1 of 3x(4x^2-1)^5 is 91/2