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

The integral from 0 to 1 of x(3x^2-1)^5 is 7/4

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

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

\frac{7}{4}

Decimal

1.75

Solution steps

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

Apply u-substitution

: \int_{- 1}^{2} \frac{u ^{5}}{6} d u

= \int_{- 1}^{2} \frac{u ^{5}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{- 1}^{2} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 1}^{2}

= \frac{1}{6} [\frac{u ^{6}}{6}]_{- 1}^{2}

Compute the boundaries:

\frac{21}{2}

= \frac{1}{6} \cdot \frac{21}{2}

Simplify

= \frac{7}{4}

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What is the integral from 0 to 1 of x(3x^2-1)^5 ?
The integral from 0 to 1 of x(3x^2-1)^5 is 7/4