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

The integral from 1 to 2 of x^2sqrt(x^3-1) is (14sqrt(7))/9

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

\int_{1}^{2} x^{2} x^{3} - 1 d x

\frac{14 7}{9}

Decimal

4.11561 \dots

Solution steps

\int_{1}^{2} x^{2} x^{3} - 1 d x

Apply u-substitution

: \int_{0}^{7} \frac{u}{3} d u

= \int_{0}^{7} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{7} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{7}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{7}

Compute the boundaries:

\frac{14 7}{3}

= \frac{1}{3} \cdot \frac{14 7}{3}

Simplify

= \frac{14 7}{9}

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What is the integral from 1 to 2 of x^2sqrt(x^3-1) ?
The integral from 1 to 2 of x^2sqrt(x^3-1) is (14sqrt(7))/9