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

The integral from 1 to 3 of x^3sqrt(x^4-1) is (160sqrt(5))/3

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

\int_{1}^{3} x^{3} x^{4} - 1 d x

\frac{160 5}{3}

Decimal

119.25695 \dots

Solution steps

\int_{1}^{3} x^{3} x^{4} - 1 d x

Apply u-substitution

: \int_{0}^{80} \frac{u}{4} d u

= \int_{0}^{80} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{80} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{80}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{80}

Compute the boundaries:

\frac{640 5}{3}

= \frac{1}{4} \cdot \frac{640 5}{3}

Simplify

= \frac{160 5}{3}

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What is the integral from 1 to 3 of x^3sqrt(x^4-1) ?
The integral from 1 to 3 of x^3sqrt(x^4-1) is (160sqrt(5))/3