What is the integral from 1 to 3 of x\sqrt[3]{x^2-1} ?

The integral from 1 to 3 of x\sqrt[3]{x^2-1} is 6

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integral from 1 to 3 of x\sqrt[3]{x^2-1}

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

6

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{3 u}{2} d u

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8}

= \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8}

Compute the boundaries:

12

= \frac{1}{2} \cdot 12

Simplify

= 6

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

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

\int_{0}^{x} x^{2} sin (x z) d z

\int_{0}^{s i n (x)} 7 t d t

\int_{1}^{8} \frac{6}{t} d t

What is the integral from 1 to 3 of x\sqrt[3]{x^2-1} ?
The integral from 1 to 3 of x\sqrt[3]{x^2-1} is 6