What is the integral from 8 to 1 of (\sqrt[3]{x+1}) ?

The integral from 8 to 1 of (\sqrt[3]{x+1}) is -(27\sqrt[3]{9})/4+3/(2^{2/3)}

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integral from 8 to 1 of (\sqrt[3]{x+1})

Solution steps

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

Apply u-substitution

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{2}^{9}

= - [\frac{3}{4} u^{\frac{4}{3}}]_{2}^{9}

Compute the boundaries:

Simplify

\int_{4}^{9} 5 d x

\int_{0}^{1} x 8 x^{2} + 1 d x

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

\int_{- 1}^{0} 1 - x d x

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

What is the integral from 8 to 1 of (\sqrt[3]{x+1}) ?
The integral from 8 to 1 of (\sqrt[3]{x+1}) is -(27\sqrt[3]{9})/4+3/(2^{2/3)}