What is the integral from-2 to 5 of \sqrt[3]{(3t+8)} ?

The integral from-2 to 5 of \sqrt[3]{(3t+8)} is (23\sqrt[3]{23}*2^{2/3}-4)/(4*2^{2/3)}

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integral from-2 to 5 of \sqrt[3]{(3t+8)}

Decimal Notation

15.72227 \dots

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{2}^{23}

= \frac{1}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{2}^{23}

Compute the boundaries:

Simplify

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

\int_{0}^{1} (x + y)^{2} d x

\int_{0}^{z} z d z

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

\int_{1}^{3} \frac{3}{x ^{3}} d x

What is the integral from-2 to 5 of \sqrt[3]{(3t+8)} ?
The integral from-2 to 5 of \sqrt[3]{(3t+8)} is (23\sqrt[3]{23}*2^{2/3}-4)/(4*2^{2/3)}