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

The integral from-2 to 6 of x^2\sqrt[3]{x+2} is 4752/35

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

\frac{4752}{35}

Solution steps

Apply u-substitution

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int_{0}^{8} u^{\frac{7}{3}} d u = \frac{1536}{5}

\int_{0}^{8} 4 u^{\frac{4}{3}} d u = \frac{1536}{7}

= \frac{1536}{5} - \frac{1536}{7} + 48

Simplify

\frac{1536}{5} - \frac{1536}{7} + 48 : \frac{4752}{35}

= \frac{4752}{35}

\int - \frac{4 cos ( t )}{3 - 10 sin ( t )} d t

\int (2 x - 5)^{\frac{1}{2}} d x

f (x) = ln (x^{6} e^{x})

What is the integral from-2 to 6 of x^2\sqrt[3]{x+2} ?
The integral from-2 to 6 of x^2\sqrt[3]{x+2} is 4752/35