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

The integral from-2 to 2 of (\sqrt[3]{x}+x^2) is 16/3

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Apply the Sum Rule:

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

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

\int_{- 2}^{2} 3 x d x = 0

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

= 0 + \frac{16}{3}

Simplify

= \frac{16}{3}

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What is the integral from-2 to 2 of (\sqrt[3]{x}+x^2) ?
The integral from-2 to 2 of (\sqrt[3]{x}+x^2) is 16/3