What is the integral from 0 to 4 of (x^2+sqrt(x)) ?

The integral from 0 to 4 of (x^2+sqrt(x)) is 80/3

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integral from 0 to 4 of (x^2+sqrt(x))

\int_{0}^{4} (x^{2} + x) d x

\frac{80}{3}

Decimal

26.66666 \dots

Solution steps

\int_{0}^{4} x^{2} + x 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_{0}^{4} x^{2} d x + \int_{0}^{4} x d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} x d x = \frac{16}{3}

= \frac{64}{3} + \frac{16}{3}

Simplify

\frac{64}{3} + \frac{16}{3} : \frac{80}{3}

= \frac{80}{3}

\int_{0}^{2} cos (\frac{nπ x}{2}) d x

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\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 1)^{2} d x

\int_{- 3}^{1} ∣ x + 2 ∣ d x

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

What is the integral from 0 to 4 of (x^2+sqrt(x)) ?
The integral from 0 to 4 of (x^2+sqrt(x)) is 80/3