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_{1}^{2} x^{2} - x^{- 2} d x

\int_{1}^{6} \frac{5}{t} d t

\int_{0}^{1} t cosh (2 t) d t

\int_{0}^{1} 3 x + 4 d x

\int_{0}^{3} ∣ 4 - 2 x ∣ 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