What is the integral from 0 to 1 of 2x^2+sqrt(x) ?

The integral from 0 to 1 of 2x^2+sqrt(x) is 4/3

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integral from 0 to 1 of 2x^2+sqrt(x)

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

\int_{0}^{1} 2 x^{2} d x = \frac{2}{3}

\int_{0}^{1} x d x = \frac{2}{3}

= \frac{2}{3} + \frac{2}{3}

Simplify

\frac{2}{3} + \frac{2}{3} : \frac{4}{3}

= \frac{4}{3}

\int_{2}^{5} x^{2} 4 x^{3} + 3 d x

\int_{2}^{3} x - 1 d x

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

\int_{- 1}^{2} x^{2} + 4 d x

\int_{0}^{a} x sin (x) d x

What is the integral from 0 to 1 of 2x^2+sqrt(x) ?
The integral from 0 to 1 of 2x^2+sqrt(x) is 4/3