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

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

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

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

1

Solution steps

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

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

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

= \frac{1}{3} + \frac{2}{3}

Simplify

\frac{1}{3} + \frac{2}{3} : 1

= 1

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

\int 10 d x

x \to 2 + lim (x)

\int \frac{21}{1 - cos ( 7 x )} d x

n = 1 \sum \infty \frac{1}{( 2 n - 1 ) !}

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