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

y^{^{''}} + y = 0, y (0) = 0

(5, 2), (- 2, - 2)

\frac{d}{d r} (4 - r^{2})

\int x (3 sec^{2} (x) + 2 tan^{2} (x)) d x

y = 3 x^{4}

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