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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Simplify

(x)^{2} : x

= \int_{0}^{1} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{1}

= [\frac{x ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\frac{\partial}{\partial x} (5 y)

\frac{d}{d x} y = (x^{2} + 3 x)^{2}

\frac{\partial}{\partial x} (6 - x^{7} - y^{5})

\frac{d}{d x} (arctan (x + 2))

\frac{d}{d x} (sin (x - y))

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