What is the integral from 0 to 4 of 2y^{1/2} ?

The integral from 0 to 4 of 2y^{1/2} is 32/3

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integral from 0 to 4 of 2y^{1/2}

\int_{0}^{4} 2 y^{\frac{1}{2}} d y

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{0}^{4} 2 y^{\frac{1}{2}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{4} y^{\frac{1}{2}} d y

Apply the Power Rule

: [\frac{2}{3} y^{\frac{3}{2}}]_{0}^{4}

= 2 [\frac{2}{3} y^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= 2 \cdot \frac{16}{3}

Simplify

= \frac{32}{3}

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What is the integral from 0 to 4 of 2y^{1/2} ?
The integral from 0 to 4 of 2y^{1/2} is 32/3