What is the integral from 0 to s of 40s^{1/2} ?

The integral from 0 to s of 40s^{1/2} is (80s^{3/2})/3

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integral from 0 to s of 40s^{1/2}

\int_{0}^{s} 40 s^{\frac{1}{2}} d s

\frac{80 s ^{\frac{3}{2}}}{3}

Solution steps

\int_{0}^{s} 40 s^{\frac{1}{2}} d s

Take the constant out:

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

= 40 \cdot \int_{0}^{s} s^{\frac{1}{2}} d s

Apply the Power Rule

= 40 [\frac{2}{3} s^{\frac{3}{2}}]_{0}^{s}

Compute the boundaries:

\frac{2}{3} s^{\frac{3}{2}}

= 40 \cdot \frac{2}{3} s^{\frac{3}{2}}

Simplify

= \frac{80 s ^{\frac{3}{2}}}{3}

What is the integral from 0 to s of 40s^{1/2} ?
The integral from 0 to s of 40s^{1/2} is (80s^{3/2})/3