What is the integral from 0 to 1 of sqrt(s+t) ?

The integral from 0 to 1 of sqrt(s+t) is (2(t+1)^{3/2}-2t^{3/2})/3

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integral from 0 to 1 of sqrt(s+t)

\int_{0}^{1} s + t d s

\frac{2 ( t + 1 ) ^{\frac{3}{2}} - 2 t ^{\frac{3}{2}}}{3}

Solution steps

\int_{0}^{1} s + t d s

Apply u-substitution

: \int_{t}^{t + 1} u d u

= \int_{t}^{t + 1} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{t}^{t + 1}

= [\frac{2}{3} u^{\frac{3}{2}}]_{t}^{t + 1}

Compute the boundaries:

\frac{2}{3} (t + 1)^{\frac{3}{2}} - \frac{2}{3} t^{\frac{3}{2}}

= \frac{2}{3} (t + 1)^{\frac{3}{2}} - \frac{2}{3} t^{\frac{3}{2}}

Simplify

= \frac{2 ( t + 1 ) ^{\frac{3}{2}} - 2 t ^{\frac{3}{2}}}{3}

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What is the integral from 0 to 1 of sqrt(s+t) ?
The integral from 0 to 1 of sqrt(s+t) is (2(t+1)^{3/2}-2t^{3/2})/3