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Q18E

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Found in: Page 699

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Calculate the iterated integral $$\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt$$

Calculating the integral we get $$\Rightarrow \int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{{16\sqrt 2 }}{{15}} - \frac{8}{{15}}$$

See the step by step solution

## Step $$1$$: Given and to find

$$\Rightarrow \int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt$$ (given)

The value of the given integral (to find)

## Step $$2$$: Find the value of the integral with respect to $$s$$

Let

$$\begin{array}{c}u = s + t\\du = ds\end{array}$$

If $$s = 1$$then

$$u = 1 + t$$ and

$$s = 0$$then

$$u = t$$

$$\begin{array}{c}\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \int\limits_0^1 {\int\limits_0^{1 + t} {{u^{1/2}}} } {\rm{ }}dudt\\ = \int\limits_0^1 {\mathop {\left( {\frac{2}{3}{u^{3/2}}} \right)}\nolimits_0^t {\rm{ }}dt} \end{array}$$

$$\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{2}{3}\int\limits_0^1 {\left( {{{\left( {1 + t} \right)}^{3/2}} - {t^{3/2}}} \right){\rm{ }}dt}$$

## Step $$3$$ :Find the value of the find integral

$$\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{2}{3}\int\limits_0^1 {\left( {{{\left( {1 + t} \right)}^{3/2}} - {t^{3/2}}} \right){\rm{ }}dt}$$

Let

$$\begin{array}{l}v = 1 + t\\dv = dt\end{array}$$

If $$t = 1$$ then $$v = 2$$

$$t = 0$$ then $$v = 1$$

$$\begin{array}{c}\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{2}{3}\int\limits_0^1 {{{\left( {1 + t} \right)}^{3/2}}\;dt{\rm{ }} - \frac{2}{3}\int\limits_0^1 {{t^{3/2}}} {\rm{ }}dt} \\ = \frac{2}{3}\int\limits_1^2 {{{\left( v \right)}^{3/2}}\;dv{\rm{ }} - \frac{2}{3}\int\limits_0^1 {{t^{3/2}}} {\rm{ }}dt} \\ = \frac{2}{3}\mathop {\left( {\frac{{{v^{5/2}}}}{{5/2}}} \right)}\nolimits_1^2 - \frac{2}{3}\mathop {\left( {\frac{{{t^{5/2}}}}{{5/2}}} \right)}\nolimits_0^1 \end{array}$$

$$\begin{array}{c}\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{4}{{15}}\left( {{2^{5/2}} - {1^{5/2}}} \right) - \frac{4}{{15}}\left( {{1^{5/2}} - 0} \right)\\ = \frac{4}{{15}}\left( {{{\left( {{2^5}} \right)}^{1/2}} - 1} \right) - \frac{4}{{15}}\left( {1 - 0} \right)\\ = \frac{4}{{15}}\left( {\sqrt {32} - 1 - 1} \right)\\ = \frac{4}{{15}}\left( {4\sqrt 2 - 2} \right)\end{array}$$

$$\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{{16\sqrt 2 }}{{15}} - \frac{8}{{15}}$$

Therefore, $$\int\limits_0^1 {\int\limits_0^1 {\sqrt {s + t} {\rm{ }}} } dsdt = \frac{{16\sqrt 2 }}{{15}} - \frac{8}{{15}}$$