Rivka Thorpe

2021-10-21

Evaluate iterated integral.
${\int }_{1}^{2}{\int }_{4}^{9}\frac{3+5y}{\sqrt{x}}dxdy$

davonliefI

We can evaluate the double integral by integrating with respect to each variable one by one according to their order in the integration term . After integrating with respect to one variable , we have to apply its limit and then we can perform integration using next variable .
The following integration formulas can be used here ,
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$
$\int kxdx=k\frac{{x}^{2}}{2}+C$
$\int \frac{1}{\sqrt{x}}dx=\int {x}^{-\frac{1}{2}}dx$
Consider the double integral ${\int }_{1}^{2}{\int }_{4}^{9}\frac{3+5y}{\sqrt{x}}dxdy$
Where $\sqrt{x}$ can be written as ${x}^{\frac{1}{2}}$ using the conversion formula of radical to exponent $\sqrt{m}\left\{{a}^{n}\right\}={a}^{\frac{n}{m}}$
The expression can be written as $\left(3+5y\right){x}^{-\frac{1}{2}}$ by taking ${x}^{\frac{1}{2}}$ to the numerator.
Thus we can write the integral as ${\int }_{1}^{2}{\int }_{4}^{9}\left(3+5y\right){x}^{-\frac{1}{2}}dxdy$
First we have to integrate with respect to x by taking y terms outside as a constant.
Use the formula $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$
${\int }_{1}^{2}\left(3+5y\right)\left[{\int }_{4}^{9}{x}^{-\frac{1}{2}}dx\right]dy={\int }_{1}^{2}\left(3+5y\right)\left[{\left(\frac{{x}^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}\right)}_{4}^{9}\right]dy$
$={\int }_{1}^{2}\left(3+5y\right)\left[{\left(\frac{{x}^{\frac{1}{2}}}{\frac{1}{2}}\right)}_{4}^{9}\right]dy$
$={\int }_{1}^{2}\left(3+5y\right)\left[2\left({9}^{\frac{1}{2}}-{4}^{\frac{1}{2}}\right)\right]dy$

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