Rivka Thorpe

## Answered question

2021-10-21

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

### Answer & Explanation

davonliefI

Skilled2021-10-22Added 79 answers

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 $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$
${\int }_{1}^{2}\left(3+5y\right)\left[{\int }_{4}^{9}{x}^{-\frac{1}{2}}dx\right]dy={\int }_{1}^{2}\left(3+5y\right)\left[{\left(\frac{{x}^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}\right)}_{4}^{9}\right]dy$
$={\int }_{1}^{2}\left(3+5y\right)\left[{\left(\frac{{x}^{\frac{1}{2}}}{\frac{1}{2}}\right)}_{4}^{9}\right]dy$
$={\int }_{1}^{2}\left(3+5y\right)\left[2\left({9}^{\frac{1}{2}}-{4}^{\frac{1}{2}}\right)\right]dy$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?