Rivka Thorpe

2021-10-21

Evaluate iterated integral.

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

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$(3+5y){x}^{-\frac{1}{2}}$ by taking $x}^{\frac{1}{2}$ to the numerator.

Thus we can write the integral as${\int}_{1}^{2}{\int}_{4}^{9}(3+5y){x}^{-\frac{1}{2}}dxdy$

First we have to integrate with respect to x by taking y terms outside as a constant.

Use the formula$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$

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

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

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

The following integration formulas can be used here ,

Consider the double integral

Where

The expression can be written as

Thus we can write the integral as

First we have to integrate with respect to x by taking y terms outside as a constant.

Use the formula