Question

evaluate the integral by reversing the order of integration. $ \displaystyle\int {0}^{{\color{red}4}} \displaystyle\int {\sqrt{x}}^{{\color{red}2}}\, \dfrac{{\color{red}3}}{y^3 1}\,dydx$

Answer 1

The integral by reversing the order of integration $\frac{1}{2}$

• Performing u-substitution with definite integrals is very similar to how it's done with indefinite integrals, but with an added step: accounting for the limits of integration.

• The limit and summation that we examined in the previous section to get the net area between a function and the x-axis are exactly what is characterized as the definite integral.

Now, Use u-substitution by given integration

$\mathrm{u}=4+8 \mathrm{x}^2$

du $=16 \mathrm{x}$...now back substitute,

$\begin{aligned}&\int_0^2 \frac{x}{\sqrt{4+8 x^2}} \mathrm{~d} x=\frac{1}{16} \int_0^2 \frac{1}{\sqrt{4+8 x^2}} \mathrm{~d}\left(4+8 x^2\right)=\left.\frac{1}{8} \sqrt{4+8 x^2}\right|_0 ^2 \\&=\frac{1}{8}(\sqrt{4+8 * 4}-\sqrt{4})=\frac{4}{8}=\frac{1}{2}\end{aligned}$

Therefore,

The integration is $\frac{1}{2}$

To learn more about the u-substitution visit: brainly.com/question/1407154

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