adiadas8o7

2022-04-30

Show that the value of a definite integral is unity

${\int}_{2}^{4}\frac{\sqrt{\mathrm{log}(9-x)}}{\sqrt{\mathrm{log}(9-x)}+\sqrt{\mathrm{log}(3+x)}}dx=1$

Volsa280

Beginner2022-05-01Added 16 answers

Proof. By making a substitution $y=6-x$ , we get

$I={\int}_{4}^{2}-\frac{f(6-y)}{f\left(y\right)+f(6-y)},dy={\int}_{2}^{4}\frac{f(6-y)}{f\left(y\right)+f(6-y)},dy$

Therefore

$2I={\int}_{2}^{4}\frac{f\left(x\right)}{f\left(x\right)+f(6-x)}dx+{\int}_{2}^{4}\frac{f(6-y)}{f\left(y\right)+f(6-y)}dy$

$={\int}_{2}^{4}\frac{f\left(x\right)}{f\left(x\right)+f(6-x)}dx+{\int}_{2}^{4}\frac{f(6-x)}{f\left(x\right)+f(6-x)}dx={\int}_{2}^{4}1,dx=2$

Therefore

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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