Question Number 204921 by mathlove last updated on 02/Mar/24
Answered by Frix last updated on 02/Mar/24
![∫_0 ^1 (dx/( (√(x+3))+(√(x+1))))=(1/2)∫_0 ^1 (√(x+3))−(√(x+1))dx= =[(((x+3)^(3/2) −(x+1)^(3/2) )/3)]_0 ^1 =3−((2(√2))/3)−(√3) 2a+3b−4c=8](https://www.tinkutara.com/question/Q204935.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:\sqrt{{x}+\mathrm{3}}+\sqrt{{x}+\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{{x}+\mathrm{3}}−\sqrt{{x}+\mathrm{1}}{dx}= \\ $$$$=\left[\frac{\left({x}+\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{3}−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}−\mathrm{4}{c}=\mathrm{8} \\ $$