Question Number 190746 by leicianocosta last updated on 10/Apr/23
Answered by Mathspace last updated on 12/Apr/23
![a)∫_1 ^4 x(√(1+x^2 ))dx=(1/2)∫_1 ^4 (2x)(x^2 +1)^(1/2) dx [(1/3)(x^2 +1)^(3/2) ]_1 ^4 =(1/3){5^(3/2) −2^(3/2) }](https://www.tinkutara.com/question/Q190796.png)
$$\left.{a}\right)\int_{\mathrm{1}} ^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{4}} \left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{1}} ^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \right\} \\ $$
Answered by Mathspace last updated on 12/Apr/23
![b) ∫_2 ^3 (x/( (√(x−1))))dx we do the changement (√(x−1))=t ⇒x−1=t^2 and ∫_2 ^3 (x/( (√(x−1))))dx=∫_1 ^(√2) ((1+t^2 )/t)(2t)dt =2∫_1 ^(√2) (1+t^2 )dt =2[t+(t^3 /3)]_1 ^(√2) =2{(√2)+((((√2))^3 )/3)−1−(1/3)}](https://www.tinkutara.com/question/Q190797.png)
$$\left.{b}\right)\:\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}}{\:\sqrt{{x}−\mathrm{1}}}{dx}\:{we}\:{do}\:{the}\:{changement} \\ $$$$\sqrt{{x}−\mathrm{1}}={t}\:\Rightarrow{x}−\mathrm{1}={t}^{\mathrm{2}} \:{and} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}}{\:\sqrt{{x}−\mathrm{1}}}{dx}=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{{t}}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt} \\ $$$$=\mathrm{2}\left[{t}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{2}\left\{\sqrt{\mathrm{2}}+\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$