Question Number 205725 by NasaSara last updated on 28/Mar/24
Answered by Berbere last updated on 29/Mar/24
![∫_0 ^2 ∫_0 ^3 [xy]dydx=^(xy=t) ∫_0 ^2 ∫_0 ^(3x) [t](dt/x)dx =∫_0 ^2 (1/x)∫_0 ^(3x) [t]dtdx= =Σ_(k=1) ^5 {∫_(k/3) ^((k+1)/3) (1/x){Σ_(j=1) ^(k−1) ∫_j ^(j+1) [t]+∫_k ^(3x) [t]}dt= isolat k=1 ;and k≥2 ∫_(1/3) ^(2/3) (1/x){3x−1}=1−[ln(2))] .Σ_(k=2) ^5 ∫_(k/3) ^((k+1)/2) (1/x){Σ_(j=1) ^(k−1) j+∫_k ^(3x) kdt}dx+1−ln(2)= Σ_(k=2) ^5 ∫_(k/3) ^((k+1)/3) (1/x){(((k−1)k)/2)+k(3x−k)}dx =Σ_(k=2) ^5 k+Σ_(k=2) ^5 ∫_(k/3) ^((k+1)/3) (1/x)(−(((1+k)k)/2))dx+1−ln(2) =15−Σ_(k=2) ^5 (((1+k)k)/2)ln(((k+1)/k))−ln(2) =15−3ln((3/2))−6ln((4/3))−10ln((5/4))−15ln((6/5))−ln(2) =15+3ln(3)+2ln(2)+4ln(4)+5ln(5)−15ln(6) =15−ln((6^(15) /(3^3 .2^2 .4^4 .5^5 )))=15−ln(((3^(12) .2^5 )/(.5^5 )))=15−ln(((17006112)/(3125)))](Q205738.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}} \left[{xy}\right]{dydx}\overset{{xy}={t}} {=}\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}{x}} \left[{t}\right]\frac{{dt}}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\mathrm{3}{x}} \left[{t}\right]{dtdx}= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left\{\int_{\frac{{k}}{\mathrm{3}}} ^{\frac{{k}+\mathrm{1}}{\mathrm{3}}} \frac{\mathrm{1}}{{x}}\left\{\underset{{j}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\int_{{j}} ^{{j}+\mathrm{1}} \left[{t}\right]+\int_{{k}} ^{\mathrm{3}{x}} \left[{t}\right]\right\}{dt}=\right. \\ $$$${isolat}\:{k}=\mathrm{1}\:;{and}\:{k}\geqslant\mathrm{2} \\ $$$$\left.\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \frac{\mathrm{1}}{{x}}\left\{\mathrm{3}{x}−\mathrm{1}\right\}=\mathrm{1}−\left[{ln}\left(\mathrm{2}\right)\right)\right] \\ $$$$.\underset{{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\int_{\frac{{k}}{\mathrm{3}}} ^{\frac{{k}+\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{{x}}\left\{\underset{{j}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}{j}+\int_{{k}} ^{\mathrm{3}{x}} {kdt}\right\}{dx}+\mathrm{1}−{ln}\left(\mathrm{2}\right)= \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\int_{\frac{{k}}{\mathrm{3}}} ^{\frac{{k}+\mathrm{1}}{\mathrm{3}}} \frac{\mathrm{1}}{{x}}\left\{\frac{\left({k}−\mathrm{1}\right){k}}{\mathrm{2}}+{k}\left(\mathrm{3}{x}−{k}\right)\right\}{dx} \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}{k}+\underset{{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\int_{\frac{{k}}{\mathrm{3}}} ^{\frac{{k}+\mathrm{1}}{\mathrm{3}}} \frac{\mathrm{1}}{{x}}\left(−\frac{\left(\mathrm{1}+{k}\right){k}}{\mathrm{2}}\right){dx}+\mathrm{1}−{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{15}−\underset{{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\frac{\left(\mathrm{1}+{k}\right){k}}{\mathrm{2}}{ln}\left(\frac{{k}+\mathrm{1}}{{k}}\right)−{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{15}−\mathrm{3}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{6}{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\mathrm{10}{ln}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)−\mathrm{15}{ln}\left(\frac{\mathrm{6}}{\mathrm{5}}\right)−{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{15}+\mathrm{3}{ln}\left(\mathrm{3}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{4}{ln}\left(\mathrm{4}\right)+\mathrm{5}{ln}\left(\mathrm{5}\right)−\mathrm{15}{ln}\left(\mathrm{6}\right) \\ $$$$=\mathrm{15}−{ln}\left(\frac{\mathrm{6}^{\mathrm{15}} }{\mathrm{3}^{\mathrm{3}} .\mathrm{2}^{\mathrm{2}} .\mathrm{4}^{\mathrm{4}} .\mathrm{5}^{\mathrm{5}} }\right)=\mathrm{15}−{ln}\left(\frac{\mathrm{3}^{\mathrm{12}} .\mathrm{2}^{\mathrm{5}} }{.\mathrm{5}^{\mathrm{5}} }\right)=\mathrm{15}−{ln}\left(\frac{\mathrm{17006112}}{\mathrm{3125}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by NasaSara last updated on 04/Apr/24
thank you so much