Question Number 6132 by sanusihammed last updated on 15/Jun/16 | ||
$${Evaluate}\:{the}\:{integral}\:{of}\:... \\ $$ $$ \\ $$ $$\left[\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{5}} }{\mathrm{2}.\mathrm{4}}−\frac{{x}^{\mathrm{7}} }{\mathrm{2}.\mathrm{4}.\mathrm{6}}+....\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} }−\frac{{x}^{\mathrm{6}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} .\mathrm{6}^{\mathrm{2}} }+....\right)\right]{dx} \\ $$ $$ \\ $$ $${for}\:\:\mathrm{0}\:<\:\:{x}\:\:<\:\:\infty \\ $$ $$ \\ $$ $${The}\:{answer}\:{is}\:{saying}\:\:............\:\:\sqrt{{e}} \\ $$ $$ \\ $$ $${How}\:{is}\:{the}\:{answer}\:\:\sqrt{{e}} \\ $$ | ||
Commented byprakash jain last updated on 15/Jun/16 | ||
$${x}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{5}} }{\mathrm{2}\centerdot\mathrm{4}}={x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}\centerdot\mathrm{4}}−+...\right) \\ $$ $$={x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} /\mathrm{2}}{\mathrm{1}!}+\frac{\left({x}^{\mathrm{2}} /\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}−\frac{\left({x}^{\mathrm{2}} /\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{3}!}\right) \\ $$ $$={xe}^{−{x}^{\mathrm{2}} /\mathrm{2}} \\ $$ $$\mathrm{similar}\:\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{evaluating}\:\mathrm{other}\:\mathrm{expression} \\ $$ $$\mathrm{and}\:\mathrm{integrate}. \\ $$ | ||