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Question-200575

Question Number 200575 by sonukgindia last updated on 20/Nov/23 Answered by AST last updated on 20/Nov/23 $$\frac{{ra}}{\mathrm{2}}=\frac{{bx}}{\mathrm{2}}\Rightarrow{ra}={bx};{x}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} };{b}={r}−{a} \\ $$$${ra}={bx}\Rightarrow{ra}=\left({r}−{a}\right)\left(\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−}…

Question-200549

Question Number 200549 by sonukgindia last updated on 20/Nov/23 Answered by witcher3 last updated on 20/Nov/23 $$\mathrm{I}_{\mathrm{12}} =\mathrm{0};\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}_{\mathrm{11}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\int_{\mathrm{1}}…

Let-u-n-k-1-n-n-n-2-k-2-for-n-N-gt-0-Show-that-u-n-n-gt-0-is-increasing-

Question Number 200521 by brahim_mekkaoui last updated on 19/Nov/23 $$\mathrm{Let}\:{u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{n}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:\:\mathrm{for}\:{n}\in\mathbb{N}_{>\mathrm{0}} \:\:. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\left({u}_{{n}} \right)_{{n}>\mathrm{0}} \:\mathrm{is}\:\mathrm{increasing}. \\ $$ Terms of Service…

Question-200395

Question Number 200395 by sonukgindia last updated on 18/Nov/23 Answered by witcher3 last updated on 18/Nov/23 $$\mathrm{I}_{\mathrm{10}} \rightarrow\mathrm{I}_{\mathrm{9}} \\ $$$$\mathrm{I}_{\mathrm{10}} =\frac{\mathrm{1}}{\mathrm{m}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{m}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\…

Question-200318

Question Number 200318 by sonukgindia last updated on 17/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $$=\int_{\mathrm{0}} ^{\pi} \frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}+\frac{{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}}…

Question-200319

Question Number 200319 by sonukgindia last updated on 17/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $$=\int_{\mathrm{0}} ^{\pi} \frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}}…