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Question Number 105026 by mohammad17 last updated on 25/Jul/20

Answered by john santu last updated on 25/Jul/20

$$Q-3$$$$\frac{dx}{dy}=\frac{y^2}{2}-\frac{1}{2y^2}=\frac{y^4-1}{2y^2}$$$$lengthofcurve$$$$=\underset{a}{\stackrel{1.5a}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy$$$$=\underset{a}{\stackrel{1.5a}{\int }}\sqrt{1+{\left(\frac{y^4-1}{2y^2}\right)}^2}dy$$$$=\underset{a}{\stackrel{1.5a}{\int }}\sqrt{\frac{y^8+4y^4-2y^4+1}{4y^4}}dy$$$$=\underset{a}{\stackrel{1.5a}{\int }}\frac{1}{2y^2}\sqrt{{(y^4+1)}^2}dy$$$$=\underset{a}{\stackrel{1.5a}{\int }}\frac{y^4+1}{2y^2}dy={[\frac{1}{6}{y}^3-\frac{1}{2y}]}_a^{1.5a}$$$$\left(JS\mathrm{♠}⧫\right)$$

Commented by mohammad17 last updated on 25/Jul/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 25/Jul/20

$$2){\int }_0^a\sqrt{4-9x^2}dx$$$$={\left[\frac{3x}{2}\sqrt{4-9x^2}\right]}_0^a-{\left[\frac{{\left(2\right)}^2}{2}{sin}^{-1}\frac{3x}{2}\right]}_0^a$$$$=\frac{3a}{2}\sqrt{4-9a^2}-2{sin}^{-1}\frac{3a}{2}$$

Commented by mohammad17 last updated on 25/Jul/20

$$thankyousir$$

Answered by john santu last updated on 25/Jul/20

$$Q-4.\left(1\right)$$$$\int {(ax+3a)}^{a+3}dx=\frac{1}{a}\int {(ax+3a)}^{a+3}d(ax+3a)$$$$[b=ax+3a]$$$$\frac{1}{a}\int b^{a+3}db=\frac{b^{a+4}}{a(a+4)}+c$$$$=\frac{{(ax+3a)}^{a+4}}{a^2+4a}+c$$

Answered by mathmax by abdo last updated on 25/Jul/20

$$4)\int {(\mathrm{ax}+3\mathrm{a})}^{\mathrm{a}+3}\mathrm{dx}=\int {\mathrm{a}}^{\mathrm{a}+3}{(\mathrm{x}+3\mathrm{a})}^{\mathrm{a}+3}\mathrm{dx}={\mathrm{a}}^{\mathrm{a}+3}\int {(\mathrm{x}+3\mathrm{a})}^{\mathrm{a}+3}\mathrm{dx}$$$$=\frac{{\mathrm{a}}^{\mathrm{a}+3}}{\mathrm{a}+4}{(\mathrm{x}+3\mathrm{a})}^{\mathrm{a}+4}+\mathrm{c}$$

Commented by bemath last updated on 26/Jul/20

$$typosir.\frac{a^{a+3}}{a+4}{(x+3)}^{a+4}+c$$

Commented by mathmax by abdo last updated on 26/Jul/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 25/Jul/20

$$2)\mathrm{I}={\int }_0^{\mathrm{a}}\sqrt{4-{9\mathrm{x}}^2}\mathrm{dx}=2{\int }_0^{\mathrm{a}}\sqrt{1-\frac{9}{4}{\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$\frac{3}{2}\mathrm{x}=\mathrm{sint}\Rightarrow \mathrm{I}=2{\int }_0^{\arcsin \left(\frac{3\mathrm{a}}{2}\right)}\sqrt{1-{\sin }^2\mathrm{t}}\times \frac{2}{3}\mathrm{cost}\mathrm{dt}$$$$=\frac{4}{3}{\int }_0^{\arcsin \left(\frac{3\mathrm{a}}{2}\right)}{\cos }^2\mathrm{t}\mathrm{dt}=\frac{2}{3}{\int }_0^{\arcsin \left(\frac{3\mathrm{a}}{2}\right)}(1+\cos \left(2\mathrm{t}\right))\mathrm{dt}$$$$=\frac{2}{3}\arcsin \left(\frac{3\mathrm{a}}{2}\right)+\frac{1}{3}{\left[\sin \left(2\mathrm{t}\right)\right]}_{\mathrm{o}}^{\arcsin \left(\frac{3\mathrm{a}}{2}\right)}$$$$=\frac{2}{3}\arcsin \left(\frac{3\mathrm{a}}{2}\right)+\frac{2}{3}{\left[\mathrm{sint}\sqrt{1-{\sin }^2\mathrm{t}}\right]}_0^{\arcsin \left(\frac{3\mathrm{a}}{2}\right)}$$$$=\frac{2}{3}\arcsin \left(\frac{3\mathrm{a}}{2}\right)+\frac{2}{3}\left\{\frac{3\mathrm{a}}{2}\sqrt{1-\frac{{9\mathrm{a}}^2}{4}}\right\}$$$$=\frac{2}{3}\arcsin \left(\frac{3\mathrm{a}}{2}\right)+\mathrm{a}\sqrt{1-\frac{{9\mathrm{a}}^2}{4}}$$

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