Integral Calculus MCQ Quiz in मल्याळम - Objective Question with Answer for Integral Calculus - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

⇒$I={\int }_0^2{\int }_0^{\sqrt{2x-x^2}}\frac{xdydx}{\sqrt{x^2+y^2}}$

Bounded region

⇒ x = 0 to x = 2 and y = 0 to y = $\sqrt{2x-x^2}$

⇒ y² = 2x - x² ⇒ $x^2-2x+y^2=0⟹(x-1{)}^2+y^2=1$

convert into polar form 

$⟹x=rcos\theta ,y=rsin\theta$

⇒    $$\begin{gathered} (x-1{)}^2+y^2=1⟹(r\cos \theta -1{)}^2+(r\sin \theta {)}^2=1 \\ {r}^2{\cos }^2\theta -2r\cos \theta +1+r^2{\sin }^2\theta =1 \\ r=0 \\ r=2\cos \theta \end{gathered}$$

⇒$0\le \theta \le \frac{\pi }{2}$

⇒ $I={\int }_0^{\frac{\pi }{2}}{\int }_0^{2cos\theta }(\frac{rcos\theta }{r})rdrd\theta$

$⟹I={\int }_0^{\frac{\pi }{2}}cos\theta {\int }_0^{2cos\theta }rdrd\theta$

$⟹I=2{\int }_0^{\frac{\pi }{2}}co{s}^3\theta d\theta$

$W_n={\int }_0^{\pi /2}{\cos }^n\theta d\theta$  

1. For even n :

${\int }_0^{\pi /2}{\cos }^n\theta d\theta =\frac{(n-1)(n-3)(n-5)\dots 1}{n(n-2)(n-4)\dots 2}\times \frac{\pi }{2}$   

​​​​2. For odd n :

${\int }_0^{\pi /2}{\cos }^n\theta d\theta =\frac{(n-1)(n-3)(n-5)\dots 2}{n(n-2)(n-4)\dots 1}$  

$⟹{\int }_0^{\pi /2}{\cos }^3\theta d\theta =\frac{2}{3}$

⇒ I = $2\times \frac{2}{3}=\frac{4}{3}$

Hence option 2 is correct

Explanation:

⇒$I={\int }_0^1{\int }_0^x(x^2+y^2)dydx$

Compute integral.

⇒   ${\int }_0^1{\int }_0^x(x^2+y^2)dydx={\int }_0^1(x^{2}y+\frac{y^3}{3}{)}_0^{x}dx$

⇒  ${\int }_0^1(x^3+\frac{x^3}{3})={\int }_0^1\frac{4x^3}{3}dx$

⇒  $\frac{4}{3}(\frac{x^4}{4}{)}_0^1=\frac{1}{3}(1^4-0^4)=\frac{1}{3}$

Hence option 3  is correct.