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 ⇒ \(x^2-2x+y^2 = 0 \implies (x-1)^2 +y^2 = 1\)

convert into polar form 

\(\implies x= rcos\theta , y = r sin\theta\)

⇒    \((x - 1)^2 + y^2 = 1 \implies (r \cos \theta - 1)^2 + (r \sin \theta)^2 = 1 \\ r^2 \cos^2 \theta - 2r \cos \theta + 1 + r^2 \sin^2 \theta = 1\\ r = 0\\ r = 2 \cos \theta\)

⇒\(0 \leq \theta \leq \frac{\pi}{2}\)

⇒ \(I = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2cos\theta} (\frac{rcos\theta}{r})rdrd\theta\)

\(\implies I = \int_{0}^{\frac{\pi}{2}}cos\theta \int_{0}^{2cos\theta}rdr d\theta\)

\(\implies I = 2\int_{0}^{\frac{\pi}{2}}cos^3\theta d\theta\)

\(W_n = \int_0^{\pi/2} \cos^n \theta \, d\theta \)  

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 :

\(\int_0^{\pi/2} \cos^n \theta \, d\theta = \frac{(n-1)(n-3)(n-5) \dots 2}{n(n-2)(n-4) \dots 1} \)  

\(\implies \int_0^{\pi/2} \cos^3 \theta \, d\theta = \frac{2}{3} \)

⇒ 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.

⇒   \( \int_{0}^{1}\int_0^x (x^2 + y^2) \, dydx = \int_{0}^{1}(x^2y+\frac{y^3}{3})_{0}^{x}dx \)

⇒  \(\int_{0}^{1}(x^3 + \frac{x^3}{3}) = \int_0^1\frac{4x^3}{3}dx\)

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

Hence option 3  is correct.