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

Last updated on Apr 10, 2025

നേടുക Integral Calculus ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Integral Calculus MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Integral Calculus MCQ Objective Questions

Top Integral Calculus MCQ Objective Questions

Integral Calculus Question 1:

Double integal \(\int_0^2\int_0^{\sqrt{2x-x^2}}\frac{xdydx}{\sqrt{x^2+y^2}}\) equals:

  1. \(\frac{2}{3}\)
  2. \(\frac{4}{3}\)
  3. \(\frac{1}{3}\)
  4. \(\frac{8}{3}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{4}{3}\)

Integral Calculus Question 1 Detailed Solution

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

Integral Calculus Question 2:

The integral \(\rm \int_0^1\int_0^x (x^2+y^2)dydx\) is

  1. \(\frac{1}{6}\)
  2. \(\frac{1}{2}\)
  3. \(\frac{1}{3}\)
  4. 1

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{3}\)

Integral Calculus Question 2 Detailed Solution

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. 

Get Free Access Now
Hot Links: online teen patti real money teen patti gold downloadable content teen patti - 3patti cards game teen patti royal - 3 patti