Integral Calculus MCQ Quiz in తెలుగు - Objective Question with Answer for Integral Calculus - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Apr 11, 2025

పొందండి Integral Calculus సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Integral Calculus MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

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. 

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