Evaluate using Special Integral Forms MCQ Quiz in मराठी - Objective Question with Answer for Evaluate using Special Integral Forms - मोफत PDF डाउनलोड करा

Last updated on Apr 10, 2025

पाईये Evaluate using Special Integral Forms उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Evaluate using Special Integral Forms एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Evaluate using Special Integral Forms MCQ Objective Questions

Top Evaluate using Special Integral Forms MCQ Objective Questions

Evaluate using Special Integral Forms Question 1:

The value of \(\int_{-1}^1\left|\tan ^{-1} x\right| d x\) is:

  1. \(\frac{\pi}{2}-\log _{\mathrm{e}} 2\)
  2. \(\frac{\pi}{2}+\log _{\mathrm{e}} 2\)
  3. \(\frac{\pi-1-\log _{\mathrm{e}} 2}{2}\)
  4. \(\frac{\pi-1+\log _{\mathrm{e}} 2}{2}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{\pi}{2}-\log _{\mathrm{e}} 2\)

Evaluate using Special Integral Forms Question 1 Detailed Solution

Calculation:

We are given the function:

I = ∫-11 |tan-1(x)| dx

Step 1: Split the integral based on the absolute value:

I = ∫-10 -tan-1(x) dx + ∫01 tan-1(x) dx

Step 2: Use the standard result for the integral of tan-1(x):

∫ tan-1(x) dx = x tan-1(x) - 1/2 ln(1 + x2)

Step 3: Compute each integral:

For ∫-10 -tan-1(x) dx, we get:

- [ x tan-1(x) - 1/2 ln(1 + x2) ]-10

At x = 0, the expression becomes 0.

At x = -1, we get:

- [-1 × (-π/4) - 1/2 ln(1 + 1)] = π/4 - 1/2 ln(2)

Thus, the value of the first integral is:

π/4 - 1/2 ln(2)

For ∫01 tan-1(x) dx, we use the same result:

[ x tan-1(x) - 1/2 ln(1 + x2) ]01

At x = 1, the expression becomes:

1 × (π/4) - 1/2 ln(2) = π/4 - 1/2 ln(2)

At x = 0, the expression is 0.

Thus, the value of the second integral is:

π/4 - 1/2 ln(2)

Step 4: Add the results of the two integrals:

I = (π/4 - 1/2 ln(2)) + (π/4 - 1/2 ln(2)) = π/2 - ln(2)

∴ The value of the integral is π/2 - ln(2).

The correct answer is Option (1): π/2 - ln(2)

Evaluate using Special Integral Forms Question 2:

Let \(\rm \beta(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx\) m, n > 0 If \(\rm \int_0^1(1-x^{10})^{20}dx=a\times \beta (b,c)\) then 100 (a + b + c) equals ______. 

  1. 1021
  2. 1120
  3. 2012
  4. 2120

Answer (Detailed Solution Below)

Option 4 : 2120

Evaluate using Special Integral Forms Question 2 Detailed Solution

Calculation:

Given, \(\rm \beta(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx\)

Let I = \(\int_0^1 1 \cdot\left(1-x^{10}\right)^{20} d x\)

Put x10 = t ⇒ x = t1/10 

⇒ \(\mathrm{dx}=\frac{1}{10}(\mathrm{t})^{-9 / 10} \mathrm{dt}\)

∴ I = \(\int_0^1(1-t)^{20} \frac{1}{10}(t)^{-9 / 10} d t\)

⇒ I = \(\frac{1}{10} \int_0^1 t^{-9 / 10}(1-t)^{20} d t\) 

\(\frac{1}{10} \int_0^1 x^{-9 / 10}(1-x)^{20} d x\)

\(\rm \frac{1}{10}\times \beta (\frac{1}{10},21)\) = \(\rm a\times \beta (b,c)\)

⇒ a = \(\frac{1}{10}\) b = \(\frac{1}{10}\) c = 21

⇒ 100(a + b + c) = 100(\(\frac{1}{10}\) + \(\frac{1}{10}\) + 21) = 10 + 10 + 2100 = 2120

∴ The value of 100(a + b + c) is 2120.

The correct answer is Option 4.

Evaluate using Special Integral Forms Question 3:

What is the value of \(\rm \int e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \)

  1. \(\rm e^x ({1\over x^2})\) + c
  2. \(\rm e^x ({-1\over x^2})\) + c
  3. \(\rm e^x ({1\over x})\) + c
  4. \(\rm e^x ({-1\over x})\) + c
  5. None of these

Answer (Detailed Solution Below)

Option 3 : \(\rm e^x ({1\over x})\) + c

Evaluate using Special Integral Forms Question 3 Detailed Solution

Concept

\(\rm \int e^x \left(f(x)+f'(x)\right)dx \) = ex f(x) + c

Calculation:

Let, \(\rm I=\int e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \)

Let f(x) = \(\rm 1\over x\)

⇒ \(\rm f'(x) = - {1\over x^2}\)

∴ \(\rm I=\int e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \)\(\rm \int e^x \left(f(x)+f'(x)\right)dx \)

ex f(x) + c

\(\rm e^x ({1\over x})\) ​​+ c

Hence, option (3) is correct.

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