Evaluate using Special Integral Forms MCQ Quiz in বাংলা - Objective Question with Answer for Evaluate using Special Integral Forms - বিনামূল্যে ডাউনলোড করুন [PDF]

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)

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 = t^1/10 

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

∴ 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.

Concept

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