Evaluate using Special Integral Forms MCQ Quiz - Objective Question with Answer for Evaluate using Special Integral Forms - Download Free PDF

Calculation: 

Let, 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Hence, the correct answer is Option 4.

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, 

Let I = 

Put x¹⁰ = t ⇒ x = t^1/10 

⇒ 

∴ I = 

⇒ I =  

= 

=  = 

⇒ a =  b =  c = 21

⇒ 100(a + b + c) = 100( +  + 21) = 10 + 10 + 2100 = 2120

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

The correct answer is Option 4.

Explanation -

Let e^x – 1 = t²

e^x dx = 2t dt 

= 

= 2 tan^–1 t 

= 

= 

= 

⇒ 

2x² – 5x + 2 = 0

Hence Option (1) is correct.

Concept:

Integrals of even and odd functions:

​​For continuous even functions such that f(−x) = f(x),

f(x)dx = 2f(x)dx

For continuous odd functions such that f(−x) = −f(x),

f(x)dx = 0

Solution:

I = 

Let f(x) = 

⇒ f(-x) = 

⇒ f(-x) = 

⇒ f(-x)= - f(x)

Hence, f(x) is odd function.

∴ The value of the given integral is 0.

Concept

  = e^x f(x) + c

Calculation:

Let, 

Let f(x) = 

⇒ 

∴ = 

= ex f(x) + c

=  ​​+ c

Hence, option (3) is correct.

Concept:

Calculation:

Let, I = 

= 

=                (∵ )

= 

= 

Hence, option (1) is correct. 

Concept:

Integration Formula

Trigonometry Formula

tan() = 1

Calculation:

Given

⇒ 

⇒ 

⇒ tan^-1(a) - tan^-1(0) = 

⇒ tan-1(a) - 0 = 

⇒ a = tan()

⇒ a = 1

The value of a is 1

Additional Information

Integral Formulas:

• ∫ dx = x + C
• ∫ a dx = ax+ C
• ∫ sin x dx = – cos x + C
• ∫ cos x dx = sin x + C
• ∫ sec2 x dx = tan x + C
• ∫ cosec2 x dx = - cot x + C
• ∫ sec x (tan x) dx = sec x + C
• ∫ cosec x ( cot x) dx = – cosec x + C
• ∫ (1/x) dx = ln |x| + C
• ∫ ex dx = ex + C

TFormula used:

Calculation:

Required Area =  , [∵ Area can never be negative]

Since, [x] represents G.I.F and  -1.8

So, The value of [x] for the given range will be - 2

⇒ Required Area =  

⇒ Required Area = 2 [-1.5 + 1.8]

⇒ Required Area = 0.6

∴ The required area is 0.6 square root.

Concept:

1. If f(x) is periodic function with period T then,

2. Fractional part of x: This is the difference between x and its greatest integer part, [x]

• The fractional part of x is given by: {x} = x – [x]
• {x} = x for 0 ≤ x
• Period of the fractional part of x is one.

Graph of the fractional part of x:

Calculation:

We have to find the value of 

                        (∵ x – [x] = {x})

As we know that period of {x} is one.

Concept:

Some standard integration formulae:

Calculation:

∴  The correct option is 3

Concept:

Integral properties:

Calculation:

To find:

Let f(x) = x cos x

Now, checking the function is odd or even

Put x = -x

⇒ f(-x) = (-x) cos (-x)

⇒ f(-x) = -x cos x (∵ cos (-x) = cos x)

So, f(x) = -f(x)

Hence, the given function is odd function.

And we know that, if f(x) is odd.

∴ 

So,  

Concept

 = ex f(x) + c

Calculation:

Given: 

Let f(x) = cos x

⇒ f'(x) = - sin x

∴ 

Concept:

Properties for definite Integral

Calculation:

Given:

                          …. (1)

Now,

From equation 1^st, we get

Concept:

Integration Formula

Trigonometry Formula

tan() = 1

Calculation:

Given

⇒ 

⇒ 

⇒ tan^-1(a) - tan^-1(0) = 

⇒ tan-1(a) - 0 = 

⇒ a = tan()

⇒ a = 1

The value of a is 1

Additional Information

Integral Formulas:

• ∫ dx = x + C
• ∫ a dx = ax+ C
• ∫ sin x dx = – cos x + C
• ∫ cos x dx = sin x + C
• ∫ sec2 x dx = tan x + C
• ∫ cosec2 x dx = - cot x + C
• ∫ sec x (tan x) dx = sec x + C
• ∫ cosec x ( cot x) dx = – cosec x + C
• ∫ (1/x) dx = ln |x| + C
• ∫ ex dx = ex + C