Definite Integrals MCQ Quiz - Objective Question with Answer for Definite Integrals - Download Free PDF

Calculation:

Given,

The function is .

We are tasked with finding the value of .

We can break the integral into two parts as follows:

For the range , . So the first part of the integral is:

For the range , . So the second part of the integral is:

Now, we calculate the values:

Combining the two parts:

Hence, the correct answer is Option 1.

Calculation:

Given,

The function is , where is the greatest integer function.

We are tasked with finding:

Decomposing the integral based on the function:

For \, x² lies between and 1, so . Therefore,

For, x²  lies between 1 and 2, so Therefore,

For , x²  lies between 2 and 3, so . Therefore,

Summing up all the results:

= .

Hence, the correct answer is Option 2. 

Calculation:

The given function is:

We handle the absolute value function |x - t|  by splitting the integral into two cases based on the value of x  relative to t.

If , then , and if x

The function can then be written as:

First, compute the first integral:

Now, compute the second integral:

The total function is:

Thus, simplifying:

To minimize the function, we differentiate f(x) and set it equal to zero.

Set f'(x) = 0:

Taking the natural logarithm:

 ⇒ x = 1.

Substitute x = 1 into f(x) to find the minimum value:

 = 2 (e - 1)

Hence, the correct answer is Option 1

Concept:

Calculation:

Concept:

Integral properties: Consider a function f(x) defined on x.

Calculation:

Let f(x) = sin x – tan x

Checking the function is odd or even,

f(-x) = sin (-x) – tan (-x)

⇒ f(-x) = – sin x + tan x

⇒ f(-x) = –{sin x – tan x}

⇒ f(-x) = –f(x)

Hence, the function is odd.

And we know that,  if f(x) is odd.

∴ 

Concept:

Definite Integral properties:

Calculation:

Let f(x) = x(1 – x)⁹

Now using property, 

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral  is 1/110.

Concept:

Calculation:

Let I = 

= 

= 

Concept:

Calculation: 

Let I =          ----(1)

Using property f(a + b – x),

I =    

As we know,  sin (2π - x) = - sin x and cos (2π - x) = cos x

I =          ----(2)       

I = -I

2I = 0

∴ I = 0

Concept:

Calculation: 

I = 

= 

= 

= 

= 

= 

= 

Concept:

Calculations:

Consider, I =              ....(1)

I = 

I =                            ....(2)

Adding (1) and (2), we have

2I = 

2I = 

2I = 

I = 

Concept:

Calculation:

Let I = 

Let (2 + ln x) = t²

Differentiating with respect to x, we get

⇒ (0 + )dx = 2tdt

⇒ dx = 2tdt

Now,

Concept:

Calculation: 

Let I =              .... (1)

Using property f(a + b – x),

I = 

As we know,  sin (π - x) = sin x and cos (π - x) = -cos x

I = -                .... (2)

I = -I

2I = 0

∴ I = 0

Concept:

f(x) = |x| will be equal to 

• x, if x > 0
• -x, if x
• 0, if x = 0

∫ dx = x + C  (C is a constant)

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

Calculation:

Let, 

Using the above concept, as x ∈ (-2, -1)

⇒ 

⇒ 

⇒   

⇒ I = -[-1 - (-2)] 

∴   = -1

Concept:

Integral property:

• ∫ xn dx = + C ; n ≠ -1
•  + C
• ∫ ex dx = ex+ C

Calculation:

I = 

Let x2 + x + 1 = t

⇒ (2x + 1) dx = dt

I = 

I = 

I = 

I = 

Putting limits

I = 

I = 

I =  = 

Concept:

Calculation:

We know, 

Here, limit of integration is same (i.e., π/4)

Hence, option (3) is correct.