Integral Calculus MCQ Quiz - Objective Question with Answer for Integral Calculus - 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: 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

Substituting the values of the base (4 units) and the height (16 units):

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

Integrating f'(x) = 4  with respect to  x :

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

Therefore, the equation of the curve is:

This is the equation of a straight line with a slope of 4, passing through the origin.

∴ The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = 

Calculation:

Here, we have to find the area of the region bounded by the curves y = , the line x = 2, x  = 0 and the x - axis

So, the area enclosed by the given curves = 

As we know that, 

Hence, option 4 is the correct answer.

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:

The area under the curve y = f(x) between x = a and x = b, is given by:

Area = 

Similarly, the area under the curve y = f(x) between y = a and y = b, is given by:

Area = 

Calculation:

Here, 

x² = y  and line y = 1 cut the parabola

∴ x² = 1

⇒ x = 1 and -1

Here, the area is symmetric about the y-axis, we can find the area on one side and then multiply it by 2, we will get the area,

This area is between y = x² and the positive x-axis.

To get the area of the shaded region, we have to subtract this area from the area of square i.e.

 square units.

Concept:

Calculation: 

I = 

Let x² + 4 = t

Differentiating with respect to x, we get

⇒ 2xdx = dt

⇒ xdx = 

Now,

I = 

= 

= 

= 

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

Calculation:

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: 

Function y = √f(x) is defined for f(x) ≥ 0. Therefore y can not be negative.

Calculation:

Given: 

y =  and x-axis

At x-axis, y will be zero

y = 

⇒ 0 = 

⇒ 16 - x² = 0

⇒ x2 = 16

∴ x = ± 4

So, the intersection points are (4, 0) and (−4, 0)

Since the curve is y = 

So, y ≥ o [always]

So, we will take the circular part which is above the x-axis

Area of the curve, A 

We know that,

=  

= 

= 8 sin^-1 (1) + 8 sin-1 (1)

= 16 sin^-1 (1)

= 16 × π/2

= 8π sq units

Concept:

Calculation:

I = 

= 

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = 

Now,

I = 

=  + c

=  + c