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

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.

Calculation

Area =

cos x is positive from 0 to π/2 and negative from π/2 to π.

So, we split the integral:

Area =

Area =

Area =

Area =

Area =

Area =

Area =

∴ The area bounded by the curve y = cos x, x = 0 and x = π is 2.

Hence option 1 is correct

Calculation:

Find the equation of line first intercept of y is 2

equation becomes:

y = x + 2

If intercept of y is n

Equation becomes

y = x + n

I = 

I = 

I = 

n = 2

I = 

I = 

Calculation

\sin x, \forall x \in \left( 0, \frac{\pi}{4} \right)\) and

Hence option 2 is correct

Calculation

Given:

Parabola: y² = 4x

Line: x = 3

Since y² = 4x, then y = (in the first quadrant, y > 0)

Required Area = 2 × (Area of the region OCAO)

⇒ Area =

⇒ Area =

⇒ Area =

⇒ Area =

⇒ Area =

⇒ Area =

∴ Required area = sq. units.

Hence option 2 is correct

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: 

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

Calculation:

Enclosed Area

Concept:

Area under a Curve by Integration

Find the area under this curve by summing vertically.

• In this case, we find the area is the sum of the rectangles, height x = f(y) and width dy.
• If we are given y = f(x), then we need to re-express this as x = f(y) and we need to sum from the bottom to top.

So, 

Calculation:

Given Curve: x = 4 - y2

⇒ y2 = 4 - x
⇒ y2 = - (x - 4)           

The above curve is the equation of the Parabola,

We know that at y-axis; x = 0

⇒ y2 = 4 - x

⇒ y2 = 4 - 0 = 4

⇒ y = ± 2

⇒ (0, 2) or (0, -2) are Point of intersection.

Area under the curve 

 Sq. unit

Explanation:

Equation of circle is given by x² + y² = a²

Let's take the strip along a y-direction and integrate it from 0 to 'a' this will give the area of the first quadrant and in order to find out the area of a circle multiply by 4

Area of first Quadrant =  = 

Area of circle = 4 × 

Given:

The parabolas y = 3x2 and x2 - y + 4 = 0

Concept:

Apply concept of area between two curves y₁ and y₂ between x = a and x = b

Calculation:

The parabolas y = 3x2 and x2 - y + 4 = 0

then 3x² = x² + 4

⇒ x² = 2

⇒ x = ± √ 2

Then the area is 

 sq unit.

Hence option (4) is correct.

Concept:

The area of the curve y = f(x) is given by:

A = 

where x1 and x2 are the endpoints between which the area is required.

Imp. Note: The net area will be the addition of the area below the x-axis and the area above the x-axis.

Calculation:

The f(x) = y = 4x³

Given the end points x1 = -2, x2 = 3

Area of the curve (A) =

⇒ A = 

⇒ A = 

⇒ A = 

⇒ A = 

⇒ A = 97

Additional Information

Integral property:

• ∫ xn dx = + C ; n ≠ -1
•  + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Explanation:

Given equation of curves are

y = x²    ---(1) and y = 16    ---(2)

By solving both equation (1) and (2) we have:

x² = 16

x = 4, -4.

∴ Points of intersection are (4, 16) and (-4, 16).

From the figure we have,

By using Integral property we have,

Alternate Method 

There is another method also by which we can solved the problem,

By considering horizontal strip and by the condition of symmetry we have:

Area = 

Concept:

The area under a Curve by Integration:

Find the area under this curve is by summing horizontally.

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

We need to sum from left to right.

∴ Area =  

Calculation: 

Here, we have to find the area of the region bounded by the curves y = x2, x-axis and ordinates x = -1 and x = 2

So, the area enclosed by the given curves is given by 

As we know that, 

Area = 

= 

= 

Area = 3 sq. units.

Explanation:

Given curves are y = x - 1 and y2 = 2x + 6 

On solving, we get, 

y² = 2(y + 1) + 6

⇒ y² - 2y - 8 = 0

⇒ (y - 4)(y + 2) = 0

⇒ y = -2, 4

Now, we can find the area by

A = 

= 

= 

= 

∴ A = 18