# RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions

Here we provide RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions

### Question 1. Using integration, find the area of region bounded between the line x = 2 and parabola y2 = 8x

Solution:

Here,

Given equations are:

x = 2       ……..(1)

y2 = 8x     ……..(2)

Here,

Equation (1) represents a line parallel to y-axis and equation (2) represents a parabola with vertex at origin and x-axis,

Here is the rough sketch

We have to find the area of shaded region. We sliced it in vertical rectangle width of rectangle = △x,

Length = (y – 0) = y

Area of rectangle = y△x

This rectangle can move horizontal from x = 0 to x = 2

Required area = Shaded region OCBO

Required area =  square units.

### Question 2. Using integration, find the area of the region bounded by the line y – 1 = x, the x-axis and the ordinates x = -2 and x = 3.

Solution:

The find area of region bounded by x-axis the ordinates x = -2 and x = 3

y – 1 = x   ………….(1)

Equation (1) is a line that meets at axes at (0, 1) and (-1, 0)

Here, is the rough sketch

Required area is enclosed between the lines.

Required area = Region ABCA + Region ADEA

### Question 3. Find the area of the region bounded by the parabola y2 = 4ax and the line x = a

Solution:

Here we have to find the area of the region that is bounded by

x = a     ………(1)

and

y2 = 4a    ………(2)

Equation (1) represents a line parallel to y-axis and equation (2) represent a parabola with vertex at origin and axis as x-axis.

Here, is the rough sketch

Here we have to find area between the region,

Thus, slice it in rectangles of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This assumed triangle can go from x = 0 to x = a.

Required area = Region OCBO

= 2 (Region OABO)

Required area =  square units

### Question 4. Find the area lying above the x-axis and under the parabola y = 4x – x2.

Solution:

We have to find here the bounded area by x-axis and parabola

y = 4x – x2

x2 – 4x +4 = -y + 4

(x – 2)2 = -(y – 4)     ……….(1)

Equation (1) represent a downward parabola with vertex (2, 4) and passing through (0, 0) and (0, 4).

Here is the rough sketch

Here the shaded region represents the required area.

We slice the region in approximation rectangles

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = a

Thus,

Required area = Region OABO

Required area =  square units

### Question 5. Draw a rough sketch to indicate the region bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region.

Solution:

We have to find area bounded by

y2 = 4x   ……..(1)

and

x = 3   ………..(2)

Equation (1) represents a parabola with vertex at origin and axis as x-axis and equation (2) represents a line parallel to y-axis

Here is the rough sketch

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

This approximation rectangle slide from x = 0 to x = 3

Required area = Region OCBO

= 2(Region OABO)

Required area =  square units

### Question 6. Make a rough sketch of the graph of the function y = 4 – x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis, and the line x = 0 and x = 2.

Solution:

Here, we will find the area enclosed by

y = 4 – x2

x2 = -(y – 4)     ……….(1)

x = 0    ………(2)

x = 2     ……….(3)

Equation (1) represents a downward parabola with vertex at (0, 4) and passing through (2, 0), (-2, 0). Equation (2) represents y-axis and equation (3) represents a line parallel to y-axis.

Here’s a rough sketch

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

length = y – 0 = y

This approximation rectangle slide from x = 0 to x = 2

Required area = Region OABO

Required area =  square units.

### Question 7. Sketch the graph of  in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the line x = 0 and x = 4.

Solution:

Here, we will find the area enclosed by x-axis and

y2 = x + 1  …….(1)

x = 0   ………(2)

x = 4    ………(3)

Equation (1) represents a parabola with vertex at (-1, 0) and passing through (0, 1) and (0, -1). Equation (2) represents y-axis and equation (3) represents a line parallel to y-axis passing through (4, 0).

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 4

Required area = Region OECDO

Required area =  square units or  square units.

### Question 8. Find the area under the curve  above x-axis from x = 0 to x = 2. Draw a sketch of the curve also.

Solution:

Here, we will find the area enclosed by x-axis

x = 0,

x = 2    ………(1)

y2 = 6x + 4     ……….(2)

Equation (1) represents y-axis and a line parallel to y-axis passing through (2, 0). Equation (2) represents a parabola with vertex at  and passes through the points (0, 2) , ( 0, -2).

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 2,

Required area = Region OABCO

Required area =   square units.

### Question 9. Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Solution:

Here, we will find the area enclosed by x-axis

y2 = x + 1     ……….(1)

and

x = 2  ……………(2)

Equation (1) is a parabola with vertex at (0, 1) and axis as x-axis.

Equation (2) represents a line parallel to y-axis passing through  (2, 0)

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 2,

Required area = Region ABCA

= 2(Region AOCA)

Required area =  square units.

### Question 10. Draw a rough sketch of the graph of the curve  and evaluate the area of the region under the curve and above the x-axis

Solution:

Here, we can observe that ellipse is symmetrical about x-axis.

Area bounded by ellipse =

### Question 11. Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area enclosed by it, using integration.

Solution:

9×2 + 4y2 = 36

Area of Sector OABCO =

Area of the whole figure = 4 x area of DOABCO

### Question 12. Draw a rough sketch of the graph of the function , x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis

Solution:

Here, we have to find the area enclosed between the curve and x-axis.

Equation (1) represents an ellipse with centre at origin and passes through (±1, 0) and (0, ±2) and x ∈ [0, 1] as represented by region between y-axis and line x = 1.

Here, is the rough sketch.

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = 1,

Thus,

Required area = Region OAPBO

Required area =  square units

### Question 13. Determine the area under the curve  included between the line x = 0 and x = 8.

Solution:

Here,

We have to find area under the curve

x2 + y2 = a  ………..(1)

between x = 0    ………(2)

x = a    ………..(3)

Equation (1) represents a circle with Centre (0, 0) and passes axes at (0, ±a), (±a, 0).

Equation (2) represents y-axis and

Equation x = a represents  a line parallel to y-axis passing through (a, 0)

Here, is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = a,

Thus,

Required area = Region OAPBO

Required area =  square units

### Question 14. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8

Solution:

Here,

We have to find area bounded by x-axis

2y + 5x = 7   ………(1)

x = 2      ……..(2)

x = 8    ………(3)

Equation (1)  represents line passing through  and  equation.

Equation (2), (3) shows line parallel to y-axis passing through (2, 0), (8, 0) respectively.

Here, is the rough sketch;

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 2 to x = 8,

Thus,

Required area = Region ABCDA

Required area = 96 square units

### Question 15. Using definite integrals, find the area of the circle x2 + y2 = a2

Solution:

Here, we have to find the area of circle,

x2 + y2 = a2

Equation (1) represents a circle with centre (0, 0) and radius a, Thus is meets the axes (±a, 0), (0, ±a).

Here, is the rough sketch;

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = a,

Thus,

Required area = Region ABCDA

= 4 ( Region ABOA)

### Question 16. Using the integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + |x + 1|, x = -2, x = 3, y =

Solution:

Here, we have to find the area enclosed by;

x = -2,

x = 3,

y = 0 and

y = 1 + |x + 1|

⇒ y = 1 + x + 1, if x + 1 0

⇒ y = 2 + x    ……….(1), if x ≥ -1

and

⇒ y = 1 – (x + 1), if x + 1 < 0

⇒ y = 1 – x – 1, if x < -1

⇒ y = -x    ………(2), if x < -1

Thus,

Equation (1) is a straight line that passes through (0, 2) and (-1 , 1).

Equation (2) is a line passing through (-1, 1) and (-2, 2) and it is enclosed by line x = 2 and x = 3 which are lines parallel to y-axis and pass through (2, 0) and (3, 0) respectively y = 0 is x-axis

Here is the rough sketch

Shaded region represents the required area.

Thus,

Required area = Region (ABECDFA)

Required area = (Region ABEFA + Region ECDFE)    ……..(1)

Region ECDFE

We slice it into approximation rectangle of

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = -2 to x = -1,

Region ABEFA

We slice it into approximation rectangle of

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = -1 to x = 3,

Required area =

Required area =  square units

### Question 17. Sketch the graph y = |x -5|. Evaluate . What does the value of the integral represent on the graph?

Solution:

Here, is the sketch of the given graph:

y = |x – 5|

Hence,

Required area =

Thus,

The given integral represents the area bounded by the curves that are,

x = 0,

y = 0,

x = 1

and

y = -(x – 5).

### Question 18. Sketch the graph of y = |x + 3| and evaluate . What does this integral represents on the graph?

Solution:

Here,

The given equation is y = |x + 3|

The corresponding values of x and y are given in the following table.

Thus,

After plotting these points,

We will get the graph of y = |x + 3|

It is shown as;

It is known that (x + 3) ≤ 0 for -6 ≤ x ≤  -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

Therefore,

Question 19. Sketch the graph y = |x + 1|. Evaluate . What does the value of this integral represent on this graph?

Solution:

Here,

Given:

y = |x + 1|=

y = x + 1   …………(1)

and

y = -x – 1   ……….(2)

Equation (1) represents a line which meets axes at (0, 1).

Equation (2) represents a line passing through (0, -1) and (-1, 0)

Here is the rough sketch

Required area = 9 square units.

### Question 20. Find the area of the region bounded by the curve xy – 3x – 2y – 10 = 0, x-axis and the lines x = 3, x = 4.

Solution:

Here,

We have to find the area bounded by

x axis,

x = 3,

x = 4

and

xy – 3x -2y – 10 = 0

⇒ y(x – 2) = 3x + 10

⇒

Here, is the rough sketch

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 3 to x = 4,

Required area = Region ABCDA

Required area = (3 + 16 log2) square units.

### Question 21. Draw a rough sketch of the curve  and find the area between x-axis, the curve and the ordinates x = 0, x = π

Solution:

Here, we have to find the bounded by

x-axis, x = 0 and x = π

Here is the table for values of

Here is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABCDO

Required area =  square units

### Question 22. Draw a rough sketch of the curve  and find the area between the x-axis, the curve and the ordinates x = 0, x = π.

Solution:

Here, we have the area between y-axis,

x = 0,

x = π

and

Thus, the table for equation (1) is

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABOA

Required area =  square units

### Question 23. Find the area bounded by the curve y = cos x between x = 0 and x = 2π

Solution:

Here from the figure we can see that

The required area = area of the region OABO + area of the region BCDB + area of the region DEFD

Therefore,

The required area =

### Question 24. Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x =  are the ratio 2:3.

Solution:

We have to find the area under the curve

y = sin x    ……..(1)

and

y = sin 2x     …………(2)

Between x = 0 and x =

Here is the rough sketch

Area under curve y = sin 2x

Shaded region represents the required area.

We slice it into approximation rectangle of

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = 0 to x = ,

Thus,

Required area = Region OPACO

We slice it into approximation rectangle of

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = 0 to x = ,

Thus,

Required area = Region OQACO

Thus,

### Question 25. Compare the area under the curves y = cos2x and y = sin2x between x = 0 and x = π

Solution:

Here to compare area under curves

y = cos2x

and

y = sin2x

Between x = 0 and x = π

This is the table for y = cos2x and y = sin2x

Area of region enclosed by

y = cos2x and axis

A1 = Region OABO + Region BCDB

= 2(Region BCDB)

Area of region enclosed by y = sin2x and axis

A2 = Region OEDO

From equation (1) and (2),

A1 = A2

Thus,

Area enclosed by y = cos2x = Area enclosed by y = sin2x.

### Question 26. Find the area bounded by the ellipse  and the ordinates x = 0 and x = ae, where, b2 = a2(1 – e 2) and e < 1.

Solution:

Thus, the required area in the figure below of the region BOB’RFSB is enclosed by the ellipse and the lines x = 0 and x = ae

Here is the area of the region BOB’RFSB

### Question 27. Find the area of the minor segment of the circle x2 + y2 = a2 cut-off by the line x =  .

Solution:

Area of the mirror segment of the circle

### Question 28. Find the area of the region bounded by the curve x = at, y = 2at between the ordinates corresponding t = 1 and t = 2.

Solution:

Area of the bounded region

### Question 29. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.

Solution:

Area of the bounded region

= -8 [0 – 1]

= 8 square units

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.