Class 12 RD Sharma Solutions – Chapter 21 Areas of Bounded Regions – Exercise 21.1 | Set 3

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 

x	0						π
	1.57	2.07	2.57                  3.07	3.57	3.07            2.57	2.07	1.57

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

x	0				π
y	0         0.66	1.25	1.88	2.5  1.88   1.25   0.66	0

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 = 

x y = sin x
y = sin 2x	0      0.8       1       0.8	0

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

x y = cos2x	0 1
y = sin2x	0	0.25          0.5	0.75      1	0.75     0.5	0.25       0

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