NCERT Solutions for Class 12 Maths Exercise Miscellaneous Chapter 8 Application of Integrals – FREE PDF Download

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex Misc.) Exercise Miscellaneous

1. Find the area under the given curves and given lines:

(i) and axis.

(ii) and axis.

Ans. (i)Equation of the curve (parabola) is

……..(i)

Required area bounded by curve (i), vertical line and axis

= sq. units

(ii)Equation of the curve

…..(i)

It is clear that curve (i) passes through the origin because from (i)

Table of values for curve for and (given)

	1	2	3	4	5
	1

Required shaded area between the curve , vertical lines and axis

= =

= 624.8 sq. units

2. Find the area between the curves and

Ans. Equation of one curve (straight line) is …..(i)

Equation of second curve (parabola) is …..(ii)

Solving eq. (i) and (ii), we get or and or

Points of intersection of line (i) and parabola (ii) are O (0, 0) and A (1, 1).

Now Area of triangle OAM

= Area bounded by line (i) and axis

= = =

= sq. units

Also Area OBAM = Area bounded by parabola (ii) and axis

= = =

= sq. units

Required area OBA between line (i) and parabola (ii)

= Area of triangle OAM – Area of OBAM

= = sq. units

3. Find the area of the region lying in the first quadrant and bounded by and

Ans. Equation of the curve (parabola) is

……….(i)

……….(ii)

Here required shaded area of the region lying in first quadrant bounded by parabola (i),

and the horizontal lines and is

= =

= = sq. units

4. Sketch the graph of and evaluate

Ans. Equation of the given curve is ……….(i)

0 for all real

Graph of curve is only above the axis i.e., in first and second quadrant only.

If 0

….(ii)

And

If 0

…….(iii)

Table of values for for

				0
	0	1	2	3

Table of values for for


	0	1	2	3

Now,

= +

= –

= $- [\frac{9}{2} - 9 - (18 - 18)] + [0 - (\frac{9}{2} - 9)]$

= $- \frac{9}{2} + 9 + 0 + 0 - \frac{9}{2} + 9$

= sq. units

5. Find the area bounded by the curve between and

Ans. Equation of the curve is ……….(i)

for i.e., graph is in first and second quadrant.

And for i.e., graph is in third and fourth quadrant.

If tangent is parallel to axis, then

Table of values for curve between and

	0
	0	1	0		0

Now Required shaded area = Area OAB + Area BCD

= $| - {(\cos x)}_{0}^{π} | + | - {(\cos x)}_{π}^{2 π} |$

= 2 + 2 = 4 sq. units

6. Find the area enclosed by the parabola and the line

Ans. Equation of parabola is ……..(i)

The area enclosed between the parabola and line is represented by shaded area OADO.

Here Points of intersection of curve (i) and line are O (0, 0) and A.

Now Area ODAMO = Area of parabola and axis

= ……….(ii)

Again Area of = Area between line and axis

= =

= = ……….(ii)

Requires shaded area = Area ODAMO – Area of

7. Find the area enclosed by the parabola and the line

Ans. Equation of the parabola is

……….(i)

Equation of the line is ……(ii) $\Rightarrow$ $y = \frac{3 x + 12}{2}$ = $\frac{3 x}{2} + 6$

In the graph, points of intersection are B (4, 12) and C.

Now, Area ABCD =

= 45 sq. units

Again, Area CDO + Area OAB =

= = 18 sq. units

Required area = Area ABCD – (Area CDO + Area OAB)

= 45 – 18 = 27 sq. units

8. Find the area of the smaller region bounded by the ellipse and the line

Ans. Equation of the ellipse is

………(i)

Here intersection of ellipse (i) with axis are

A (3, 0) and A’ and intersection of ellipse (i) with axis are B (0, 2) and B’.

Also, the points of intersections of ellipse (i) and line are A (3, 0) and B (0, 2).

Area OADB = Area between ellipse (i) (arc AB of it) and axis

= = sq. units……….(ii)

Again Area of triangle OAB = Area bounded by line AB and axis

= $| \int_{0}^{3} \frac{2}{3} (3 - x) d x |$

= sq. units ….(iii)

Now Required shaded area = Area OADB – Area OAB

= sq. units

9. Find the area of the smaller region bounded by the ellipse and the line

Ans. Equation of ellipse is ……..(i)

Area between arc AB of the ellipse and axis

= ………..(ii)

Also Area between chord AB and axis

Now Required area

= Area between arc AB of the ellipse and axis – Area between chord AB and axis

= sq. units

10.Find the area of the region enclosed by the parabola the line and axis.

Ans. Equation of parabola is …………(i)

Equation of line is …………(ii)

Here the two points of intersections of parabola (i) and line (ii) are A and B (2, 4).

Area ALODBM = Area bounded by parabola (i) and axis

= =

= = = 3 sq. units

AlsoArea of trapezium ALMB = Area bounded by line (ii) and axis

= =

= sq. units

Now Required area = Area of trapezium ALMB – Area ALODBM

= sq. units

11.Using the method of integration, find the area enclosed by the curve

Ans. Equation of the curve (graph) is

……….(i)

The area bounded by the curve (i) is represented by the shaded region ABCD.

The curve intersects the axes at points A (1, 0), B (0, 1), C and D

It is observed clearly that given curve is symmetrical about axis and axis.

Area bounded by the curve

= Area of square ABCD

= 4 x OAB

= = 2 sq. units

12.Find the area bounded by the curves .

Ans. The area bounded by the curves is represented by the shaded region.

It is clearly observed that the required area is symmetrical about axis.

Required area

= Area between parabola and axis between limits and

= =

= ………..(i)

And Area of ray and axis,

= = = ………..(ii)

Required shaded area in first quadrant

= Area between ray for and axis – Area between parabola and axis in first quadrant

= Area given by eq. (ii) – Area given by eq. (i) = sq. units

Similarly, shaded area in second quadrant = $\frac{1}{6}$ sq. units

Therefore, Total area of shaded region in the above figure

= $\frac{1}{6} + \frac{1}{6} = 2 \times \frac{1}{6} = \frac{1}{3}$ sq. units

13.Using the method of integration, find the area of the triangle ABC whose vertices are A (2, 0), B (4, 5) and C (6, 3).

Ans. Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).

Equation of side AB is

Equation of side BC is

Equation of side AC is

Now, Required shaded area = Area + Area of trapezium BLMC – Area

= 5 + 8 – 6 = 7 sq. units

14.Using the method of integration, find the area of the region bounded by the lines: and

Ans. Equation of one line is ,

Equation of second line is

And Equation of third line is

Here, vertices of triangle ABC are A (2, 0), B (4, 3) and C (1, 2).

Now, Required area of triangle

= Area of trapezium CLMB – Area Area

= sq. units

15.Find the area of the region .

Ans. Equation of parabola is ……(i)

And equation of circle is ……(ii)

Here, the two points of intersection of parabola (i) and circle (ii) are A and B

Required shaded area OADBO (Area of the circle which is interior to the parabola)

= 2 x Area OADO = 2 [Area OAC + Area CAD]

= $2 [{2. \frac{x^{\frac{3}{2}}}{\frac{3}{2}}}_{0}^{\frac{1}{2}} + {\frac{x \sqrt{\frac{9}{4} - x^{2}}}{2} + \frac{9 / 4}{2} \sin^{- 1} \frac{x}{3 / 2}}_{\frac{1}{2}}^{\frac{3}{2}}]$