Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.8 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.

Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.

Solution:

3x – 2y + 6 = 0

2y = 3x + 6

y = \(\frac { 1 }{ 2 }\)(3x + 6)

Question 2.

Find the area of the region bounded by 2x – y + 1 = 0, y = -1, y = 3 and y axis.

Solution:

Given straight line is 2x – y + 1 = 0

y = 2x + 1, x = \(\frac { y-1 }{ 2 }\)

= 1 + 1

= 2 sq. units

Area required = 2 sq. units

Question 3.

Find the area of the region bounded by the curve 2 + x – x² + y = 0, x axis, x = -3 and x = 3

Solution:

Given curve is

2 + x – x² + y = 0

y = x² – x – 2

y = (x – 2)(x + 1)

Question 4.

Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.

Solution:

First, we find the point of intersection of

y = 2x + 5 and y = x² – 2x

x² – 2x = 2x + 5

x² – 4x – 5 = 0

(x – 5) (x + 1) = 0

x = 5, x = – 1

when x = 5, y = 15

x = -1. y = 3

(5, 15) (-1, 3) are intersecting points.

54 – 18 = 36

Area required = 36 sq. units

Question 5.

Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π

Solution:

First find the intersecting point of two curves

sin x = cos x

tan x = 1

x = \(\frac { π }{ 4 }\)

Area required = 2√2 sq. units

Question 6.

Find the area of the region bounded by y = tan x, y = cot x and the lines x = 0, x = \(\frac { π }{ 2}\), y = 0.

Solution:

First find the intersecting point of y = tan x and y = cot x

tan x = cot x

\(\frac { tan x }{ cot x }\) = 1

tan²x = 1

tan x = 1

= log √2 + log √2

= 2 log √2

= log(√2)²

= log 2 sq. units

Question 7.

Find the area of the region bounded by the parabola y² = x and the line y = x – 2.

Solution:

First find the intersecting point of y² = x and y = x – 2

y = y² – 2

y² – y – 2 = 0

y = 2, y = -1

Intersecting points are (4, 2), (1, -1)

Required Area = \(\frac { 9 }{ 2 }\) sq. units

Question 8.

Father of a family wishes to divide his square field bounded by x = 0, x = 4, y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.

Solution:

Given curve y² = 4x and x² = 4y

Draw these two curves

Also draw the square bounded by the lines

x = 0, x = 4, y = 4 and y = 0

To prove Area A_{1} = Area A_{2} = Area A_{3}

Now the point of intersection of the curves y² = 4x and x² = 4y is given by

(\(\frac { y^2 }{ 4 }\))² = 4y

y^{4} = 64y ⇒ y (y³ – 64) = 0

y = 0, y = 4

when y = 0 ⇒ x = 0

y = 4 ⇒ x = 4

Point of intersection are O (0, 0) and B (4, 4)

Now, the area of the region bounded by the curves y² = 4x and x² = 4y is

A_{2} = \(\int_{0}^{4}\)(\(\sqrt { 4x }\) – \(\frac { x^2 }{ 4 }\)) dx

Now the area of the region bounded by the curves x² = 4y, x = 4 and x axis is

Similarly the area of the region bounded by the curve y² = 4x, y axis and y = 4 is

Hence we see that

A_{1} = A_{2} = A_{3} = \(\frac { 16 }{ 3 }\) sq. units

Question 9.

The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.

Solution:

Given curve is y = (x – 2)² + 1

(i.e) (y – 1) = (x – 2)²

Vertex of the parabola is (2, 1)

Minimum point P is (2, 1)

Slope of PQ is 2.

Equation of PQ is y – 1 = 2 (x – 2)

y – 1 = 2x – 4

y = 2x – 3

Intersecting point of y = 2x – 3 and y = (x – 2)² + 1

2x – 3 = (x – 2)² + 1

2x – 4 = (x – 2)²

2(x – 2) = (x – 2)²

x – 2 = 2

x = 4

when x = 4, y = 5

Required Area = \(\frac { 4 }{ 3 }\) sq. units

Question 10.

Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x

Solution:

First find the intersecting point of the curves

x² + y² = 16 and y² = 6x

x² + 6x = 16

x² + 6x – 16 = 0

(x + 8) (x – 2) = 0

x = -8, x = 2

x = -8 is impossible

x = 2, y = 2√3

Radius of the circle x² + y² = 16 is 4

Area OABC = 2 (Area of OAB)

= 2 (Area of the curve y² = 6x in [0, 2] + Area of the curve x² + y² = 16 in [2, 4])

Required Area = \(\frac { 4 }{ 3 }\) [4π + √3] sq. units