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## 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:

Question 2.

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

Solution:

2x – y + 1 = 0

To find further limit put x = 0, we get y = 1

Question 3.

Find the area of the region bounded by the curve 2 + x – x^{2} + y = 0, x – axis, x = – 3 and x = 3.

Solution:

Question 4.

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

Solution:

To find point of intersection of the curves

y = 2x + 5 and y = x^{2} – 2x we get (-1, 3) and (5, 15)

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:

To find the points of intersection,

Question 6.

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

Solution:

To find the points of intersection of these two curves between 0 to \(\frac{\pi}{2}\) is \(\frac{\pi}{4}\)

Question 7.

Find the area of the region bounded by parabola y^{2} = x and the line y = x – 2

Solution:

To find the points of intersection solve the two equations y^{2} = x and y = x – 2

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 y^{2} = 4x and x^{2} = 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:

To find the points of intersection of the two curves, y^{2} = 4x and x^{2} = 4y are (0, 0) and (4, 4).

Area of the square field = 4 × 4 = 16 sq. units

So, the remaining each of the two parts must be \(\frac{16}{3}\) sq.units.

∴ Yes, It is possible to divide the square field into three equal parts.

Question 9.

The curves = (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:

y = (x – 2)^{2} + 1

∴ x = 2 is a minimum point

∴ The point P is (2, 1)

But slope of PQ is 2

∴ Equation of the chord PQ

y – y_{1} = m(x – x_{1})

y – 1 = 2 (x – 2)

y – 1 = 2x – 4

y = 2x – 3

On solving the curve and line we get the point Q(4, 5)

Question 10.

Find the area of the region common to the circle x^{2} + y^{2} = 16 and the parabola y^{2} = 6x.

Solution:

To find points of intersection of x^{2} + y^{2} = 16 and y^{2} = 6x are (2, \(2 \sqrt{3})\)) and (2, –\(2 \sqrt{3})\))

Due to symmetrical property,

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

Question 1.

Find the area of the region enclosed by y^{2} = x and y = x – 2.

Solution:

The points of intersection of the parabola y^{2} = x and the line y = x – 2 are (1, -1) and (4, 2)

To compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect toy no splitting is necessary.

Question 2.

Find the area bounded by the curve y = x^{3} and the line y = x.

Solution:

The line y = x lies above the curve y = x^{3} in the first quadrant and y = x^{3} lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x^{3}, y = x ⇒ x^{3} = x. We get x = {0, ± 1}

Question 3.

Find the area of the loop of the curve 3ay^{2} = x (x – a)^{2}.

Solution:

Put y = 0; we get x = 0, a

It meets the x – axis at x = 0 and x = a

∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 4.

Find the area between the line y = x + 1 and the curve y = x^{2} – 1.

Solution:

To get the points of intersection of the curves we should solve the equations y = x +1 and y = x^{2} – 1.

we get, x^{2} – 1 = x + 1

x^{2} – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0

x = – 1 or x = 2

∴ The line intersects the curve at x = – 1 and x = 2.