Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.1

Question 1.

Using Interation, find the area of the region bounded the line given is 2y + x = 8, the x axis and the lines x = 2, x = 4.

Solution:

The equation of the line given is 2y + x = 8

⇒ 2y = 8 – x ⇒ y = \(\frac { 8-x }{2}\)

y = 4 – \(\frac { x }{2}\)

Also x varies from 2 to 4

The required Area

= (16 – 4) – (8 – 1)

= 12 – 7 = 5 sq.units

Question 2.

Find the area bounded by the lines y – 2x – 4 = 0, y = 0, y = 3 and the y-axis.

Solution:

The equation of the line given is y – 2x – 4 = 0

⇒ 2x = y – 4 ⇒ x = \(\frac { y-4 }{2}\)

∴ x = \(\frac { y }{2}\) – 2

Also y varies from 1 to 3

Required Area

= 2 – 4 = -2

Area can’t be in negative.

∴ Area = 2 sq.units

Question 3.

Calculate the area bounded by the parabola y² = 4ax and its latus rectum.

Solution:

Given parabola is y^{2} = 4ax

Its focus is (a, 0)

Equation of latus rectum is x = a

The parabola is symmetrical about the x-axis

Question 4.

Find the area bounded by the line y = x and x-axis and the ordinates x = 1, x = 2

Solution:

The equation of the given line is y = x and x varies from 1 to 2

Question 5.

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

Solution :

The equation of given line is y – 1 = x

y = x + 1

The line y = x + 1 meets the x-axis at x = -1

Since x varies from -2 to 3

Hence a part of lies below the x-axis and the other part lies above the x-axis.

Question 6.

Find the area of the region lying in the first quadrant bounded by the region y = 4x², x = 0, y = 0 and y = 4

Solution:

The equation of a parabola given is y = 4x²

Area of the region lying in the first quadrant

Question 7.

Find the area bounded by the curve y = x² and the line y = 4

Solution:

Equation of the curve y = x² ………. (1)

Equation of the line y = 4 ………. (2)

Solving equation (1) & (2)

x² = 4 ⇒ x = ± 2