12th Commerce Maths 1 Chapter 7 Exercise 7.1 Answers Maharashtra Board

Application of Definite Integration Class 12 Commerce Maths 1 Chapter 7 Exercise 7.1 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Ex 7.1 Questions and Answers.

Std 12 Maths 1 Exercise 7.1 Solutions Commerce Maths

Question 1.
Find the area of the region bounded by the following curves, the X-axis, and the given lines:
(i) y = x 4 , x = 1, x = 5
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1i

(ii) y = \(\sqrt{6 x+4}\), x = 0, x = 2
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1ii

(iii) \(\sqrt{16-x^{2}}\), x = 0, x = 4
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1iii

Maharashtra-Board-Solutions

(iv) 2y = 5x + 7, x = 2, x = 8
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1iv

(v) 2y + x = 8, x = 2, x = 4
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1v

(vi) y = x 2 + 1, x = 0, x = 3
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1vi

(vii) y = 2 – x 2 , x = -1, x = 1
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q1vii

Maharashtra-Board-Solutions

Question 2.
Find the area of the region bounded by the parabola y 2 = 4x and the line x = 3.
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q2
Required area = area of the region OABO
= 2(area of the region OACO)
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q2.1

Question 3.
Find the area of the circle x 2 + y 2 = 25.
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q3
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the circle, y 2 = 25 – x 2 .
In the first quadrant y > 0
∴ y = \(\sqrt{25-x^{2}}\)
∴ area of the circle = 4(area of region OABO)
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q3.1

Maharashtra-Board-Solutions

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q4
By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 2.
From the equation of the ellipse,
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q4.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-7-Application-of-Definite-Integration-Ex-7.1-Q4.2