Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

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

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

(I) Choose the correct alternatives:

Question 1.
Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) \(\frac{3}{26}\) sq units
(c) 26 sq units
(d) 3 sq units
Answer:
(a) \(\frac{26}{3}\) sq units

Question 2.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) \(\frac{3}{28}\) sq units
Answer:
(c) \(\frac{28}{3}\) sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{3}{56}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer:
(c) \(\frac{56}{3}\) sq units

Question 4.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer:
(b) \(\frac{3124}{5}\) sq units

Question 5.
Using definite integration area of circle x² + y² = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer:
(c) 25π sq units

(II) Fill in the blanks:

Question 1.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is _________
Answer:
\(\frac{3124}{5}\) sq units

Question 2.
Using definite integration area of the circle x² + y² = 49 is ___________
Answer:
49π sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is _________
Answer:
\(\frac{56}{3}\) sq units

Question 4.
The area of the region bounded by the curve x² = y, the X-axis and the lines x = 3 and x = 9 is _________
Answer:
234 sq units

Question 5.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer:
\(\frac{28}{3}\) sq units

(III) State whether each of the following is True or False.

Question 1.
The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x d y=\int_{y=c}^{y=d} g(y) d y\)
Answer:
True

Question 2.
The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left|\int_{a}^{b} f(x) d x-\int_{b}^{a} g(x) d x\right|\)
Answer:
False

Question 3.
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
True

Question 4.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer:
False

Question 5.
The area of the portion lying above the X-axis is positive.
Answer:
True

(IV) Solve the following:

Question 1.
Find the area of the region bounded by the curve xy = c², the X-axis, and the lines x = c, x = 2c.
Solution:

= c² log(\(\frac{2 c}{c}\))
= c² . log 2 sq units.

Question 2.
Find the area between the parabolas y² = 7x and x² = 7y.
Solution:

For finding the points of intersection of the two parabolas,
we equate the values of y² from their equations.
From the equation x² = 7y, y² = \(\frac{x^{4}}{49}\)
∴ \(\frac{x^{4}}{49}\) = 7x
∴ x⁴ = 343x
∴ x⁴ – 343x = 0
∴ x(x³ – 343) = 0
∴ x = 0 or x³ = 343, i.e. x = 7
When x = 0, y = 0
When x = 7, 7y = 49
∴ y = 7
∴ the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) – (area of the region ODABO)
Now, area of the region ODACO = area under the parabola y² = 7x
i.e. y = √7 √x

Area of the region ODABO = Area under the parabola
x² = 7y
i.e. y = \(\frac{x^{2}}{7}\)

∴ required area = \(\frac{98}{3}-\frac{49}{3}=\frac{49}{3}\) sq units.

Question 3.
Find the area of the region bounded by the curve y = x² and the line y = 10.
Solution:

By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution:
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.

Question 5.
Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y d x\), where y = x²
= \(\int_{1}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{1}^{4}=\frac{4^{3}}{3}-\frac{1}{3}=\frac{64-1}{3}\)
= 21 sq units.

Question 6.
Find the area of the region bounded by the curve x² = 25y, y = 1, y = 4, and the Y-axis.
Solution:

Question 7.
Find the area of the region bounded by the parabola y² = 25x and the line x = 5.
Solution:

Given the equation of the parabola is y² = 25x
∴ y = 5√x …… [∵ IIn first quadrant, y > 0]
Required area = area of the region OQRPO
= 2(area of the region ORPO)