Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

   

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d)

I.

Question 1.
Fine the area of the region enclosed by the given curves.
i) y = cos x, y = 1 – \(\frac{2x}{\pi}\)
Solution:
Equations of the given curves are
y = cos x ………….. (1)
y = 1 – \(\frac{2x}{\pi}\) ………….. (2)
Eliminating y from eq’s (1) and (2)
cos x = 1 – \(\frac{2x}{\pi}\)
When x = \(\frac{\pi}{2}\), cos x = cos\(\frac{\pi}{2}\) = 0
1 – \(\frac{2}{\pi}\), x = \(\frac{2}{\pi}\) . \(\frac{\pi}{2}\) = 1 – 1 = 0
When x = 0, cos x = cos 0 = 1
1 – \(\frac{2x}{\pi}\) = 1 – 0 = 1
∴ Point of intersection are A = (\(\frac{\pi}{2}\), 0) B = [π – 1]

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 1

Question 2.
y =cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 3

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 3.
y = x³ + 3, y = 0, x = -1, x = 2.
Solution:
Required area PABQ
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 4

Question 4.
y = ex, y = x, x = 0, x = 1.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 5

Question 5.
y = sin x, y = cos x; x = 0, x = \(\frac{\pi}{2}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 6
Between 0 and \(\frac{\pi}{4}\).
cos x > sin x
Between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\)
cos x < sin x
Required area
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 7

Question 6.
x = 4 – y², x = 0.
Solution:
The given parabola x = 4 – y² meets, the x – axis at A(4, 0) and Y – axis at P(0, 2) and Q(0, -2).

The parabola is symmetrical about X – axis
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 8
Required area = 2 Area of OAP
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 9

Question 7.
Find the area enclosed with in the curve |x| + |y| = 1
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 10

II.

Question 1.
x = 2 – 5y – 3y², x = 0.
Solution:
Solving the equation of given curves
2 – 5y – 3y² = 0
3y² + 5y – 2 = 0
(y + 2)(3y – 1) = 0 y = -2 or \(\frac{1}{3}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 11

Question 2.
x² = 4y, x = 2, y = 0.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 12

Question 3.
y² = 3x, x = 3.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 13
The parabola is symmetrical about X – axis
Required area = 2\(\int_0^3\)√3. √x dx
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 14

Question 4.
y = x², y = 2x.
Solution:
Given equation are y = x² ………….. (1)
y = 2x …………. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 15
Eliminating y, we get x² = 2x
x² – 2x = 0
x(x – 2) = 0
x = 0 or x = 2
y = 0, or y = 4

Point of intersection are O(0, 0), A(2, 4)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 16

Question 5.
y = sin 2x, y = √3 sin x, x = 0, x = \(\frac{\pi}{6}\).
Solution:
Given equation are y = sin 2x ………… (1)
y = √3 sin x …………. (2)
sin 2x = √3 sin x
2 sin x. cos x = √3 sin x
sin x = 0 or 2 cos x = \(\frac{\sqrt{3}}{2}\)
x = 0, cos x = \(\frac{\sqrt{3}}{2}\) ⇒ x = \(\frac{\pi}{6}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 17
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 18

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 6.
y = x², y = x³.
Solution:
Given equations are y = x² ………….. (1)
y = x³ ………. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 19
From equation (1) and (2) x² = x³
x³ – x² = 0
x²(x – 1) = 0
x = 0 or 1
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 20

Question 7.
y = 4x – x², y = 5 – 2x
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 21
y = 4x – x² ………… (i)
y = 5 – 2x …………..(ii)
y = -([x – 2]²) = 4
y – 4 = -(x – 2)²
Solving equations (i) and (ii) we get
4x – x² = 5 – 2x
x² – 6x + 5 = 0
(x – 5)(x – 1 ) = 0
x = 1, 5
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 22

Question 8.
Find the area in Sq. units bounded by the x – axis, part of the curve y = 1 + \(\frac{8}{x^2}\) and the ordinates x = 2 and x = 4.
Solution:
Given equations y = 1 + \(\frac{8}{x^2}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 23

Question 9.
Find the area of the region bounded by the parabolas y² = 4x and x² = 4y.
Solution:
Equations of the given curve are
y² = 4x
x² = 4y
(\(\frac{x^2}{4}\))² = 4x
\(\frac{x^4}{16}\) = 4x
x4 = 64x ⇒ x4 = 0 or x³ = 64, x = 4
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 24

Question 10.
Find the area bounded by the curve y = lnx the X – axis and the straight line x = e.
Solution:
Equation of the curve is y = lnx
x = 1 ⇒ y = 0
The curve y = lnx meets
X – axis at C(1, 0)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 25
Required area = \(\int_1^e\)lnx dx
= (x.lnx)e1 – \(\int_1^e\)x.\(\frac{1}{x}\) dx
= (e.ln e – 1.ln 1) – (x)e1
= e – (e – 1)
= e – e + 1 = 1 Sq.unit.

III.

Question 1.
y = x² + 1, y = 2x – 2, x = -1, x = 2.
Solution:
Equation of the curve are
y = x² + 1 …………. (1)
y = 2x – 2 ………….. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 26
Area between the given curves
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 27

Question 2.
y² = 4x, y² = 4(4 – x).
Solution:
Equations of the curve are y² = 4x ………… (1)
y² = 4(4 – x) …………. (2)
Eliminating y, we get
4x = 4(4 – x)
2x = 4 ⇒ x = 2
Substituting in equation (1), y² = 8
y = ± 2√2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 28
Points of intersection are
A(2, 2√2), B(2, -2√2)
Required area is symmetrical about X – axis
Area OACB
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 29

Question 3.
y = 2 – x², y = x².
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 30
y = 2 – x² …………. (1)
y = x² …………. (2)
x² = -(y – 2)
From equation (2)
2 – x² = x²
2 = 2x² or x² = 1
x = ±1
Area bounded by two curve be
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 31

Question 4.
Show that the area enclosed between the curve y² = 12(x + 3) and y² = 20(5 – x) is 64\(\sqrt{\frac{5}{3}}\).
Solution:
Equation of the curve are
y² = 12(x + 3) ……….. (1)
y² = 20(5 – x) ……….. (2)
Eliminating y
12(x + 3) = 20(5 – x)
3x + 9 = 25 – 5x
8x = 16
x = 2
y² = 12(2 + 3) = 60
y = √60 = ±2√15
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 32
Points of intersection are B’ (2, 2√15)
B’ (+2, -2√15)
The required area is symmetrical about X – axis
Area ABCB’
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 33

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 5.
Find the area of the region {(x, y)/x² – x – 1 ≤ y ≤ -1}
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 34

Question 6.
The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both the parts.
Solution:
Equations of the curves are
x² + y² = 8 ………… (1)
2y = x² ………… (2)
Eliminating y between equations (1) and (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 35
Let x² = t
4t + t² = 32
t² + 4t – 32 = 0
(t + 8)(t – 4) = 0
t = -8 (not possible) x² = 4 ⇒ x = ±2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 36
As curve is symmetric about Y – axis, total area be
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 37

Question 7.
Show that the area of the region bounded \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is π ab. also deduce the area of the cricle x² + y² = a².
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 38
The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4.\(\frac{\pi}{4}\) ab
Equation of elliple = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 39
Substituting b = a, we get the circle
x² + y² = a²
Area of the circle = πa(a) = πa² sq. units.

Question 8.
Find the area of region enclosed by the curves y = sin πx, y = x² – x, x = 2.
Solution:
Required area
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 40
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 41

Question 9.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with OA = a, OB = b. Then show that the area bounded the chord AB and the arc AB of the elliple is \(\frac{(\pi-2) a b}{4}\).
Solution:
Let OA = a, OB = b
Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{y}{b}\) = 1 – \(\frac{x}{a}\)
y = b(1 – \(\frac{x}{a}\))
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 42
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 43

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 10.
Prove that curves y² = 4x and x² = 4y divide the area of the square bounded by the lines x = 0, x = 4, y = 4 and x = 0 into three equal parts.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 44
The given equations are y² = 4x …………. (1)
x² = 4y …………. (2)
The points of intersection are O(0, 0) A(4, 4)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 45

Leave a Comment