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## Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Miscellaneous Exercise

Question 1.
Find area under gives curves and line:
(i) y = x2, x = I, x = 2 and x – axis
(ii) y = x2, x = 1, x = 5 and x – axis
(i) y = x2 is upward parabola

(ii) y = x4 is upward parabola

Question 2.
Find the area between the curves y = x and y = x2.
y = x is line with pts (0, 0) (-1, 1) (1, 1) (2, 2) etc
y = x2 is parabola with vertex (0,0) up ward parabola
They meet at pts (0, 0) and (1, 1)

⇒ y = x putting in y = x2
x = x2 ⇒ x = 0 ⇒ y = 0
x =1 ⇒ y = 1
Area is (shaded one) = (Area under line x = 0 to x = 1)
– (Area under parabola x = 0 to x = 1)

Question 3.
Find the area of the region lying in the first quadrant and bounded by
y = 4x2, x = 0, y = 1 and y = 4.
(1) y = 4x2 is an upward parabola with vertex (0, 0)

x = 0 is line with pts (0,1) (0,0) etc….
y = 1 is line with pts (0, 1) (1, 1) etc….
y = 4 is line with pts (0,4) (1, 4) etc….
parabola meets at pts (0,0) with line x = 0
(1/2,1) with line y = 1 (1,4) with line y = 4
Here we need to do along ‘y’ axis as we don’t know eq: of line with respect to ‘x’. Required area (shaded one) = Area under parabola from y = 1 to y = 4

Question 4.
Sketch the graph of y = |x+3| and evaluate $$\int _{ -6 }^{ 0 } |x+3|dx$$
y = |x+3|
points are    x = 0, y = 3 ; x = 1, y = 4 etc
x = -1, y = 2 ; x = -4, y = (-1) =   1
so point are (0, 3), (1,4), (2, 5), (-1,2), (-2, 1) (-3, 0), (-4,1) etc.

Question 5.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π
Sin curve is

y = |sin x| is a sin curve
x = 0 is a line pts (0,1) (0, 2) etc
x = 2 π is line through pts 2 π
Here in this we take n multiples in ‘x’ axis and 1 and -1 in ‘y’ axis

Question 6.
Find the area enclosed between the parabola y2 = 4ax and the line y = mx.
y2 = 4ax is right sided parabola vertex (O, O). y
y = mx is straight line through (O, O)

Question 7.
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.
4y = 3x2 is parabola upward
line 2y = 3x +12

 X 0 2 -2 y 6 0 3

They meet at pts (4,12) and (-2,3)
4y = 3x2
putting 2y = 3x +12 in it
2(3x +12) = 3x2

Required area (shaded) = Area under line 2y = 3x + 12 from x = -2 to x = 4
– Area under parabola 4y = 3x2 from x = -2 to x = 4

Question 8.
Find the area of the smaller region bounded by the ellipse
$$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \text { and the line } \frac{x}{3}+\frac{y}{2}=1$$

Required area (shaded area) = Area under ellipse from x = 0 to x = 3)
-(Area under line from x = 0 to x = 3)

Question 9.
Find the area of the smaller region bounded by the ellipse
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { and the line } \frac{x}{a}+\frac{y}{b}=1$$

Required area (shaded area) = Area under ellipse from x = 0 to x = 3)
-(Area under line from x = 0 to x = 3)

OR
Required area (Shaded) = Area of ellipse in 1st quadrant – Area of Δ ABC

Question 10.
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and the
x – axis.
Ans:
x2 = y is an upward parabola
line y = x +2

They meet at pts (2, 4) and (-1, 1)
By solving the equation
x2 = y, y = x + 2
⇒ x2– x – 2 = 0
⇒ x = 2 or -1
y = 4 or 1
$$Required\quad area=\int _{ -1 }^{ 2 } Area\quad of\quad line-\int _{ -1 }^{ 2 } Area\quad of\quad parabola$$

Question 11.
Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required is bounded by lines x + y = 1, – x + y = 1 and -x -y = 1].
|x|+ |y| = 1

so points are (0, 1) (0, -1) (1, 0) (-1, 0)
but |x| + |y| =  1 can be written as: follows: ⇒ -x + y = 1 (2nd quadrant); x + y =1
(1st quadrant); x – y = 1 (4th qudrant); -x -y = 1 (3rd quadrant) .

Not by Integration
= 4 x Area of any one part Δ
= 4 x Area Δ ABC
$$=4 \times \frac{1}{2} \times(\mathrm{AB}) \times|\mathrm{BC}|$$
$$=4 \times \frac{1}{2} \times 1 \times 1=2 \mathrm{sq.units.}$$

By integration
Area required
= 4 x Area of ΔABC
= 4 x Area under line AC (x + y = 1)

Question 12.
Find the area bounded by curves {(x, y): y ≥ x2 and y = |x| }.
y ≥ x2 and y ≤ |x|

consider then as curves y = x2 (upward parabola vertex (0, 0)
y= |x| (line)
As y ≥ x2, so area above parabola
y ≤ |x|, so area below modulus
∴ required area is shaded one
= 2x [Area under line in 1st quadrant)
– (Area under parabola in 1 st quadrant)]

Question 13.
Using the method of integration find the area of the triangle ABC, coordinates of whose
vertical are A(2,0), B(4 ,5) and C (6 , 3)
By 2 points form, equation of the line is

Question 14.
Using the method of integration find the area of the region bounded by lines:
2x + y = 4,3x – 2y = 6 and x – 3y + 5 = 0

Required area (shaded) = Area under line AB – Area under AC -Area under BC
$$=\int_{1}^{4} \text { line } \mathrm{AB}-\int_{1}^{2} \text { line } \mathrm{AC}-\int_{2}^{4} \text { line } \mathrm{BC}$$

Question 15.
Find the area of the region {(x, y) : y2< 4x, 4×2 + 4y2< 9}
Curve y2 = 4x is a parabola right sided
Curve 4x2 + 4y2 = 9 is a circle with radius $$\frac{3}{2}$$ and centre (0, 0)

They meet at points
$$\left(\frac{1}{2}, \sqrt{2}\right)$$
They meet at points
y2 = 4x in 4x2 + 4y2 = 9
⇒ 4x2 + 4 (4x) = 9
⇒ 4x2+16x – 9 = 0
$$\Rightarrow x=\frac{1}{2}, \frac{-9}{2} \text { but } x \ 0, \text { so } \frac{-9}{2} \text { not possible }$$
So, required area (shaded) = 2 x half of the area we get above ‘x’ axis
= 2 x [Area under curve y2= 4x + Area under curve 4x2 + 4y2 = 9]

Choose the correct answer in the following Exercise from 16 to 20.

Question 16.
Area bounded by the curve y = x3, the x-axis and the ordinates x = -2 and x = 1 is
(A) -9
(B)$$\frac{-15}{4}$$
(C)$$\frac{15}{4}$$
(D)$$\frac{17}{4}$$
y = x3

= Area under curve y = x3 with respect to ‘x’ axis from x = -2 to x = 1

Question 17.
The area bounded by the curve y = x |x| , x – axis and the ordinates x = -1 and x = 1 is given by
(A) 0
(B)$$\frac{1}{3}$$
(C)$$\frac{2}{3}$$
(D)$$\frac{4}{3}$$
[Hint: y = x2 if x > 0 and y = – x2 if x < 0].

NOTE: As explained in summary of this chapter
So correct answer is (C) $$\frac{2}{3}$$

Question 18.
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A) $$\frac{4}{3}(4 \pi-\sqrt{3})$$
(B) $$\frac{4}{3}(4 \pi+\sqrt{3})$$
(C) $$\frac{4}{3}(8 \pi-\sqrt{3})$$
(D) $$\frac{4}{3}(8 \pi+\sqrt{3})$$

x2 + y2 = 16 is circle with radius 4 centre (0, 0)
y2 = 6x is is right side parabola, vertex (0, 0)
They both meet at pts:
x2 + y2 = 16, But given y2 = 6x
so x2 + 6x – 16 = 0 ⇒ x = -8, but x <0 is eliminated
x = 2, $$y=2 \sqrt{3}$$
Required area (shaded) = Area of circle – shaded area (as ext. to parabola in circle)
Shaded area = 2 x [(Area of parabola from x = 0 to x = 2) +
(Area of circle from x = 2 to x = 4)]

Question 19.
The area bounded by the y – axis, y = cos x and y = sin x when $$0 \leq x \leq \frac{\pi}{2} \text { is }$$
(A)$$2(\sqrt{2-1})$$
(B)$$\sqrt{2}-$$
(C)$$\sqrt{2}+1$$
(D)$$\sqrt{2}$$

2nd PUC  Maths Application of Integrals Miscellaneous Exercise Additional Questions and Answers

Question 1.
Find the area of ABC, where vertices are A (4, 1) B (6, 6), C (8, 4) (CBSE 2010)

Question 2.
Find the area of the circle 4x2 + 4y2 = 9, which is interior to the parabola x2 = 4y.

Question 3.
Sketch the graph of y=|x+3| and evaluate x-axis and between x = – 6 and x = 0.
(CBSE – 2011)

Question 4.
Find the area of the region {(x, y) : x2 + y2 = 4, x + y ≥ 2}  (CBSE 2012)

Circle x2 + y2 = 4 his centre (0, 0) and radius 2
line x + y = 2 pts are (1,1) (2, 0) (0, 2) etc………
circle and line intersect at pts.
It can be found by solving the equation
x2 + y2 = 4 put (y = 2 – x)
⇒ x2 + (2 -x)2 = 4 ⇒ 2x2 – 4x + 0 = 0
⇒ 2x2 – 4x = 0 ⇒ x2 – 2x = 0
⇒ (x -2) x = 0  ⇒ x = 0 or x = 2
so y = 2 – x , x = 0 , y = 2    (0, 2)
x = 2 , y = 0 , (2,0)
so they meet at (0, 2) and (2, 0)