Students can Download Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

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

Question 1.

Find area under gives curves and line:

(i) y = x^{2}, x = I, x = 2 and x – axis

(ii) y = x^{2}, x = 1, x = 5 and x – axis

Answer:

(i) y = x^{2} is upward parabola

Area = shaded one)

(ii) y = x^{4} is upward parabola

Area = shaded one)

Question 2.

Find the area between the curves y = x and y = x^{2}.

Answer:

y = x is line with pts (0, 0) (-1, 1) (1, 1) (2, 2) etc

y = x^{2} is parabola with vertex (0,0) up ward parabola

They meet at pts (0, 0) and (1, 1)

⇒ y = x putting in y = x^{2
}x = x^{2} ⇒ 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 = 4x^{2}, x = 0, y = 1 and y = 4.

Answer:

(1) y = 4x^{2} 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 \(\int _{ -6 }^{ 0 } |x+3|dx\)

Answer:

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

Answer:

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 y^{2} = 4ax and the line y = mx.

Answer:

y^{2} = 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 = 3x^{2} and the line 2y = 3x + 12.

Answer:

4y = 3x^{2} 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 = 3x^{2}

putting 2y = 3x +12 in it

2(3x +12) = 3x^{2
}

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

– Area under parabola 4y = 3x^{2} from x = -2 to x = 4

Question 8.

Find the area of the smaller region bounded by the ellipse

\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \text { and the line } \frac{x}{3}+\frac{y}{2}=1\)

Answer:

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

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { and the line } \frac{x}{a}+\frac{y}{b}=1\)

Answer:

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 x^{2} = y, the line y = x + 2 and the

x – axis.

Ans:

x^{2} = y is an upward parabola

line y = x +2

They meet at pts (2, 4) and (-1, 1)

By solving the equation

x^{2} = y, y = x + 2

⇒ x^{2}– 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].

Answer:

|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 (2^{nd} quadrant); x + y =1

(1^{st} quadrant); x – y = 1 (4^{th} qudrant); -x -y = 1 (3^{rd} quadrant) .

|x|+ |y|=1 in different quadrants

Not by Integration

so area required (shaded)

= 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 ≥ x^{2} and y = |x| }.

Answer:

y ≥ x^{2} and y ≤ |x|

consider then as curves y = x^{2} (upward parabola vertex (0, 0)

y= |x| (line)

As y ≥ x^{2}, 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)

Answer:

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

Answer:

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) : y^{2}< 4x, 4×2 + 4y^{2}< 9}

Answer:

Curve y^{2} = 4x is a parabola right sided

Curve 4x^{2} + 4y^{2} = 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

y^{2} = 4x in 4x^{2} + 4y^{2} = 9

⇒ 4x^{2} + 4 (4x) = 9

⇒ 4x^{2}+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 y^{2}= 4x + Area under curve 4x^{2} + 4y^{2} = 9]

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

Question 16.

Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = -2 and x = 1 is

(A) -9

(B)\( \frac{-15}{4}\)

(C)\(\frac{15}{4}\)

(D)\(\frac{17}{4}\)

Answer:

y = x^{3
}

Required area (shaded)

= Area under curve y = x^{3 }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 = x^{2} if x > 0 and y = – x^{2} if x < 0].

Answer:

NOTE: As explained in summary of this chapter

So correct answer is (C) \( \frac{2}{3}\)

Question 18.

The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 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) \(\frac{4}{3}(8 \pi+\sqrt{3})\)

Answer:

x^{2} + y^{2} = 16 is circle with radius 4 centre (0, 0)

y^{2} = 6x is is right side parabola, vertex (0, 0)

They both meet at pts:

x^{2} + y^{2} = 16, But given y^{2} = 6x

so x^{2} + 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)

= πr^{2} – shaded area

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)\(\sqrt{2}\)

Answer:

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)

Answer:

Question 2.

Find the area of the circle 4x^{2} + 4y^{2} = 9, which is interior to the parabola x^{2} = 4y.

Answer:

Question 3.

Sketch the graph of y=|x+3| and evaluate x-axis and between x = – 6 and x = 0.

(CBSE – 2011)

Answer:

Question 4.

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

Answer:

Circle x^{2} + y^{2} = 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

x^{2} + y^{2} = 4 put (y = 2 – x)

⇒ x^{2} + (2 -x)^{2} = 4 ⇒ 2x^{2} – 4x + 0 = 0

⇒ 2x^{2} – 4x = 0 ⇒ x^{2} – 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)

Required area is shaded area:-

(Area under circle x^{2} + y^{2} = 4 from x = 0 to x = 2)

– (Area under line x+y = 2 from x = 0 to x = 2)