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Sunday, September 12, 2021

Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF Download

Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF Download: Students of Standard 12 can now download Plus Two Maths Chapter 8 Application of Integrals chapter wise question and answers pdf from the links provided below in this article. Plus Two Maths Chapter 8 Application of Integrals Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus Two Maths Chapter 8 Application of Integrals exams.


Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers

Plus Two Maths Chapter 8 Application of Integrals question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus Two Maths Chapter 8 Application of Integrals chapter wise questions and answers to help them secure good marks in class tests and exams.


Board

Kerala Board

Study Materials

Chapter wise Question and Answers

For Year

2021

Class

12

Subject

Mathematics

Chapters

Maths Chapter 8 Application of Integrals

Format

PDF

Provider

Spandanam Blog


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Plus Two Maths Chapter 8 Application of Integrals Question and Answers PDF Download

We have provided below the question and answers of Plus Two Maths Chapter 8 Application of Integrals Chapter wise study material which can be downloaded by you for free. These Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and answers will contain important questions and answers and have been designed based on the latest Plus Two Maths Chapter 8 Application of Integrals, books and syllabus. You can click on the links below to download the Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF. 

Question 1.
Consider the following figure.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q1

  1. Find the point of intersection (P) of the given parabola and the line. (2)
  2. Find the area of the shaded region. (2)

Answer:
1. We have, y = x2 and y = x ⇒ x = x2 ⇒
⇒ x2 – x = 0 ⇒ x(x – 1) = 0 ⇒ x = 0, 1
When x = 0, y =0 and x = 1, y = 1.
Therefore the points of intersections are (0, 0) and(1, 1).

2. Required area
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q1.1

Question 2.
1. Find the point of intersection ‘p’ of the given parabola and the line. (2)
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q2
2. Find the area of the shaded region. (2)
Answer:
1. Given, y = x2, y = 2x
⇒ 2x = x2 ⇒ x2 – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, 2
We have, y = 2x
⇒ when x = 0 ⇒ y = 0, when x = 2 ⇒ y = 4
‘P’ has co-ordinate (2, 4)

2. Area = \(\int_{0}^{2} 2 x d x-\int_{0}^{2} x^{2} d x=\left(x^{2}\right)^{2}-\left(\frac{x^{3}}{3}\right)_{0}^{2}=4-\frac{8}{3}=\frac{12-8}{3}=\frac{4}{3}\).

Question 3.
Consider the following figure.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q3

  1. Find the point of Intersection P of the circle x2 + y2 = 32 and the line y = x. (1)
  2. Find the area of the shaded region. (3)

Answer:
1. x2 + x2 = 32 ⇒ 2x2 = 32 ⇒ x2 = 16 = 4
Therefore the point of intersection P is (4, 4).

2. We have x2 + y2 = 32 ⇒ y = \(\sqrt{32-x^{2}}\).
The required area =
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q3.1
= 8 + [8π – 8 – 4π] = 4π.

Question 4.

  1. Shade the area enclosed by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)
  2. Find the area of the region bound by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q4

2. Area
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q4.1

Question 5.

  1. Draw a rough sketch of the graph of the function y2 = 4x. (2)
  2. Find the area by the curve and the line x= 2. (2)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q5

2. Area
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 4M Q5.1

Plus Two Maths Application of Integrals Six Mark Questions and Answers

Question 1.

  1. Draw the graph of y2 = 4x and y = x. (2)
  2. Find the points of intersection of y2 = 4x and y = x. (2)
  3. Find the area bounded by the graphs.(2)

Answer:
1. y2 = 4x is a parabola and y = x is a straight line passing through the origin.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q1

2. x2 = 4x ⇒ x2 – 4x = 0 ⇒ x(x – 4) = 0 ⇒ x = 0, 4
When x = 0, y = 0 and when x = 4, y =4. Therefore the points of intersection are (0, 0) and (4, 4).

3. Area bounded by the graphs = Area under the parabola in the first quadrant – Area under the line.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q1.1

Question 2.

  1. Draw the graph of the function y = x2 and x = y2 in a coordinate axes. (2)
  2. Find the point of intersection of the above graphs. (2)
  3. Find the area of the region bounded by the above two curves. (2)

Answer:
1. The two function are parabolas as shown in the figure.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q2

2. We have, y = x2 and x = y2
x = (x2)2 ⇒ x – x4 = 0 ⇒ x(1 – x3) = 0 ⇒ x = 0, 1
When x= 1, y= 1 and x = 0, y = 0.
Therefore the point is (0, 0) and (1, 1).

3. The required area = \(\int_{0}^{1} \sqrt{x} d x-\int_{0}^{1} x^{2} d x\)
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q2.1

Question 3.
Using the figure answer the following questions
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q3

  1. Find the area of the shaded region as the sum of the area of two triangles. (2)
  2. Define the function of the given graph. (2)
  3. Verify the area of the shaded region using integration. (2)

Answer:
1. Area = Area of ∆OAD + Area of ∆ ABC
\(\frac{1}{2}\) × 3 × 3 + \(\frac{1}{2}\) × 2 × 2 = \(\frac{9}{2}\) + 2 = \(\frac{13}{2}\).

2. Area = \(\int_{0}^{5}\)f(x) dx = \(\int_{0}^{3}\)(-x + 3) dx + \(\int_{3}^{5}\)(x – 5) dx
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q3.1
Therefore verified.

Question 4.
The figure given below contains a straight line L with slope \(\sqrt{8}\) and a circle.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q4

  1. Find the equation of the line L and circle. (1)
  2. Find the point of intersection P. (2)
  3. Find the area of the shaded region. (3)

Answer:
1. The line L passes through origin and have slope 3, therefore its equation is y = \(\sqrt{8}\) x. The circle passes through origin and have radius 3, therefore its equation is x2 + y2 = 9.

2. We have, y =3x and x2 + y2 = 9
⇒ x2 + (\(\sqrt{8}\)x)2 = 9 ⇒ 9x2 = 9
⇒ x = 1
∴ y = \(\sqrt{8}\) × 1 = \(\sqrt{8}\).
Therefore, coordinate of ‘P’ is (1, \(\sqrt{8}\)).

3. Area of the shaded region
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q4.1

Question 5.
Using the given figure answer the following
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q5

  1. Define the equation of the circle and ellipse in the figure. (1)
  2. Find the area of the ellipse using integration. (4)
  3. Find the area of the shaded region. (Use formula to find the area of the circle.) (1)

Answer:
1. From the figure equation of the circle is x2 + y2 = 4 and that of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\).

2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y2 = 1 ⇒ y = \(\frac{1}{2} \sqrt{4-x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q5.1

3. Area of the circle of radius 2 = π (2)2 = 4π
∴ Area of the shaded region = Area of the circle – Area of the ellipse
= 4π – 2π = 2π.

Question 6.

  1. Find the area bounded by the curve y = sin x with X – axis, between x = 0 and x = 2π. (2)
  2. Find the area of the region bounded by the curve y = x2 and y = |x| (4)

Answer:
1. Area of y = sin x in each loop is same. Therefore;
2\(\int_{0}^{\pi}\)sin xdx = \(-2(\cos x)_{0}^{\pi}\) = -2 (cos π – cos0)
= -2(-1 – 1) = 4

2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q6
Area
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q6.1

Question 7.
Using integration, find the area of the region bounded by the triangle whose vertices are(-1, 1), (0, 5) and (3, 2). (6)
Answer:
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q7
Equation of BC is \(\frac{y-5}{1-5}=\frac{x-0}{-1-0}\)
⇒ y – 5 = 4x ⇒ 4x – y + 5 = 0 ⇒ y = 4x + 5
Equation of AB is x + y – 5 = 0 ⇒ y = 5 – x
Equation of AC is x – 4y + 5 = 0 ⇒ y = \(\frac{x}{4}+\frac{5}{2}\)
The required area = Area of the region PABCQP – Area of the region PACQP
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q7.1

Question 8.
Consider the functions f(x) = sin x and g(x) = cosx in the interval [0, 2π]

  1. Find the x coordinates of the meeting points of the functions. (1)
  2. Draw the rough sketch of the above functions. (2)
  3. Find the area enclosed by these curves in the given interval. (3)

Answer:
1. f(x) = sin x and g(x) = cos x meet at multiples of \(\frac{\pi}{4}\)
x = \(\frac{\pi}{4}\), \(\frac{5 \pi}{4}\).

2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q8

3. Area = 2{Area under f(x) = sinx from \(\frac{\pi}{4}\) to π – Area under g(x) = cosx from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)}
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q8.1

Question 9.

Evaluate \(\int_{0}^{r} \sqrt{r^{2}-x^{2}} d x\), where r is a fixed positive number. Hence prove the area of the circle of radius r is π r2. (2)
Find the area of the circle, x2 + y2 = 16 which is exterior to parabola y2 = 6x. (4)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q9

Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q9.1

2. Given, x2 + y2 = 16 and y2 = 6x ⇒ x2 + 6x = 16 ⇒ x2 + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0 ⇒ x = -8, 2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q9.2
Area = Area of the circle – Interior area of the parabola.
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q9.3

Question 10.
Using the figure answer the following questions
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q10

  1. Define the equation of the ellipse and circle in the given figure. (1)
  2. Find the area of the ellipse using integration. (4)
  3. Find the area of the shaded region. (Area of the circle can be found using direct formula). (1)

Answer:
1. Equation of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\) and circle is x2 + y2 = 1.

2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y2 = 1 ⇒ y = \(\frac{1}{2} \sqrt{4-x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx
Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals 6M Q10.1

3. Area of the circle = πr2 = π × 1 = π
Required area = Area of ellipse – area of the circle = 2π – π = π.


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