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 

Provider 
How to check Plus Two Maths Chapter 8 Application of Integrals Question and Answers?
 Visit our website  https://spandanamblog.com
 Click on the 'Plus Two Question and Answers'.
 Look for your 'Plus Two Maths Chapter 8 Application of Integrals Question and Answers'.
 Now download or read the 'Class 12 Maths Chapter 8 Application of Integrals Question and Answers'.
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.
 Find the point of intersection (P) of the given parabola and the line. (2)
 Find the area of the shaded region. (2)
Answer:
1. We have, y = x^{2} and y = x ⇒ x = x^{2} ⇒
⇒ x^{2} – 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
Question 2.
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. Given, y = x^{2}, y = 2x
⇒ 2x = x^{2} ⇒ x^{2} – 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 coordinate (2, 4)
2. Area = \(\int_{0}^{2} 2 x d x\int_{0}^{2} x^{2} d x=\left(x^{2}\right)^{2}\left(\frac{x^{3}}{3}\right)_{0}^{2}=4\frac{8}{3}=\frac{128}{3}=\frac{4}{3}\).
Question 3.
Consider the following figure.
 Find the point of Intersection P of the circle x^{2} + y^{2} = 32 and the line y = x. (1)
 Find the area of the shaded region. (3)
Answer:
1. x^{2} + x^{2} = 32 ⇒ 2x^{2} = 32 ⇒ x^{2} = 16 = 4
Therefore the point of intersection P is (4, 4).
2. We have x^{2} + y^{2} = 32 ⇒ y = \(\sqrt{32x^{2}}\).
The required area =
= 8 + [8π – 8 – 4π] = 4π.
Question 4.
 Shade the area enclosed by x^{2} = 4y, y = 2, y = 4 and the yaxis in the first quadrant. (2)
 Find the area of the region bound by x^{2} = 4y, y = 2, y = 4 and the yaxis in the first quadrant. (2)
Answer:
1.
2. Area
Question 5.
 Draw a rough sketch of the graph of the function y^{2} = 4x. (2)
 Find the area by the curve and the line x= 2. (2)
Answer:
1.
2. Area
Plus Two Maths Application of Integrals Six Mark Questions and Answers
Question 1.
 Draw the graph of y^{2} = 4x and y = x. (2)
 Find the points of intersection of y^{2} = 4x and y = x. (2)
 Find the area bounded by the graphs.(2)
Answer:
1. y^{2} = 4x is a parabola and y = x is a straight line passing through the origin.
2. x^{2} = 4x ⇒ x^{2} – 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.
Question 2.
 Draw the graph of the function y = x^{2} and x = y^{2} in a coordinate axes. (2)
 Find the point of intersection of the above graphs. (2)
 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.
2. We have, y = x^{2} and x = y^{2}
x = (x^{2})^{2} ⇒ x – x^{4} = 0 ⇒ x(1 – x^{3}) = 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 = \(\int_{0}^{1} \sqrt{x} d x\int_{0}^{1} x^{2} d x\)
Question 3.
Using the figure answer the following questions
 Find the area of the shaded region as the sum of the area of two triangles. (2)
 Define the function of the given graph. (2)
 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
Therefore verified.
Question 4.
The figure given below contains a straight line L with slope \(\sqrt{8}\) and a circle.
 Find the equation of the line L and circle. (1)
 Find the point of intersection P. (2)
 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 x^{2} + y^{2} = 9.
2. We have, y =3x and x^{2} + y^{2} = 9
⇒ x^{2} + (\(\sqrt{8}\)x)^{2} = 9 ⇒ 9x^{2} = 9
⇒ x = 1
∴ y = \(\sqrt{8}\) × 1 = \(\sqrt{8}\).
Therefore, coordinate of ‘P’ is (1, \(\sqrt{8}\)).
3. Area of the shaded region
Question 5.
Using the given figure answer the following
 Define the equation of the circle and ellipse in the figure. (1)
 Find the area of the ellipse using integration. (4)
 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 x^{2} + y^{2} = 4 and that of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\).
2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y^{2} = 1 ⇒ y = \(\frac{1}{2} \sqrt{4x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx
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.
 Find the area bounded by the curve y = sin x with X – axis, between x = 0 and x = 2π. (2)
 Find the area of the region bounded by the curve y = x^{2} 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.
Area
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:
Equation of BC is \(\frac{y5}{15}=\frac{x0}{10}\)
⇒ 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
Question 8.
Consider the functions f(x) = sin x and g(x) = cosx in the interval [0, 2π]
 Find the x coordinates of the meeting points of the functions. (1)
 Draw the rough sketch of the above functions. (2)
 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.
3. Area = 2{Area under f(x) = sinx from \(\frac{\pi}{4}\) to π – Area under g(x) = cosx from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)}
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 r^{2}. (2)
Find the area of the circle, x^{2} + y^{2} = 16 which is exterior to parabola y^{2} = 6x. (4)
Answer:
1.
2. Given, x^{2} + y^{2} = 16 and y^{2} = 6x ⇒ x^{2} + 6x = 16 ⇒ x^{2} + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0 ⇒ x = 8, 2.
Area = Area of the circle – Interior area of the parabola.
Question 10.
Using the figure answer the following questions
 Define the equation of the ellipse and circle in the given figure. (1)
 Find the area of the ellipse using integration. (4)
 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 x^{2} + y^{2} = 1.
2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y^{2} = 1 ⇒ y = \(\frac{1}{2} \sqrt{4x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx
3. Area of the circle = πr^{2} = π × 1 = π
Required area = Area of ellipse – area of the circle = 2π – π = π.
Plus Two Mathematics All Chapters Question and Answers
 Plus Two Mathematics Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 1 Relations and Functions Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 2 Inverse Trigonometric Functions Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 3 Matrices Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 4 Determinants Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 6 Application of Derivatives Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 7 Integrals Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 9 Differential Equations Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 10 Vector Algebra Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 11 Three Dimensional Geometry Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 12 Linear Programming Chapter Wise Question and Answers PDF
 Plus Two Maths Chapter 13 Probability Chapter Wise Question and Answers PDF
Benefits of the Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF
The Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF that has been provided above is extremely helpful for all students because of the way it has been drafted. It is designed by teachers who have over 10 years of experience in the field of education. These teachers use the help of all the past years’ question papers to create the perfect Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF.
0 comments:
Post a Comment