### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Application of Integrals Chapter 8 Class 12

Application of Integrals  Class 12 Chapter 8
Method of finding the area under the curve, explanation with different examples

Introduction:
In geometry, we have learnt formulas to calculate areas of various geometrical figures including triangles, rectangles, trapezium and circles. However they are inadequate for calculating the areas enclosed by curves.
Now we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabola and ellipses.
Method of taking the limits:
If limit is taken on the x-axis, then find the value of y in terms of x. If  limit is taken on the y- axis, then find the value of x in terms of y.
Algorithm
• First of all find the limits on the x-axis or on the y-axis.
• If limit is on the x-axis, then find the value of y in terms of x.
• If limit is on the y-axis, then find the value of x in terms of y.
• Find the area under the curve by integrating the given function in the respective limits.
• We always take the absolute value of area. It means that area is always taken positive.
The formula which is mostly used in these questions
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}$
Some important graphs used in this chapter are as follows
Graph of the parabola x2 = 4ay  is symmetric about y-axis as shown below.
Graph of the parabola y2 = 4ay  is symmetric about x-axis as shown below.
Explanation of the topic by taking some examples
Example 1: [Q. No. 2 Exercise 8.1]
Find the are of the region bounded by  y2 = 9x, x = 2, x = 4 and the x- axis in the first quadrant.
Given equation of the curve is  $y^{2}=9x\: \: \Rightarrow \: \: y=\sqrt{9x}=3\sqrt{x}$
Required area = Area of ABCD$=\int_{2}^{4}ydx=\int_{2}^{4}3\sqrt{x}dx=3\int_{2}^{4}x^{\frac{1}{2}}dx$$3\left [ \frac{x^{3/2}}{3/2} \right ]_{2}^{4}=3\times \frac{2}{3}\left [ x^{3/2} \right ]_{2}^{4}=2\left [ 4^{3/2}-2^{3/2} \right ]$$=2[8-\sqrt{8}]=[16-4\sqrt{2}]sq.units.$
Example 2 [Question 2 Exercise 8.2]
Find the area of the region enclosed between the two circles
x2 + y2 = 1 and (x - 1)2 + y2 = 1
Solution
The equations of two circles are $x^{2}+y^{2}=1\: \: .............\: \: (1)\\(x-1)^{2}+y^{2}=1\: \: ..........\; (2)$Centre of the circle (1) is (0, 0) and radius is 1.
Centre of the circle (2) is (1, 0) and radius is 1
First of all we draw graphs for these circles as shown in the figure and then find the point of intersection of these two circles.
From eqn.(1) - eqn.(2) we get $x^{2}+y^{2}-[x^{2}+1-2x+y^{2}]=0$$\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$Putting x = 1/2 in eqn. (1) we get $\left (\frac{1}{2} \right )^{2}+y^{2}=1\Rightarrow y=\pm \frac{\sqrt{3}}{2}$Points of intersection of circles (1) and (2) are $P\left ( \frac{1}{2},\frac{\sqrt{3}}{2} \right )\: and\: Q\left ( \frac{1}{2},-\frac{\sqrt{3}}{2} \right )$
Now from both the equations of circles we should find the value of y in terms of x as follows $From\: equation\: (1)\: \: y=\sqrt{1-x^{2}}\\From equation (2)\: \: y=\sqrt{1-(x-1)^{2}}$
Required area = Area of region OQAP = 2(area of region OMAP) = 2(area of region OMPO + area of region MAPM) $=2\left [\int_{0}^{1/2} \sqrt{1-(x-1)^{2}}dx+\int_{1/2}^{1} \sqrt{1-x^{2}}dx \right ]$ $2\left [ \frac{(x-1)\sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2}sin^{-1}(x-1) \right ]_{0}^{\frac{1}{2}}+\left [ \frac{x\sqrt{1-x^{2}}}{2}+\frac{1}{2}sin^{-1}x \right ]_{\frac{1}{2}}^{1}$$= 2\left [\frac{\left ( \frac{1}{2} -1\right )\sqrt{1-\left ( \frac{1}{2}-1 \right )^{2}}}{2} +\frac{1}{2}sin^{-1}\left ( \frac{1}{2}-1 \right ) \right ]\\ -2\left [ \frac{\left (0-1\right )\sqrt{1-\left ( 0-1 \right )^{2}}}{2} +\frac{1}{2}sin^{-1}\left ( 0-1 \right ) \right ]$$+2\left [0+\frac{1}{2}sin^{-1}(1)-\frac{\frac{1}{2}\sqrt{1-(\frac{1}{2})^{2}}}{2}-\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right ) \right ]$$=2\left [ \frac{-\frac{1}{2}\times \frac{\sqrt{3}}{2}}{2}+\frac{1}{2}sin^{-1}\left ( -\frac{1}{2} \right )-0-\frac{1}{2}sin^{-1}(-1) \right ]$$+2\left [ \frac{1}{2}sin^{-1}(1)-\frac{\frac{1}{2}\times \frac{\sqrt{3}}{2}}{2}-\frac{1}{2}sin^{-1}\frac{1}{2} \right ]$$=-\frac{\sqrt{3}}{4}+\left ( -\frac{\pi }{6} \right )-\left ( -\frac{\pi }{2} \right )+\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{6}$$=\left ( -\frac{\pi }{6}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\pi }{6} \right )-2\times \frac{\sqrt{3}}{4}$$=\left (\frac{2\pi }{3}-\frac{\sqrt{3}}{2} \right )\: sq.units$
Example 3
Using the method of integration find the area of the region bounded by the lines.
3x - 2y +1 = 0,  2x + 3y - 21 = 0 and x - 5y + 9 = 0
Solution

The equations of the side CA is 3x - 2y + 1 = 0,  .....................  (1)
Equation of the side AB is  2x + 3y - 21 = 0,  ..........................   (2)
Equation of the side BC is  x - 5y + 9 = 0,   .............................   (3)
First of all we find the point of intersection of all the lines.
Solving eqn.(1) and eqn.(2) we get the point of intersection A(3, 5)
Solving eqn.(2) and eqn.(3) we get the point of intersection B(6, 3)
Solving eqn.(3) and eqn.(1) we get the point of intersection C(1, 2)
Draw all these points on the axis and make triangle ABC as shown in the figure
Now from all the equations find the value of y in terms of x
From equation (1) $y=\frac{3x+1}{2}$
From equation (2) $y=\frac{-2x+21}{3}$
From equation (3) $y=\frac{x+9}{5}$
Required Area $=\int_{1}^{3}\left (\frac{3x+1}{2} \right )dx+\int_{3}^{6}\left (\frac{-2x+21}{3} \right )dx-\int_{1}^{6}\left (\frac{x+9}{5} \right )dx$$=\frac{1}{2}\int_{1}^{3}(3x+1)dx+\frac{1}{3}\int_{3}^{6}(-2x+21)dx-\frac{1}{5}\int_{1}^{6}(x+9)dx$$=\frac{1}{2}\left [ \frac{3x^{2}}{2}+x \right ]_{1}^{3}+\frac{1}{3}\left [ -x^{2}+21x \right ]_{3}^{6}-\frac{1}{5}\left [ \frac{x^{2}}{2}+9x \right ]_{1}^{6}$$=\frac{1}{2}\left [ \left (\frac{27}{2}+3 \right )-\left ( \frac{3}{2}+1 \right ) \right ]+\frac{1}{3}\left [ (-36+126)-(-9+63) \right ]\\-\frac{1}{5}\left [ (18+54)-\left ( \frac{1}{2}+9 \right ) \right ]$$=\frac{1}{2}\left ( \frac{33}{2}-\frac{5}{2} \right )+\frac{1}{3}(90-54)-\frac{1}{5}\left ( 72-\frac{19}{2} \right )$$=\frac{1}{2}(14)+\frac{1}{3}(36)-\frac{1}{5}\left ( \frac{125}{2} \right )$$=7+12-\frac{25}{2}=\frac{13}{2}=6.5\: sq.units$