**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 x^{2} = 4ay is symmetric about y-axis as shown below.

Graph of the parabola y^{2} = 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 y**^{2} = 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**

** x**^{2} + y^{2} = 1 and (x - 1)^{2} + y^{2}
= 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 \[=\int_{1}^{3}\left (\frac{3x+1}{2} \right )dx+\int_{3}^{6}\left (\frac{-2x+21}{3} \right )dx-\int_{1}^{6}\left (\frac{x+9}{5} \right )dx\]\[=\frac{1}{2}\int_{1}^{3}(3x+1)dx+\frac{1}{3}\int_{3}^{6}(-2x+21)dx-\frac{1}{5}\int_{1}^{6}(x+9)dx\]\[=\frac{1}{2}\left [ \frac{3x^{2}}{2}+x \right ]_{1}^{3}+\frac{1}{3}\left [ -x^{2}+21x \right ]_{3}^{6}-\frac{1}{5}\left [ \frac{x^{2}}{2}+9x \right ]_{1}^{6}\]\[=\frac{1}{2}\left [ \left (\frac{27}{2}+3 \right )-\left ( \frac{3}{2}+1 \right ) \right ]+\frac{1}{3}\left [ (-36+126)-(-9+63) \right ]\\-\frac{1}{5}\left [ (18+54)-\left ( \frac{1}{2}+9 \right ) \right ]\]\[=\frac{1}{2}\left ( \frac{33}{2}-\frac{5}{2} \right )+\frac{1}{3}(90-54)-\frac{1}{5}\left ( 72-\frac{19}{2} \right )\]\[=\frac{1}{2}(14)+\frac{1}{3}(36)-\frac{1}{5}\left ( \frac{125}{2} \right )\]\[=7+12-\frac{25}{2}=\frac{13}{2}=6.5\: sq.units\]**

**THANKS FOR YOUR VISIT**

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