mathematics-class-12-unit-07-chapter-05-definite-integral-l-5-7-ox65qukgttu

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
<hr>
<main>

<div style='float: left; width:50%;font-size:30px;text-align:left;'>
	
- Compute the area bounded between $y^2=4 a x$ and $y=m x,(a>0, m>0)$.

- <ins>solution</ins>

- $[\sqrt{4a x}-m x] d x=$ elementary area

- $ \text { Required } \left.A=\int (\sqrt{4 a x}-m x\right) d x$

- $ m^2 x^2=4 a x$

- $ \left(m^2 x-4 a\right) x=0$     
- $x=\frac{4 a}{m^2}, x=0$

</div>
<div style='float: right; width:40%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Definite integral lecture-5 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ \int_0^{\frac{4 a}{m^2}}(\sqrt{4 a x}-m x) d x$      
- $ =   \left.\sqrt{4 a} \frac{x^{3 / 2}}{3 / 2}\right|_0 ^{\frac{4 a}{m^2}}-\left.\frac{m}{2} x^2\right|_0 ^{4 a / m^2}$       
- $ =   \frac{2 \sqrt{4 a}}{3} \frac{4 a}{m^2} \sqrt{\frac{4 a}{m^2}}-\frac{m}{2} \frac{16 a^2}{m^4}$      
- $ =   \frac{32 a^2}{3 m^3}-\frac{8 a^2}{m^3}=\frac{8 a^2}{3m^3}$

</div>
<div style='float: right; width:01%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Definite integral lecture-5 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
<hr>
<main>

<div style='float: left; width:75%;font-size:30px;text-align:left;'>
	
- Area bounded between $x$-axis,

- $ y=2\left(x^3-x^2-2 x\right) \text { and } x=-1 \text { to } x=2 \text {.}$

- solution:
- $ y=2 x\left(x^2-x-2\right)$       
- $ y=2 x(x+1)(x-2)$     
- $ -1<x<0 \quad(x+1)>0, x<0$    
- $ (x-2)<0$

- $ y>0$

- $ 2>x>0 \quad x+1>0 \quad x-2<0$       
- $ y<0,0<x<2$

</div>
<div style='float: right; width:25%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ y=2(x^3-x^2-2x)$        
- $ A_1=\int_{-1}^0 2\left(x^3-x^2-2 x\right) d x $         
 
- $ =2\left(\frac{x^4}{4}-\frac{x^3}{3}-2 {\frac{x^2}{2}}\right)_{-1}^0 $        
- $ =-2\left(+\frac{1}{4}+\frac{1}{3}-1\right)$    
- $=-2\left(\frac{7}{12}-1\right)$
- $=\frac{-2(-5)}{12} $	
- $ =\frac{5}{6} $        
- $ A_2=\int_0^2 2\left(x^3-x^2-2 x\right) d x $         
     
- $ =\left.2\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2\right)\right|_0 ^2=2\left(4-\frac{8}{3}-4\right)=-\frac{16}{3} $         
	
- Total Area= $ A_1 + |A_2|$     
- $ =\frac{5}{6}+\frac{16}{3}=\frac{37}{6} $    
</div>
<div style='float: right; width:50%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
<hr>
<main>

<div style='float: left; width:70%;font-size:30px;text-align:left;'>
	
- Find out area bounded between 
- $y=2-x^2$ and $y=-x$ 
- $-x=2-x^2$      
- $ x^2-x-2=0$       
- $ x=\frac{1 \pm \sqrt{1+8}}{2}$     
- $ x=2,-1$

- $\int_ {-1}^2\left[2-x^2-(-x)\right] d x=  \int_{-1}^2[2-x^2+x] d x$     
- = $ 2 x-\frac{x^3}{3}+\left.\frac{x^2}{2}\right|_{-1} ^2 =  4-\frac{8}{3}+2-\left(-2+\frac{1}{3}+\frac{1}{2}\right)$

- $ =8-\frac{8}{3}-\frac{5}{6} =8-\frac{16+5}{6} =8-\frac{21}{6} =\frac{9}{2}$

</div>
<div style='float: right; width:30%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
<hr>
<main>

- Area bounded between $y=x \sqrt{4-x^2}$,       
- $x-axis$ and $x=-2$ and $x=+2$.

- solution:

- $ y=0, \quad x=0$   
- $ y=0, \quad x= \pm 2$     
- $ (0,2), y>0 $     
- $ (-2,0), y<0$

- $ A=\left|A_1\right| +A_2$     
- $A_2=\int_0^2 x \sqrt{4-x^2} d x $

</div>
<div style='float: right; width:30%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

<div style='float: left; width:100%;font-size:30px;text-align:left;'>
	
- $ A_1=-A_2$   
- $ A_2   =\int_0^2 x \sqrt{4-x^2} d x $    
- [ $x^2   =t$  $\Rightarrow 2 x d x=d t$     
- $\Rightarrow x=0, t=0;$   $ x=2, t=4$     
- $ x d x   =\frac{1}{2} d t$  ]

- $\Rightarrow A_2  =\frac{1}{2} \int_0^4 \sqrt{4-t} d t=\frac{1}{2} {\frac{(4-t)^{3 / 2}} {-3/2}}|_0 ^4$     
- $=-\frac{1}{3}\left[0-(4)^{3 / 2}\right]=+\frac{1}{3} \times 8=\frac{8}{3}$        
- $ A_2=\frac{8}{3}$ $\Rightarrow A_1=-\frac{8}{3}$

- $ \text { Required } =\left|-\frac{8}{3}\right|+\frac{8}{3}=\frac{16}{3}$     
</div>
<div style='float: right; width:01%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
<hr>
<main>

- Area bounded between two parabolas

- $y=x^2$ \& $y=8-x^2$

- solution:
- [$ x^2=8-x^2 $     
- $ 2 x^2=8$       
- $ x^2=4, x= \pm 2$     
- $ \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x$

- $ \int_ {-2}^2\left(8-x^2-x^2\right) d x  = \int_{-2}^2\left(8-2 x^2\right) d x$      
- $= 2 \int_ 0^2\left(8-2 x^2\right) d x= \left.2\left(8 x-\frac{2}{3} x^3\right)\right|_0 ^2$       
- $=2\left(16-\frac{2 \times 8}{3}\right)=32\left(1-\frac{1}{3}\right)=\frac{64}{3}$.
</div>
<div style='float: right; width:20%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Problem-6 $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
<hr>
<main>

- Find out area between $y=(1-\cos x) \sin x$      
- $x$-axis, $x=0$ and $x=\pi$.

- $ y={(1-\cos x)}{\sin x}$      
- $ x=0 \quad y=0$     
- $ x=\pi \quad y=0$      
- $ \sin x \geqslant 0 \quad[0, \pi]$      
- $ 1-\cos x \geqslant 0 \quad[0, \pi]$

- $\int_0^\pi {[(1-\cos x) \sin x d x} $       
- $ \int_0^\pi(\sin x-\sin x \cos x) d x=\int_0^\pi\left(\sin x-\frac{1}{2} \sin 2 x\right) d x$      
- $ =-\cos x+\left.\frac{\cos 2 x}{4}\right|_0 ^\pi$

- $=-(-1-1)+\frac{1}{4}(1-1)=2$

</div>
<div style='float: right; width:30%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
<hr>
<main>

- Area bounded between $y=\cos ^2 x$ and

- $y=1, x=0$ and $x=\pi$

- solution:
- $ A =\int_0^\pi\left[1-\cos ^2 x\right] d x$      
	
- $ =\int_0^\pi \sin ^2 x d x$

- $=\int_0^\pi \frac{1-\cos 2 x}{2} d x$        
- $A = \frac{x}{2}-\frac{sin2x}{4}|_0 ^ \pi$ 
- =$\frac{\pi}{2} - (0-0)=\frac{\pi}{2}$
- $\int_a^b f(x) d x$

</div>
<div style='float: right; width:20%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Remark $\rightarrow$  </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Remark </primary>
<hr>
<main>

- $f(x)$ is continuous on $[a, b]$    
- $a, b$ are finite

- 1) What if $f(x)$ is discontinuous on $[a, b]$

- 2) What if interval of integration is not bounded, i.e, $[a, \infty),(-\infty, a]$ OR $(-\infty, \infty)$.

</div>
<div style='float: right; width:40%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Example </primary>
<hr>
<main>

<div style='float: left; width:50%;font-size:30px;text-align:left;'>
	
- $f(x)$ is discontinuous on $[a, b]$ but $f(x)$ is piecewise continuous.

- $ \int_{-2}^2[x] d x$      
- $ = \int_{-2}^{-1}(-2) d x+\int_{-1}^0 (-1) d x$ + $\int_0^1 0 d x+\int_1^2 1 d x$

- =$(-2)(x)|_ {-2}  ^ {-1} $ + (-1) $x|_ {-1} ^ 0$ +0+ $x|_{1}^2$

- $ =(-2)(-1+2)+(-1)(0-(-1))+2-1$       
- $=-2-1+1=-2$       
	
</div>
<div style='float: right; width:30%; max-width:200px;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Example </primary>
<hr>
<main>

- Function is discontinous but interval is finite $(say  [a, b])$.

- (discontinous - Infinite Values Some in the interval of $\Rightarrow$ integration).

- Functions which are not continuous 
- also interval of integration is infinite.

- For. $\int_{-\infty}^{\infty} \frac{1}{x^2} d x \quad \frac{1}{x^2}$ is not continuous at $x=0$

- and interval of integration is also unbounded.

</div>
<div style='float: right; width:50%;'>

</main>
<hr>
<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Problem-9 $\rightarrow$ Problem-10 </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-9 </primary>
<hr>
<main>

<div style='float: left; width:50%;font-size:30px;text-align:left;'>
	
- $ I =\int_0^{\infty} \frac{1}{1+x^2} d x$

- solution:
- $ =\lim _{a \rightarrow \infty} \int_0^a \frac{1}{1+x^2} d x$     
- $ =\left.\lim _{a \rightarrow \infty} \tan ^{-1} x\right|_0 ^a$        
- $ =\lim _{a \rightarrow \infty}\left[\tan ^{-1} a-\tan ^{-1} 0\right]$       
- $ =\tan ^{-1} \infty=\pi / 2 .$       
</div>
<div style='float: right; width:50%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Problem-9 </span> $\rightarrow$ Problem-10 $\rightarrow$ Thankyou </footer>

### <Success style="font-size:25px"> Definite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-10 </primary>
<hr>
<main>

- $ \int_0^{ 1} \frac{d x}{\sqrt{x}}$         
- $ =\lim _{t \rightarrow 0} \int_t^ {1} \frac{d x}{\sqrt{x}}$      
- $ = \left.\lim _{t \rightarrow 0} \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right|_t ^1$     
- $ = \lim _{t \rightarrow 0}[2-2 \sqrt{t}]=2$

</div>
<div style='float: right; width:30%;'>

</div>

</main>
<hr>
<footer style = "font-size: 18px">Example $\rightarrow$ Problem-9 $\rightarrow$ <span style = "color:blue"> Problem-10 </span> $\rightarrow$ Thankyou $\rightarrow$ </footer>