mathematics-class-12-unit-08-chapter-03-integral-calculus-l-3-6-kvwmr4mwdqo

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Area between two curves </primary>
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- The area between the curves $y=f(x)$,$y=g(x)$ and $x=a$ and $x=b$ is given by  
- $\int_a^b[f(x)-g(x)] d x \quad$ if $f(x) \geqslant g(x)$

- in general,

- area

- = $\int_a^b|f(x)-g(x)| d x$

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<footer style = "font-size: 18px">Integral calculus lecture-3 $\rightarrow$ Area enclosed by curves $\rightarrow$ <span style = "color:blue"> Area between two curves </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- Find the area enclosed by the curves

- $y=\sin x+\cos x \text { and } y=|\cos x-\sin x|$

- over the interval $[0, \frac{\pi}{2}]$.

- solution: Let $f(x)=\sin x+\cos x $

- $ g(x)=|\cos x-\sin x|$

- for $ x \in[0, \frac{\pi}{2}] $

- Here, $g(x)=|\cos x-\sin x| \leq|\cos x|+|\sin x|$

- =$\cos x+\sin x \quad$ for $ x \in[0, \frac{\pi}{2}] $ = f(x)

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<footer style = "font-size: 18px">Area enclosed by curves $\rightarrow$ Area between two curves $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ |\cos x-\sin x|= \sin x-\cos x \qquad$  if $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$

- Area $= \int_0^{\pi / 4}[(\sin x+\cos x)-(\cos x-\sin x)] d x $ $+\int_{\pi / 4}^{\pi / 4}[(\sin x+\cos x)-(\sin x-\cos x)] d x$
- $ = \int_0^{\pi / 4} 2 \sin x d x+ \int_{\pi / 4}^{\pi /2 }2\cos x d x $ $-2 \cos x\mid_0 ^{\pi / 4}+(2 \sin x)\mid_{\pi / 4} ^{\pi / 2}$ $=-2(\frac{1}{\sqrt{2}}-1)+2(1-\frac{1}{\sqrt{2}})$ $=4(1-\frac{1}{\sqrt{2}})=4-2 \sqrt{2}$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Upper curve $f(x)$ =  $\begin{cases} x^2, \&  1 \leq x \leq 2 \\\\ \frac{8}{x}, \& 2 \leq x \leq 8\end{cases}$

- Lower curve, $g(x)=1,1 \leq x \leq 8$

- Area $=\int_1^8[f(x)-g(x)] d x$

- $ =\int_1^2(x^2-1) d x+\int_2^8(\frac{8}{x}-1) d x $

- $=(\frac{x^3}{3}-x)|_1 ^2+(8 \ln \mid x\mid-x)|_2 ^8$

- $=(\frac{8}{3}-2)-(\frac{1}{3}-1)$ $+(8 \ln 8-8)-(8 \ln 2-2)$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\frac{7}{3}-1+8 \ln 4-6=16 \ln 2-\frac{14}{3}$.

- Another way:

- Area $ =\int_1^4(\frac{8}{y}-\sqrt{y}) d y$
 
- $ =8 \ln |y||_1 ^4-\frac{2}{3} y^{3 / 2}|_1 ^4$

- $ =8 \ln 4-\frac{2}{3}(4^{3 / 2}-1)$

- $ =16 \ln 2-\frac{2}{3}(8-1)=16 \ln 2-\frac{14}{3}. $

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \Rightarrow 2 \alpha^4-4 \alpha^2+1=0$
 
- $\int_0^{1 / 2}(x-x^3) dx$

- $=\frac{1}{8}-\frac{(\frac{1}{2})^4}{4}$ $=\frac{1}{8}-\frac{1}{64}$ $=\frac{7}{64}<\frac{1}{2}$ $\Rightarrow \frac{1}{2}<\alpha<1 $

- $\therefore$ Option (B) is correct.

- Another way: $ 2 \alpha^4-4 \alpha^2+1=0$ $\Rightarrow \quad \alpha^2=\frac{4 \pm \sqrt{16-8}}{4}=1 \pm \frac{1}{\sqrt{2}}$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Let $f:[-1,2] \rightarrow(0, \infty)$

- be a continuous function

-  such that : $f(x)=f(1-x)$ for all $x \in[-1,2]$. 
 
- Let, $ R_1=\int_{-1}^2 x f(x) d x$ and, $R_2$ be the area of the region bounded by

- $y=f(x), x=-1, x=2$ and the $x$-axis.

-  Then,
 
- (A) $R_1=2 R_2 \qquad$ (B) $R_1=3 R_2$

- (C) $ 2 R_1=R_2  \qquad$  (D) $3 R_1=R_2$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let, $f(x)=\sqrt{|x+3|}$

- =$\sqrt{x+3}\quad \text {if}  x \geqslant-3 $

- $\sqrt{-(x+3)} \quad \text{ if } x<-3$

- $y=\sqrt{|x+3|}$ intersects the line

- $x=6$ at $(6,3)$

- $y=\sqrt{(x+1)}$ intersects the line

- $y=\frac{x+9}{5}$ at

- the points $(-4,1)$ and $(1,2)$.

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So, the region is as below

- Area of the region

- = Area of trapezium $ A B C D-\int_{-4}^{-3} \sqrt{-(x+3)} d x $

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\frac{-2}{3}(-x-3)^{3 / 2}\mid_{-4} ^{-3}$ $=\frac{-2}{3}(0-1)=\frac{2}{3}$

- Area of trapezium ABCD

- $=\frac{1}{2} \times(A D+B C) \times C D$ $=\frac{1}{2}(1+2) \times 5=\frac{15}{2}$

- $\therefore$ Area of the region $=\frac{15}{2}-(\frac{16}{3}+\frac{2}{3})$ $=\frac{15}{2}-6=\frac{3}{2}$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- Let, $F(x)=\int_x^{x^2+ \frac{\pi}{6}} 2 \cos ^2 t d t,$ $x \in R$ and

- $f:[0, \frac{1}{2}] \rightarrow[0, \infty)$ be a continuous function.
 
-  For each $a \in[0, \frac{1}{2}]$, if $F^{\prime}(a)+2$ is the area of the region bounded by $x=0, y=0,$
 
- $ y=f(x)$ and $x=a$, then find the value of $f(0).$

- Solution: Given,

- $F^{\prime}(a)+2=\int_0^a f(x) d x$ $F(x)  =\int_x^{x^2+\frac{\pi}{6}} 2 \cos ^2 t d t $
 
- $\Rightarrow F^{\prime}(x)  =2 \cos ^2(x^2+\frac{\pi}{6}) \cdot 2 x-2 \cos ^2 x \cdot 1$

- $ =4 x \cos ^2(x^2+\frac{\pi}{6})-2 \cos ^2 x $

- $\therefore \quad F^{\prime}(a)+2 $ $=4 a \cos ^2(a^2+\frac{\pi}{6})-2 \cos ^2 a+2$

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

### <Success style="font-size:25px"> Integral Calculus L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- differentiating w.r.t.  a,
 
- $f(a)=4 \cos ^2(a^2+\frac{\pi}{6}) $

- $+4 a \cdot 2 \cos (a^2+\frac{\pi}{6}) \cdot \sin (a^2+\frac{\pi}{6}) \cdot 2 a $ $ +4 \sin a \cos a$

- Put $a=0$ :

- $f(0)  =4 \cos ^2 \frac{\pi}{6}+0+0 $ $ =4(\frac{\sqrt{3}}{2})^2=3 $

- $\therefore \quad f(0)=3$

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>