### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Definite integral lecture-6 </primary> <hr> <main> <div style='float: left; width:99%; font-size: 30px; text-align left;'> </div> <div style='float: right; width:00%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Definite integral lecture-6 </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-1 </primary> <hr> <main> <div style='float: left; width:50%; font-size: 30px; text-align left;'> - Find out area of the region described by - $\{(x, y): y^2 \leqslant 2 x, y \geqslant 4 x-1\}$ - Solution:- - $ y^2 \leqslant 2 x . $ - $ y^2=2 x $ - $\ 2 x=-2 $ - $ y^2=0 $ - $ 0>-2$ - $ y \geqslant 4 x-1$ </div> <div style='float: right; width:50%;'>  <!----> </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Definite integral lecture-6 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%; font-size: 30px; text-align left;'> - $y=4 x-1 $ - $ y \geqslant 4 x-1 $ - $ 4 x-1=-1 $ - y=1 - 1>-1 - $ 4 x-\left.1\right|_ {x=0}=-1 \left.y\right|_{-2}=-2 $ - -2<-1 </div> <div style='float: right; width:50%;'>  <!----> </div> </main> <hr> <footer style = "font-size: 18px">Definite integral lecture-6 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%; font-size: 30px; text-align left;'> - $ y^2=2 x, \quad y=4 x-1 $ - $ y^2 \leqslant 2 x, \quad y \geqslant 4 x-1$ - $\begin{aligned} & y^2=2 x, y=4 x-1, \\\\ & y=4 \frac{y^2}{2}-1\end{aligned}$ - $\ 2 y^2-y-1=0, \\\\ y=\frac{1 \pm \sqrt{1+8}}{4}=\frac{1 \pm 3}{4}=1,-\frac{1}{2} $ - $\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y=$Required Area </div> <div style='float: right; width:50%;'>  <!----> </div> </main> <hr> <footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%; font-size: 30px; text-align left;'> - $\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) dy$ - $=\frac{1}{4}(\frac{y^2}{2}+y)|_ {-1 / 2} ^1-\frac{1}{2}$ $\frac{y^3}{3}|_{-1 / 2} ^1$ - $=\frac{1}{4}\left(\frac{1}{2}+1-\left(\frac{1}{8}-\frac{1}{2}\right)\right)-\frac{1}{2} \frac{1}{3}\left(1-\left(-\frac{1}{8}\right)\right)$ - $ =\frac{1}{4}\left(\frac{1}{2}+1-\frac{1}{8}+\frac{1}{2}\right)-\frac{1}{6}\left(1+\frac{1}{8}\right) $ - $ =\frac{1}{4}\left(\frac{15}{8}\right)-\frac{9}{48} \\\\ =\frac{15}{32}-\frac{3}{16}=\frac{9}{32}$ </div> <div style='float: right; width:00%;'> <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-2 </primary> <hr> <main> <div style='float: left; width:60%;font-size: 30px; text-align left;'> - $\quad y=\sin x+\cos x,\\\\ y=|\cos x-\sin x|$ - Find out area enclosed by above two curves and the lines $x=0$ and $x=\frac{\pi}{2}$. - solution: - $y = \sin x + \cos x$ - $\quad y(0)=1, y\left(\frac{\pi}{2}\right)=1$ - $y^{\prime}=\cos x-\sin x>0,\left[0,\frac{\pi}{4}\right)$ - $y^{\prime}<0 \quad\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$ - $y^{\prime}\left(\frac{\pi}{4}\right)=0,y\left(\frac{\pi}{4}\right)=\sqrt{2}$ </div> <div style='float: right; width:40%; max-width:250px;'>   <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%;font-size: 30px; text-align left;'> - $y=|\cos x-\sin x|\quad\left[0, \frac{\pi}{4}\right]$ - $y=\cos x-\sin x,\quad\left[0, \frac{\pi}{4}\right]$ - $y^{\prime}=-\sin x-\cos x < 0,[0, \frac{\pi}{4}]$ - $\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow y=-(\cos x-\sin x)$ - $y^{\prime}=\sin x+\cos x>0,\left[\frac{\pi}{4}, \frac{\pi}{2}\right] $ - $y(0)=1, y\left(\frac{\pi}{4}\right)=0 $ - $y\left(\frac{\pi}{2}\right)=1$ </div> <div style='float: right; width:45%; max-width:250px;'>   <!----> </div> </main> <hr> <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"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:60%;font-size: 22px; text-align left;'> - $ y=\sin x+\cos x $ - $ y=|\cos x-\sin x| $ - $A=\int_0^{\pi / 4} \{[\sin x+\cos x}{-(\cos x-\sin x) dx}$ - $+\int_\{\pi / 4}^{\pi / 2}[\sin x+\cos x-(-\cos x + \sin x) d x]$ - $ =2 \int_0^{\pi / 4} \sin x d x+2 \int_{\pi / 4}^- {\pi / 2} \cos x d x $ - $ =\left.2(-\cos x)\right|_ 0 ^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}$ - $ =2\left(-\frac{1}{\sqrt{2}}+1\right)+2\left(1-\frac{1}{\sqrt{2}}\right)=4\left(1-\frac{1}{\sqrt{2}}\right)$ - Required Area $=4\left(1-\frac{1}{\sqrt{2}}\right)$Required Area $=2A$ </div> <div style='float: right; width:40%;'>  <!----> <!----> <!----> <!----> </div> </main> <hr> <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"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-3 </primary> <hr> <main> <div style='float: left; width:50%; font-size: 25px; text-align left;'> - Find all possible values of $b$ so that area of the bounded region enclosed between parabolas. - $ y=x-b x^2 \text { and } y=\frac{x^2}{b} \text { is Maximum, } $ - $\text{where} \ b > 0$ </div> <div style='float: right; width:25%;'>  <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%; font-size: 30px; text-align left;'> - $\begin{aligned} & \begin{array}{ll}y=x-b x^2 \\\\ y=-b\left(x^2-\frac{x}{b}\right)\\\\ y=-b\left[\left(x-\frac{1}{2} b\right)^2-\frac{1}{4} b^2\right]\end{array} \\\\ & y=-b\left(x-\frac{1}{2 b}\right)^2+\frac{1}{4 b} \\\\ & \left(y-\frac{1}{4 b}\right)=-b\left(x-\frac{1}{2 b}\right)^2 \\\\ & Y=-b X^2, \quad b>0 \\ & \end{aligned}$ </div> <div style='float: right; width:50%; '> <img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/9b0590d5-f997-49c7-4f37-4801f6b69b00/public"> <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%;font-size: 25px; text-align left;'> - $ y=x-b x^2, \quad b>0 $ - $y=\frac{x^2}{b}, \quad b>0 $ - $\frac{x^2}{b}=x-b x^2 $ - $\frac{x^2}{b}+b x^2=x \quad x=0 $ - $ x^2 (\frac{1+b^2}{b}) =x$ $x=\frac{b}{1+b^2} $ - Area Required - $=\int_0^{b / 1+b^2}\left[x-b x^2-\frac{x^2}{b}\right] d x$ </div> </main> <hr> <footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%;font-size: 25px; text-align left;'> - $ \text { Required \ Area }=\frac{x^2}{2}-\frac{b}{3} x^3-\left.\frac{x^3}{3 b}\right|_0 ^{b/1+b^2} $ - $ =\frac{b^2}{2\left(1+b^2\right)^2}-\frac{b}{3} \frac{b^3}{\left(1+b^2\right)^3}-\frac{b^3}{3 b\left(1+b^2\right)^3} $ - $ =\frac{b^2}{2\left(1+b^2\right)^2}-\frac{b^2}{3\left(1+b^2\right)^3}\left(1+b^2\right)$ - = $ \frac{b^2}{2\left(1+b^2\right)^2}-\frac{b^2}{3\left(1+b^2\right)^2} $ - = $ \frac{3-2}{6} \frac{b^2}{\left(1+b^2\right)^2}=\frac{b^2}{6\left(1+b^2\right)^2}$ </div> <div style='float: right; width:35%;'>  <!----> <!----> <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size: 30px; text-align left;'> - $A=\frac{1}{6} .\frac{b^2}{\left(1+b^2\right)^2}, \quad b>0$ - for which value of $b$ Area is maximum. - $\begin{aligned} \frac{d A}{d b} & =\frac{1}{6} \frac{\left(1+b^2\right)^2 2 b-b^2 2\left(1+b^2\right) 2 b}{\left(1+b^2\right)^4} \\\\ & =\frac{1}{6} \frac{1+b^2}{\left(1+b^2\right)^4}\left[2 b+2 b^3-4 b^3\right] \\\\ & =\frac{1}{6} \frac{1}{\left(1+b^2\right)^3}\left(2 b-2 b^3\right) \\\\ & =\frac{1}{3} b \frac{\left(1-b^2\right)}{\left(1+b^2\right)^3} \end{aligned} $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

<h3 id="success-stylefont-size25px-definite-integral-l-6-success"><Success style="font-size:25px"> Definite Integral L-6 </Success></h3> <h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3> <hr> <main> <div style='float: left; width:50%;font-size: 25px; text-align left;'> <ul> <li> <p>$\frac{d A}{d b}=\frac{1}{3} b \frac{\left(1-b^2\right)}{\left(1+b^2\right)^3} $</p> </li> <li> <p>$ \frac{d A}{d b}=0, b=0, b=-1, b=+1, \quad b>0$</p> </li> <li> <p>So only possible value of " $b$ " $=1$</p> </li> <li> <p>$ b > 1, \frac{d A}{d b} < 0$</p> </li> <li> <p>$ 0 < b < 1,\frac{d A}{d b} > 0$</p> </li> <li> <p>at $b=1$ Area is maximum</p> </li> <li> <p>$\frac{d A}{d b}=\frac{1}{3} b\left(\frac{(1-b^2)}{\left(1+b^2)\right.}\right) 3$</p> </li> </ul> </div> <div style='float: right; width:50%;'> <img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/9013d55b-df08-45cb-a7d3-ce33d1067f00/public"> <!----> <!----> <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-4 </primary> <hr> <main> <div style='float: left; width:99%;font-size: 30px; text-align left;'> - $ I=\int_2^4 \frac{\log x^2 d x}{\log x^2+\log \left(36-12x+x^2\right)} $ - solution: - $ =\int_2^4 \frac{2 \log x d x}{2 \log x+\log (6-x)^2} $ - $ I =\int_2^4 \frac{2 \log x d x}{2\log x+2 \log (6-x)}\quad \cdots \cdots (1)$ - $ \int_a^b f(x) d x=\int_a^b f(a+b-x) d x $ - $I=\int_2^4 \frac{\log (6-x) d x}{\log (6-x)+\log (6-(6-x))}$ - $I=\int_2^4 \frac{\log (6-x) d x}{\log (6-x)+\log x} \quad \cdots \cdots (2)$ </div> <div style='float: right; width:01%;'> <!--   --> </div> </main> <hr> <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"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size: 30px; text-align left'> - $\begin{aligned} & 2 I=\int_2^4 \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x \\\\ & 2 I=\int_2^4 d x=4-2=2 \\\\ & I=1 \end{aligned}$ </div> <div style='float: right; width:01%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-5 </primary> <hr> <main> <div style='float: left; width:50%;font-size: 30px; text-align left'> - $\begin{aligned} & \int_{-\pi / 2}^{\pi / 2}\left[x^2+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x d x \\\\ & =\int_{-\pi / 2}^{\pi / 2} x^2 \cos x d x+\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x d x \\\\ & =2 \int_0^{\pi / 2} x^2 \cos x d x+0\end{aligned}$ - $ h(x)=\log \left(\frac{\pi-x}{\pi+x}\right) \cos x \quad \$ - $ h(-x)=\operatorname{log}\left(\frac{\pi+x}{\pi-x}\right) \cos (-x) $ - $=-\log \left(\frac{\pi-x}{\pi+x}\right) \cos x $ </div> </main> <hr> <footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:50%;font-size: 30px; text-align left'> - $=-h(x) \int_{-\pi / 2}^{\pi / 2} h(x) d x=0$ - $ I =2 \int_0^{\pi / 2} x^2 \cos x d x$ - $ =2\left[\left.x^2(\sin x)\right|_0 ^{\pi / 2}-\int_0^{\pi / 2} 2 x \sin x d x\right] $ - $ =2\left[\frac{\pi^2}{4}-0-\left.2 x(-\cos x)\right|_0 ^{\pi / 2}+\int_0^{\pi / 2} 2(-\cos x) d x\right. $ - $ =2\left[\frac{\pi^2}{4}-\left.2 \sin x\right|_0 ^{\pi / 2}\right] =2\left[\frac{\pi^2}{4}-2\right]=\frac{\pi^2}{2}-4$ </div> <div style='float: right; width:01%;'> <!--  --> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Problem-6 </primary> <hr> <main> <div style='float: left; width:70%;font-size:30px;'> - $\int_{\sqrt{\log (2)}} ^{\sqrt{\log (3)}} \frac{x \sin \left(x^2\right)}{\sin x^2+\sin \left[\log 6-x^2\right]} $ - $ x^2=t, \quad 2 x d x=d t $ - $ =\int_{t=\log 2}^{t=\log 3} \frac{\sin t\left(\frac{1}{2} d t\right)}{\sin t+\sin [\log 6-t]}$ - $ I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t d t}{\sin t+\sin [\log 6-t]} $ </div> </main> <hr> <footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution $\rightarrow$ Thankyou </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:70%;font-size:30px;'> - $ \because \int_a^b f(x) dx = \int _a^b f(a+b-x) dx $ - $\begin{aligned} & I= \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+ \sin t} d t \\\\ & 2I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6-t)}{\sin (\log 6-t)+ \sin t} d t \\\\ & 2 I=\frac{1}{2} \int_{\log 2}^{\log 3} d t=\frac{1}{2}[\log 3-\log 2] \\\\ & 2 I=\frac{1}{2} \log (\frac{3}{2}), \\\\ & I=\frac{1}{4} \log (\frac{3}{2}) .\end{aligned}$ </div> <div style='float: right; width:01%;'> <!--    --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Definite Integral L-6 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> </main> <hr> <footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>