### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Integral calculus lecture-5 </primary> <hr> <main> <div style='float: right; width:0%;'>  </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Integral calculus lecture-5 </span> $\rightarrow$ First order ordinary differential equations $\rightarrow$ Methods of solving first order ODEs </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> First order ordinary differential equations </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - **Explicit form:** - $\frac{d y}{d x}=f(x, y)$, - where $x$ is the independent variable and $y$ is dependent on $x$. - $f$ is a given function of two variables. - **Implicit form:** - $F\left(x, y, \frac{d y}{d x}\right)=0$ - where $F$ is a given function of three variables </div> <div style='float: right; width:0%;'>  </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Integral calculus lecture-5 $\rightarrow$ <span style = "color:blue"> First order ordinary differential equations </span> $\rightarrow$ Methods of solving first order ODEs $\rightarrow$ Homogeneous ODEs </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Methods of solving first order ODEs </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - (1) Variables separable method - Suppose the given ODE can be written in the form $f(x) d x=g(y) d y$. - Then we can simply integrate both sides - $\int f(x) d x=\int g(y) d y$ - This gives solutions in implicit form. If possible, we will write $y$ as a function of $x$. </div> <div style='float: right; width:0%;'>  </div> </main> <hr> <footer style = "font-size: 18px">Integral calculus lecture-5 $\rightarrow$ First order ordinary differential equations $\rightarrow$ <span style = "color:blue"> Methods of solving first order ODEs </span> $\rightarrow$ Homogeneous ODEs $\rightarrow$ Linear list order ODEs </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Homogeneous ODEs </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - An ODE of the form $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$ - In such cases, we put $\frac{y}{x}=u$ ie, $y=u x$ - Then $\frac{d y}{d x}=u+x \frac{d u}{d x}$ - $\therefore u+x \frac{d u}{d x}=f(u)$ - $\Leftrightarrow \quad x \frac{d u}{d x}=f(u)-u$ - $\Leftrightarrow \quad \frac{d u}{f(u)-u}=\frac{d x}{x}$. Now integrate both sides </div> <div style='float: right; width:0%;'>  </div> </main> <hr> <footer style = "font-size: 18px">First order ordinary differential equations $\rightarrow$ Methods of solving first order ODEs $\rightarrow$ <span style = "color:blue"> Homogeneous ODEs </span> $\rightarrow$ Linear list order ODEs $\rightarrow$ Linear list order ODEs </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Linear list order ODEs </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - $\frac{d y}{d x}+p(x) y=g(x) \text {, where }$ - $p(x)$ and $g(x)$ are given functions of $x$ only. If we multiply the give equation by $e^{\int p(x) d x}$ then the L.H.S. becomes - $ e^{\int p(x) d x} \frac{d y}{d x} +p(x) e^{\int p(x) d x} y $ - $ =\frac{d}{d x} \left[e^{\int p(x) d x} \cdot y\right]$ </div> </main> <hr> <footer style = "font-size: 18px">Methods of solving first order ODEs $\rightarrow$ Homogeneous ODEs $\rightarrow$ <span style = "color:blue"> Linear list order ODEs </span> $\rightarrow$ Linear list order ODEs $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Linear list order ODEs </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - So, the ODE becomes - $ \frac{d}{d x}\left[e^{\int p(x) d x} \cdot y\right]=e^{\int p(x) d x} \cdot g(x)$ - $ \Rightarrow \quad e^{\int p(x) d x} y=\int e^{\int p(x) d x} g(x) d x+C$ - Thus we get the solution. - $e^{\int p(x) d x}$ is called an integrating factor. </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Homogeneous ODEs $\rightarrow$ Linear list order ODEs $\rightarrow$ <span style = "color:blue"> Linear list order ODEs </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-1 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - If $y=y(r)$ satisfies the differential equation $8 \sqrt{x} \sqrt{9+\sqrt{x}}$ - $ dy=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x \quad$ for $x>0$, - and $y(0)=\sqrt{7}$, then find the value of $y(256)$. </div> </main> <hr> <footer style = "font-size: 18px">Linear list order ODEs $\rightarrow$ Linear list order ODEs $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Solution- $\quad \int d y=\int \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} \cdot \frac{1}{8 \sqrt{x} \sqrt{9+\sqrt{x}}} d x$ - Put $u=4+\sqrt{9+\sqrt{x}}$ - The $d u=\frac{1}{2 \sqrt{9+\sqrt{x}}} \cdot \frac{1}{2 \sqrt{x}} d x$ - $y=\int \frac{1}{\sqrt{u}} \cdot \frac{d u}{2}=\sqrt{u}+c$ - $s \Rightarrow \quad y=\sqrt{4+\sqrt{9+\sqrt{x}}}+c$ - solve $y(0)=\sqrt{7}$, we get $C=0$ - $ \therefore \quad y(r)=\sqrt{4+\sqrt{9+\sqrt{x}}}$ - $ \therefore \quad y(256)=\sqrt{4+\sqrt{9+16}}=\sqrt{4+5}=3$ </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Linear list order ODEs $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-2 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$. - If $y=f(x)$ satisfies - $\frac{d y}{d x}=(2+5 y)(5 y-2)$, then find the value of $\lim _{x \rightarrow-\infty} f(x)$. </div> <div style='float: right; width:0%;'>  </div> </main> <hr> <footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - **Solution:** $ \int \frac{1}{(5 y+2)(5 y-2)} d y=\int d x$ - $ \Rightarrow \int \frac{1}{4}\left(\frac{1}{5 y-2}-\frac{1}{5 y+2}\right) d y=\int d x$ - $ \Rightarrow \frac{1}{4}\left[\frac{1}{5} \ln |5 y-2|-\frac{1}{5} \ln |5 y+2|\right]=x+c$ - $ \Rightarrow \quad \ln \left|\frac{5 y-2}{5 y+2}\right|=20 x+20 c$ - $ y(0)=0 \quad \Rightarrow \quad \ln \left|\frac{0-2}{0+2}\right|=0+20 c$ $\Rightarrow \quad c=0$ </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"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - $ \therefore \quad \ln \left|\frac{5 y-2}{5 y+2}\right|=20 x $ - $\Rightarrow \quad\left|\frac{5 y-2}{5 y+2}\right|=e^{20x}$ - Suppose $L=\lim _{x \rightarrow-\infty} f(x)$. Then - $ \left|\frac{5 L-2}{5 L+2}\right| =\lim _{x \rightarrow-\infty} e^{20 x}=0$ $ \Rightarrow \quad 5 L-2 =0 \Rightarrow L=\frac{2}{5},$ - $\therefore \lim _{x \rightarrow-\infty} f(x)=\frac{2}{5}$ </div> <div style='float: right; width:0%;'>    </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"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-3 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - A curve passes through the point $\left(1, \frac{\pi}{6}\right)$ Let the slope of the tangent to curve at any point $(x, y)$ be - $\frac{y}{x}+\sec \left(\frac{y}{x}\right)$ for $x>0$. Then the equation of the curve is - (A) $\quad \sin \left(\frac{y}{7}\right)=\ln x+\frac{1}{2}$ - (B) $\operatorname{cosec}\left(\frac{y}{x}\right)=\ln x+\frac{1}{2}$ - (C) $\quad \sec \left(\frac{2 y}{x}\right)=\ln x+2$ - (D) $\cos \left(\frac{2 y}{x}\right)=\ln x+\frac{1}{2}$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Solution:$ {\text{ Give}} \quad \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$ - This is a homogenears form - put $y=u x$. Then $u+x \frac{d u}{d x}=u+\sec u$ - $ \Rightarrow \quad x \frac{d u}{d x}=\sec u$ - $ \Rightarrow \quad \int \cos u d u=\int \frac{d x}{x}$ - $ \Rightarrow \quad \sin u=\ln x+c$ - $\Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c$ - $ \text{when} x=1, y=\frac{\pi}{6} \quad \therefore \quad \sin \frac{\pi}{6}=\ln 1+c$ - $\Rightarrow c=\frac{1}{2}$ - 1) $\sin \left(\frac{y}{x}\right)=\ln x+\frac{1}{2}$ </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-4 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Q4 Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ ard $f(1) \neq 1$. - Then - (A) $\lim _{x \rightarrow 0+} f^{\prime}\left(\frac{1}{x}\right)=1$ - (B) $\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right)=2$ - (C) $\lim _{x \rightarrow 0+} x^2 f^{\prime}(x)=0$ - (D) $|f(x)| \leq 2$ for all $x \in(0,2)$. </div> </main> <hr> <footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Solution:, write $y=f(x)$ - Then $\frac{d y}{d x}=2-\frac{y}{x}$ $\Rightarrow \quad \frac{d y}{d x}+\frac{1}{x} y=2 \leftarrow$ linear ODE. - $\Rightarrow \quad x \frac{d y}{d x}+y=2 x$ - $ \Rightarrow \quad \frac{d}{d x}(x y)=2 x$ - $ \Rightarrow \quad x y=x^2+c$ - $ \Rightarrow \quad y=\frac{x^2+c}{x c}=x+\frac{c}{x}$ - Equation $\quad f(x)=x+\frac{c}{x}$ $\Rightarrow f(1)=1+c$ - Since $f(1) \neq 1, c \neq 0$. </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - $ \therefore \quad f(x)=x+\frac{c}{x} \text { for some } c \neq 0$ - $ \Rightarrow f^{\prime}(x)=1-\frac{c}{x^2} \Rightarrow f^{\prime}\left(\frac{1}{x}\right)=1-c x^2$ - $ \text { i. Option (A) is correct. }$ - $ x f\left(\frac{1}{x}\right)=x\left[\frac{1}{x}+c x\right]=1+c x^2 \rightarrow 1 \text { as } x \rightarrow 0+ $ - $\therefore$ (B) is wrong. $ x^2 f^{\prime}(x)=x^2\left(1-\frac{c}{\pi^2}\right)=x^2-c$ - $\therefore \quad \lim _{x \rightarrow 0} x^2 f^{\prime}(x)=-c \neq 0$ </div> </main> <hr> <footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Note that since $f^{\prime}(x)=2-\frac{f(x)}{x}$ $x^2 f^{\prime}(x)=2 x^2-x f(x)$ - $\text{Mistake:} \lim _{x \rightarrow 0+} x f(x)=0$ - $\lim _{x \rightarrow 0+} f(x)$ need not be finite. - $f(x)=x+\frac{c}{x}, c \neq 0$ Since $\left|\frac{c}{x}\right| \rightarrow+\infty$ as $x \rightarrow 0+$ - $\therefore$ $f(x)$ is not bounded on $(0,2)$. - $\therefore$ Optor (D) is wrong. </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-5 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x)$, $x \in \mathbb{R}$, - $y(0)=0$, where $g(x)$ is a given now constant differentiable function on $\mathbb{R}$ write $g(0)=g(2)=0$. Then the value of $y(2)$ is - <ins>Solution</ins>: $\frac{d y}{d x}+g^{\prime}(x) y=g(x) g^{\prime}(x) \leftarrow \text { linear }$ - $\text {Integral factor} =e^{\int g^{\prime}(x) d x}=e^{g(x)}$ - $\Rightarrow \quad y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x$ </div> <div style='float: right; width:0%;'>   </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - By integertion by parts, - $ \quad \int g(x)\left(e^{g(x)} g^{\prime}(x) d x\right)=\int g(x) d\left(e^{g(x)}\right)$ - $ =g(x) e^{g(x)}-\int g^{\prime}(x) \cdot e^{g(x)} d x$ - $ =g(x) e^{g(x)}-e^{g(x)}+c$ - $ \therefore \quad y e^{g(x)}=[g(x)-1] e^{g(x)}+c$ - $ \Rightarrow \quad y=g(x)-1+c e^{-g(x)}$ - $ y(0)=0 \Rightarrow 0 =g(0)-1+c e^{-g(0)}$ - $ =0-1+c \Rightarrow c=1$ - $ \therefore y =g(x)-1+e^{-g(x)}$ - $ y(2)=g(2)-1+e^{-g(2)}=0-1+e^0=-1+1=0$ - $ \therefore y(2)=0 $ </div> <div style='float: right; width:0%;'>    </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"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Problem-6 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Q6. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a continuous for such that - $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t \quad$ for all $x \in[0, \infty)$ - Then (A) The curve $y=f(x)$ passes throng $(1,2)$ - (B) The curve $y=f(x)$ passes through $(2,-1)$ - (C) The area of the region $R=\{(x, y) \in[0,1] \times \mathbb{R}: f(x) \leq y \leq \sqrt{1-x}\}$ is $\frac{\pi-2}{4}$. - (D) The area of $R$ is $\frac{\pi-1}{4}$. </div> </main> <hr> <footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - Solution: $\quad f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ - $\text {putting x=0, we get f(0)=1}$ $f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t$ - Differentiating with respect to $x$, we get - $ f^{\prime}(x) =-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x)$ - $ \Rightarrow f^{\prime}(x) =-2+\int_0^x e^{x-t} f(t) d t+f(x)$ $ =-2+[f(x)-1+2 x]+f(x)$ - $ \Rightarrow \quad f^{\prime}(x)-2 f(x)=2 x-3 \quad \text { Liner ODE.}$ </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - $ \text { Integrating factor }=e^{\int-2 d x}=e^{-2 x}$ - $ \Rightarrow \quad \frac{d}{d x}\left[e^{-2 x} f(x)\right]=e^{-2 x}(2 x-3)$ - $ \Rightarrow \quad e^{-2 x} f(x)=\int(2 x-3) e^{-2 x} d x $ - $ =(2 x-3) \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x$ - $ =-\frac{1}{2}(2 x-3) e^{-2 x}+\frac{e^{-2 x}}{-2}+c $ - $ \Rightarrow f(x)=-\frac{1}{2}(2 x-3)-\frac{1}{2}+c e^{2 x}$ - $ \Rightarrow f(x)=c e^{2 x}-x+1$ </div> </main> <hr> <footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px;text-align:left'> - $ f(0)=1 \Rightarrow 1= c+1 \Rightarrow c=0 $ - $ \therefore f(x)=1-x$ - $ \quad x(1)=1-1=0$ - $ f(2)=1-2=-1 \quad \therefore \text { (B) is correct }$ - $ R=\\{(x, y) \in[0,1] \times \mathbb{R}: 1-x \leq\sqrt{1-x^2}\\}$ </div> <div style='float: right; width:0%;'>    </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:70%;font-size:30px;text-align:left'> - Area $(R)=$ ? - $ \operatorname{Area}(R) =\frac{1}{4} \pi(1)^2-\frac{1}{2} \times 1 \times 1$ - $ =\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4}$ - $\therefore$ (C) is corret - (D) is wrong. </div> <div style='float: right; width:30%;'>  <!----> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Integral Calculus L-5 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> <div style='float: right; width:0%;'>  </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>