mathematics-class-12-unit-08-chapter-04-integral-calculus-l-4-6-qdoymoizafu

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  Solution: Area S $\geqslant$ area of rectangle 0 ABC $=1 \times \frac{1}{e}$ $S \geqslant \frac{1}{e}$

-  Also, $e^{-x^2} \geqslant e^{-x}$ for all $x \in[0,1]$ because $x^2 \leqslant x$ for $x \in[0,1]$
 
- $\therefore \quad \int_0^1 e^{-x^2} d x \geqslant \int_0^1 e^{-x} d x=1-\frac{1}{e}$
 
- $ \therefore$ Option (B) is correct. By the figure,
 
- $S \leqslant \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$
 
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<footer style = "font-size: 18px">Integral calculus lecture-4 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  Now,
 
- $\left(1-\frac{1}{e}\right)-\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
 
- $=\frac{3}{4}-\frac{1}{e}-\frac{1}{4 \sqrt{e}}=\frac{1}{4 e}(3 e-4-\sqrt{e})$
 
-  We know that 2<e<3
 
- $\therefore \quad 3 e-4-\sqrt{e}>3 \times 2-4-\sqrt{3}=2-\sqrt{3}>0$
 
- $\therefore \quad 1-\frac{1}{e}>\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
 
-  Since $S \geqslant 1-\frac{1}{e}, S>\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
 
- $ \therefore$ Option (C) is Wrong
 
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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-4 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- Let $f:\left[\frac{1}{2}, 1\right] \rightarrow[0, \infty)$ be a now constant differentiable function such that
 
- $f^{\prime}(x)<2 f(x)$ and $f\left(\frac{1}{2}\right)=1$
 
-  Then the value of $\int_{1 / 2}^1 f(x) d x$ lies in the interval
 
-  (A) $(2 e-1,2 e)$
 
-  (B) $(e-1,2 e-1)$
 
-  (C) $\left(\frac{e-1}{2}, e-1\right)$
 
-  (D) $\left(0, \frac{e-1}{2}\right)$
 
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<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"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $   f^{\prime}(x)<2 f(x)$
	
- $   \Rightarrow f^{\prime}(x)-2 f(x)<0 \quad \forall x \in\left[\frac{1}{2}, 1\right]$

- $   \Rightarrow e^{-2 x} \left[f^{\prime}(x)-2 f(x)\right]<0 \\ \Rightarrow $

- $   \Rightarrow \frac{d}{d x}\left[e^{-2 x} f(x)\right]<0 \quad \forall x \in\left[\frac{1}{2}, 1\right] \\ \Rightarrow $

- $   \Rightarrow e^{-2 x} f(x) \text { is a decreasing function  } \\ in \left[\frac{1}{2}, 1\right] \therefore $
 
- $   \therefore \quad e^{-2 x} f(x)<e^{-2\left(\frac{1}{2}\right)} f\left(\frac{1}{2}\right) \quad \forall x>\frac{1}{2}$
 
- $ = \frac{1}{e}$
 
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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-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $   \Rightarrow \quad f(x)<e^{2 x-1} \quad$ for $x \in\left(\frac{1}{2}, 1\right)$
 
- $   \begin{aligned} \Rightarrow \quad \int_{1 / 2}^1 f(x) d x<\int_{1 / 2}^1 e^{2 x-1} d x & =\left.\frac{e^{2 x-1}}{2}\right|_{1 / 2} ^1 \\ & =\frac{e}{2}-\frac{1}{2}=\frac{e-}{2}\end{aligned}$
 
- $   \therefore \quad \int_{1 / 2}^1 f(x) d x<\frac{e-1}{2}$
 
- $   Also, since f(x)>0, \int_{1 / 2}^1 f(x) d x>0$

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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-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore \quad e^{-f(1)}=e^{-g(1)}+c \Rightarrow e^{-8(1)}+c=e^{-1}-(i)$
 
- $e^{-f(2)}=e^{-g(2)}+c=e^{-1}+c \quad-$ (ii)
 
-  from (i) & (ii)
 
- $e^{-f(2)}-\frac{1}{e}=\frac{1}{e}-e^{-g(1)}$
 
- $\Rightarrow \quad e^{-f(2)}+e^{-g(1)}=\frac{2}{e}$
 
- $\therefore \quad e^{-f(2)}<\frac{2}{e} \quad$ & $e^{-g(1)}<\frac{2}{e}$
 
- $\Rightarrow \quad e^{f(2)}>\frac{e}{2} \quad$ & $e^{g(1)}>\frac{e}{2}$
 
- $\Rightarrow \quad f(2)>\ln (\frac{e}{2})=1-\ln 2 \quad$ & $\quad g(1)>1-\ln 2$
 
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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- <ins>Solution</ins>
 
- $g\left(\frac{1}{2}\right)=\int_0^1 t^{-1 / 2}(1-t)^{-1 / 2} d t$

- $   =\int_0^1 \frac{1}{\sqrt{t(1-t)}} d t$

- $=\int_0^1 \frac{1}{\sqrt{\frac{1}{4}-\left(t-\frac{1}{2}\right)^2}} d t$
 
- $=\left.\sin ^{-1}\left(\frac{t-\frac{1}{2}}{1 / 2}\right)\right|_0 ^1$
 
 
 
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<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-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $   \therefore g(a)=g(1-a) \quad \forall a \in(0,1)$
 
-  Differentiating, We get
 
- $   g^{\prime}(a)=-g^{\prime}(1-a)$
 
-  Putting $   a=\frac{1}{2}$, $   g^{\prime}\left(\frac{1}{2}\right)=-g^{\prime}\left(\frac{1}{2}\right)$
 
- $   \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=0$
 
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Another way:

- $g(a)=\int_0^1 t^{-a}(1-t)^{a-1} d t$

- $  g^{\prime}(a)= \int_0^1 \frac{\partial}{\partial a}\left[t^{-a}(1-t)^{a-1}\right] dt$
 
- $  =\int_0^1\left[-t^{-a} \ln t(1-t)^{a-1}+t^{-a}(1-t)^{a-1} \ln(1-t)\right] dt$
 
- $   \text {0 at a} =\frac{1}{2}$
 
- $   \therefore \quad g^{\prime}\left(\frac{1}{2}\right)=0$

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<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-4 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- Find the value of $   \int_0^1 4x^3 \frac{d^2}{d x^2}\left[\left(1-x^2\right)^5\right] d x$
 
- <ins>Solution</ins> Direct way: First calculate $  \frac{d^2}{d x^2}\left[\left(1-x^2\right)^5\right]$ and the integrate. But this involves let of computations.

-  Smarter way: Use integration by parts formula.
 
- $   \int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$ or,
 
- $   \int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x) \int g(x) d x d x$
 
- $   \int_0^1 4 x^3 f^{\prime \prime}(x) d x$ $  =\left.4 x^3 f^{\prime}(x)\right|_0 ^1-\int_0^1 12 x^2 f^{\prime}(x) d x$

- $   =\left.4 x^3 f^{\prime}(x)\right|_0 ^1-\left[\left.12 x^2 f(x)\right|_0 ^1-\int_0^1 24 x f(x) d x\right]$
 
- $   f^{\prime}(1)=0$
 
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<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-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Solution $   I_n=\int_0^\pi\left[\frac{\sin (n x)}{\left(1+\pi^x\right) \sin x}+\frac{\sin (-n x)}{\left(1+\pi^{-x}\right) \sin (-x)}\right] d x$
 
-  because
 
- $  \int_{-a}^a f(x) d x=\int_0^a[f(x)+f(-x)] d x$
 
- $  \therefore  I_n=\int_0^\pi \frac{\sin (n x)}{\sin x} d x$
 
- $  I_0 = 0$, $  I_1=\int_0^\pi \frac{\sin x}{\sin x} d x=\pi$
 
- $  I_0 \neq I_1 \Rightarrow(D)$ is wrong.
 
-  Now, $  I_{n+2}-I_n=\int_0^\pi \frac{\sin (n+2) x-\sin (n x)}{\sin x} d x$ $  =\int_0^\pi \frac{2 \cos (n+1) x \cdot \sin  x}{\sin x} d x$ 
 
- $  = \left| 2 \frac{\sin (n+1)^x}{n+1} \right| _ 0 ^{\pi}=0   \therefore \quad I_ {n+2}=I_n \quad \forall n$
 
-  Since, $  \begin{aligned} I_0=0, \quad I_{2 m}=0 \quad \forall m\end{aligned}$ $  \begin{aligned}I_1=\pi, \quad I_{2 m+1}=\pi \quad \forall m\end{aligned}$
 
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- The total number of distinct $x \in[0,1]$ for which $\int_0^x \frac{t^2}{1+t^4} d t=2 x-1$ is
 
- Solution Let $f(x)=\int_0^x \frac{t^2}{1+t^4} d t-(2 x-1)$ for $x \in[0,1]$
 
-  clearly, f is continuous and Differentiable function
 
-  Also,
 
- $f(0)=0-(-1)=1>0$
 
- $f(1)=\int_0^1 \frac{t^2}{1+t^4} d t-1$
 
-  For $0<t<1, \quad \frac{t^2}{1+t^4}<1$
 
- $\Rightarrow \quad \int_0^1 \frac{t^2}{1+t^4} d t<1$
 
- $\Rightarrow \quad f(1)<0$
 
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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Integral Calculus L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  By the intermediate value theorem, there exists at least one $x \in(0,1)$ for which $f(x)=0$.

- Also, $f^{\prime}(x)=\frac{x^2}{1+x^4}-2<1-2=-1<0$
- $\Rightarrow f$ is strictly decreasing in $(0,1)$.

- $\Rightarrow f$ can have at most one zero in $(0,1)$
- $\therefore$ The number of $x$ for which $f(x)=0$ is one.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>