mathematics-class-12-unit-08-chapter-02-integral-calculus-l-2-6-gyrhyib3sbu

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Definite integrals as limits of sums </primary>
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- $\text { Let } x_0=a, x_1=a+h, x_2=a+2 h, \cdots, x_n=a+n h=b $

- $\text { Then } n h=x_n-x_0=b-a $

- $\therefore \quad h=\frac{b-a}{n}$

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<footer style = "font-size: 18px"> $\rightarrow$ Integral calculus lecture-2 $\rightarrow$ <span style = "color:blue"> Definite integrals as limits of sums </span> $\rightarrow$ Definite Integrals as Limits of Sums $\rightarrow$ Definite Integrals as Limits of Sums </footer>

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Definite Integrals as Limits of Sums </primary>
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- $\int_a^b f(x) d x$

- $=\lim_{n \rightarrow \infty}$ $\sum_{k=1}^n {h.f(a+k h)} $

- $=\lim_{n \rightarrow \infty} \sum_{k=1}^n{(\frac{b-a}{n}).f(a+k \frac{(b-a)}{n})}$

- In particular, if $a=0$ and $b=1$

- $\int_0^1 f(x) d x=\lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n} f\left(\frac{k}{n}\right)$

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<footer style = "font-size: 18px">Integral calculus lecture-2 $\rightarrow$ Definite integrals as limits of sums $\rightarrow$ <span style = "color:blue"> Definite Integrals as Limits of Sums </span> $\rightarrow$ Definite Integrals as Limits of Sums $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Definite Integrals as Limits of Sums </primary>
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- If we take the values of the function at the left end points of each sub interval instead of the right end points, we still get $\int_a^b f(x) d x$.

- $\int_a^b f(x) d x=\lim_{n \rightarrow \infty} \sum_{k=0}^{n-1}\left(\frac{b-a}{n}\right) f\left(a+k\left(\frac{b-a}{n}\right)\right)$

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<footer style = "font-size: 18px">Definite integrals as limits of sums $\rightarrow$ Definite Integrals as Limits of Sums $\rightarrow$ <span style = "color:blue"> Definite Integrals as Limits of Sums </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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-  For $a \in \mathbb{R},|a|>1$, let

- $\lim_{n \rightarrow \infty} \frac{1+\sqrt[3]{2}+3 \sqrt{3}+\cdots+\sqrt[3]{n}}{n^{7 / 3}\left(\frac{1}{(a n+1)^2}+\frac{1}{(a n+2)^2}+\cdots+\frac{1}{(a n+n)^2}\right)}=54$

- Then the possible values of $a$ are

- (A) -9
- (B) -6
- (C) 7
- (D) 8

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<footer style = "font-size: 18px">Definite Integrals as Limits of Sums $\rightarrow$ Definite Integrals as Limits of Sums $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\text { Sol: We have }$

- $\lim_{n \rightarrow \infty} \frac{\sum_{r=1}^n r^{1 / 3}}{n^{7 / 3} \sum_{r=1}^n \frac{1}{\left(a_n+r\right)^2}}$

- $=\lim_{n \rightarrow \infty} \frac{n^{1 / 3} \sum_{r=1}^n\left(\frac{r}{n}\right)^{1 / 3}}{n^{7 / 3} \cdot \frac{1}{n^2} \sum_{r=1}^n \frac{1}{\left(a+\frac{r}{n}\right)^2}}$

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<footer style = "font-size: 18px">Definite Integrals as Limits of Sums $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\lim_{n \rightarrow \infty} \frac{n^{1 / 3} \sum_{r=1}^n\left(\frac{r}{n}\right)^{1 / 3}}{n^{7 / 3} \cdot \frac{1}{n^2} \sum_{r=1}^n \frac{1}{\left(a+\frac{r}{n}\right)^2}}$

- $=\frac{\lim_{n \rightarrow \infty} \sum_{r=1}\left(\frac{r}{n}\right)^{1 / 3}}{\lim_{n \rightarrow \infty} \sum_{r=1}^n\frac{1}{\left(a+\frac{r}{n}\right)^2}}$

- $=\frac{\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=1}\left(\frac{r}{n}\right)^{1 / 3}}{\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=1}^n\frac{1}{\left(a+\frac{r}{n}\right)^2}}$

- $=\frac{\int_0^1 x^{1 / 3} d x}{\int_0^1 \frac{1}{(a+n)^2} d x} = \frac{\left.\frac{3}{4} x^{4 / 3}\right|_0 ^1}{\left.\frac{-1}{a+x}\right|_0 ^1}=\frac{3}{4} a(a+1)$

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $y_n =\frac{1}{n}[(n+1)(n+2) \cdots(n+n)]^{1 / n} $

- $=\left[\left(\frac{n+1}{n}\right)\left(\frac{n+2}{n}\right) \cdots\left(\frac{n+n}{n}\right)\right]^{1 / n} $

- $=\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{1 / n}$

- $\Rightarrow \quad \ln y_n=\frac{1}{n} \sum_{k=1}^n \ln \left(1+\frac{k}{n}\right) $

- $\Rightarrow \quad \lim_{n \rightarrow \infty} \ln y_n =\lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n} \ln \left(1+\frac{k}{n}\right) $

- $=\int_0^1 \ln (1+x) d x$

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

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So, $\lim_{n \rightarrow \infty} \ln y_n=\ln \left(\frac{y}{e}\right)$

- Taking exponential, $\lim_{n \rightarrow \infty} y_n=\frac{4}{e}$

- $\therefore \quad L =\frac{4}{e} \quad \text { since } 2<e<3, $

- $\frac{4}{3}<\frac{4}{e}<2 $

- $\Rightarrow [L]=1 \quad$ Ans.

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

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \cdots\left(x+\frac{n}{n}\right)}{n !\left(x^2+n^2\right)\left(x^2+\frac{n^2}{2}\right) \cdots\left(x^2+\frac{n^2}{n^2}\right)} $

- $= \frac{n^n \cdot n\left(1+\frac{x}{n}\right) \cdot \frac{n}{2}\left(1+\frac{2 x}{n}\right) \cdots \frac{n}{n}\left(1+\frac{n x}{n}\right)}{n ! n^2\left(1+\frac{x^2}{n^2}\right) \cdot \frac{n^2}{2^2}\left(1+\frac{2^2 x^2}{n^2}\right) \cdots \frac{n^2}{n^2}\left(1+\frac{n^2 x^2}{n^2}\right)}$

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\frac{\frac{n^{2 n}}{n y}\left(1+\frac{x}{n}\right)\left(1+\frac{2 x}{n}\right) \cdots\left(1+\frac{n x}{n}\right)}{n! \frac{n^{2 n}}{(n!)^2}\left(1+\frac{x^2}{n^2}\right)\left(1+\frac{2^2 x^2}{n^2}\right) \cdots\left(1+\frac{n^2 x^2}{n^2}\right)} $

- $\therefore \quad f(x)=$

- $\lim_{x \rightarrow \infty}\left[\frac{\left(1+\frac{x}{n}\right)\left(1+\frac{2 x}{n}\right) \cdots\left(1+\frac{n x}{n}\right)}{\left(1+\left(\frac{x}{n}\right)^2\right)\left(1+\left(\frac{2 x}{n}\right)^2\right) \cdots\left(1+\left(\frac{n x}{n}\right)^2\right.}\right]^{x / n}$

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\ln f(x)= \lim_{n \rightarrow \infty} \frac{x}{n} \sum_{k=1}^n \ln \left(1+\frac{k x}{n}\right) $

- $-\sum_{k=1}^n \ln (1+(\frac{k x}{n})^2] $

- $= \lim_{n \rightarrow \infty} \frac{x}{n} \sum_{k=1}^n \ln \left(1+\frac{k x}{n}\right) $

- $-\lim_{n \rightarrow \infty} \frac{x}{n} \sum_{k=1}^n \ln \left(1+\left(\frac{k x}{n}\right)^2\right)$

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\int_0^x \ln (1+y) d y-\int_0^x \ln \left(1+y^2\right) d y$

- $\Rightarrow \quad \ln f(x)=\int_0^x \ln \left(\frac{1+y}{1+y^2}\right) d y$

- differentiating w.r.t $x$, we get

- $\frac{f^{\prime}(x)}{f(x)}=\ln \left(\frac{1+x}{1+x^2}\right), x>0$

- $\Rightarrow \quad f^{\prime}(x)=f(x) \ln \left(\frac{1+x}{1+x^2}\right)$

- Note that $f(x)>0$

- and $\ln \left(\frac{1+x}{1+x^2}\right)>0 \quad$ if $0<x<1$
- $<0 \quad$ if $x>1$

- $\therefore \quad f^{\prime}(x)>0 \quad$ if $0<x<1$

- $<0 \quad$if $x>1$

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\quad \frac{f^{\prime}(x)}{f(x)}=\ln \left(\frac{1+x}{1+x^2}\right) $

- $\therefore \quad \frac{f^{\prime}(3)}{f(3)}=\ln \left(\frac{4}{10}\right)=\ln \left(\frac{2}{5}\right)$

- $\frac{f^{\prime}(2)}{f(2)}=\ln \left(\frac{3}{5}\right)$

- $\Rightarrow \frac{f^{\prime}(3)}{f(3)}<\frac{f^{\prime}(2)}{f(2)}$

- Ans: (B), (C)

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<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"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Let $f(x)=$

- $\int_{1 / x}^x \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t$ for $x \leftarrow (0, \infty)$.

- Then

- (A) $f$ is increasing on $(1, \infty)$

- (B) $f$ is decreasing on $(0,1)$

- (C) $f(x)+f\left(\frac{1}{x}\right)=0$ for all $x \leftarrow(0, \infty)$

- (D) $f\left(2^x\right)$ is an odd function of $x$ on $\mathbb{R}$

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

### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Sol: $\text { Recall: }$

- $\frac{d}{d x}\left[\int_{a(x)}^{b(x)} f(t) d t\right]=f(b(x)) \cdot b^{\prime}(x)-f(a(x)) a^{\prime}(x)$

- $f(x)=\int_{1 / x}^x \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t \Rightarrow f^{\prime}(x)= \frac{e^{-\left(x+\frac{1}{x}\right)}}{x} \cdot 1-\frac{e^{-\left(\frac{1}{x}+x\right)}}{1 / x} \cdot\left(\frac{-1}{x^2}\right)$ $=\frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}>0 \text { for all } x>0$ $\Rightarrow f$ is strictly increasing on $(0, \infty)$

- Another ways: We see that the internal $\left(\frac{1}{x}, x\right)$ increases as $x$ increases.

- Also, the integrand is positive

- $\therefore \quad f(x)$ is increasing function on $(0, \infty)$

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### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Now,

- $f\left(\frac{1}{x}\right)=\int_x^{1 / x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t $       
- $\text { put } t=\frac{1}{y} \cdot$ then $ d t=\frac{-1}{y^2} d y $       
- $\text { when } t=x, y=\frac{1}{x} $       
- $\text { when } t=\frac{1}{x}, y=x $

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### <Success style="font-size:25px"> Integral Calculus L-2 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So,

- $f\left(\frac{1}{x}\right)=\int_{1 / x}^x \frac{e^{-\left(\frac{1}{y}+y\right)}}{1 / y} \cdot\left(\frac{-1}{y^2} d y\right) $

- $= -\int_{1 / x}^x \frac{e^{-\left(y+\frac{1}{y}\right)}}{y} d y=-f(x) $

- $\Rightarrow \quad f(x)+f\left(\frac{1}{x}\right)=0 \quad \text { for all } x>0$

- Also, if $g(x) = f(2^x)$ , then

- $g(-x)=f\left(2^{-x}\right)=f\left(\frac{1}{2^x}\right)=-f\left(2^x\right)=-g(x)$

- $\Rightarrow \quad g(x)$ is an odd function

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