mathematics-class-12-unit-07-chapter-01-definite-integral-1-7-ipt2khsab0

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Definite integral </primary>
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- $ \underline{\text{Definite Integral:-}}$

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

- $a=$ lower limit

- $b=$ upper limit

- Methods to compute Definite Integral (A): 
* Limit of finite sums
* Anti Derivatives

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<footer style = "font-size: 18px">Example 3 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Definite integral </span> $\rightarrow$ Limit of finite sums $\rightarrow$ Limit of finite sums </footer>

<h3 id="success-stylefont-size25px-definite-integral-l-1-success"><Success style="font-size:25px"> Definite Integral L-1 </Success></h3>
<h3 id="primary-stylefont-size24px-limit-of-finite-sums-primary"><primary style="font-size:24px"> Limit of finite sums </primary></h3>
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<p>Ex. Area bounded between</p>
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<p>$y=0$,</p>
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<p>$y=1+x^2$,</p>
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<p>$x=0$ and $x=1$.</p>
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<p>$ A=\int_0^1 (1+x^2)  d x$</p>
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<p>$ A>L_2=R_1+R_2$</p>
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<p>$ =\frac{1}{2} \times(1+0)+\frac{1}{2} \times\left(1+\frac{1}{4}\right)$</p>
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<p>$ =\frac{1}{2}\left[1+1+\frac{1}{4}\right]\\=1+\frac{1}{8}=\frac{9}{8}=1.125$</p>
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<footer style = "font-size: 18px">Example 4 $\rightarrow$ Definite integral $\rightarrow$ <span style = "color:blue"> Limit of finite sums </span> $\rightarrow$ Limit of finite sums $\rightarrow$ Limit of finite sums </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Limit of finite sums </primary>
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- $ U_2  =\bar{R}_1+\bar{R}_2$      
- $ =\frac{1}{2} \times\left(1+\frac{1}{4}\right)+\frac{1}{2} \\\\ \times(1+1)$      
- $ =\frac{1}{2}\left(2+\frac{5}{4}\right)$      
- $ =1+\frac{5}{8}=\frac{13}{8}=1.625$

- $L_2= 1 \cdot 125$
- $ U_2=\text { Upper Sum }$       
- $ L_2  =\text { Lower sum }$    
- $ L_2  <\mathrm{A}<U_2$

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<footer style = "font-size: 18px">Definite integral $\rightarrow$ Limit of finite sums $\rightarrow$ <span style = "color:blue"> Limit of finite sums </span> $\rightarrow$ Limit of finite sums $\rightarrow$ Limit of finite sums </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Limit of finite sums </primary>
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- Lower sum
- $ L_4  =\frac{1}{4}(1+0)$

- $ +\frac{1}{4}\left(1+\frac{1}{16}\right)$

- $ +\frac{1}{4}\left(1+\frac{1}{4}\right)$

- $ +\frac{1}{4}\left(1+\frac{9}{16}\right)$

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### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Limit of finite sums </primary>
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- $ A<U _4  <U_2$

- $ U_4= {\frac{1}{4}} \times(1+\frac{1}{16})$

- $ +\frac{1}{4} \times \left(1+\frac{1}{4}\right)$

- $ +\frac{1}{4}\left(1+\frac{9}{16}\right)$

- $ +\frac{1}{4}(1+1)$

- $ U_4= \frac{1}{4}[4+\frac{1+4+9+16}{16}]$

- $ =1 + \frac{30}{4\times 16}= {\frac{47}{32}} \\\\=1.46875$

- $U_2= 1.625$

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<footer style = "font-size: 18px">Limit of finite sums $\rightarrow$ Limit of finite sums $\rightarrow$ <span style = "color:blue"> Limit of finite sums </span> $\rightarrow$ Limit of finite sums $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- **Ex.** Let $f(x)$ be a continous function on $[a, b]$.

- $ f(x)>0$  
- $f(x)$ is increasing.

- $ A= Area =\int_a^{b} f(x) d x$

- $ x_0=a$      
- $ x_n=b$     
- $ \frac{b-a}{n}=h$      
- $ x_k=x_0+hk$

- $ {k=1,2, \ldots n}$      
    
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<footer style = "font-size: 18px">Limit of finite sums $\rightarrow$ Limit of finite sums $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- **Lower sums**

- $ L_n = hf(x_0) +h f\left(x_1\right) +\cdots+h f\left(x_{n-1}\right)$

- $ L_n= $ $\sum_{k=0}^{n-1}$$f(x_k) h$      
- $ U_n=h f\left(x_1\right)+h f\left(x_2\right)+\cdots+h f\left(x_n\right)$

- $ U_n = {\sum_{k=1}^n} f(x_k) h$

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<footer style = "font-size: 18px">Limit of finite sums $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- $ L_n =\sum_{k=0}^{n-1} f\left(x_k\right) h,$
	
- $U_n=\sum_{k=1}^n f\left(x_k\right) h$,       
	
- $ L_n <A<U_n$    
	
- $ \underset{n \rightarrow \infty} {\lim} L_ n = \underset{n \rightarrow \infty}{\lim} \sum_{k=0}^{n-1} f(x_k) h$

- $=\int_a^b f(x) dx$,

- $ \underset{n \rightarrow \infty} {\lim} U_ n = \underset{n \rightarrow \infty}{\lim} \sum_{k=1}^{n} f(x_k) h$    
	
- $=\int_a^b f(x) d x$

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

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- **Ex.** 
- Area between  
- $y=0, y=1+x^2, x=0$ and $x=1$

- Solution:  
- $x_0=0 ; x_n=1$

- Since we are dividing $[0,1]$ in n-equal intervals,

- $ \frac{1-0}{n}=h$

- $x_k=x_0+k h, ~ x_k=0+\frac{1}{n} k$      
- $x_k=\frac{k}{n}$

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

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- $L_n =h\left[\underbrace{1+1+\cdots+1}_{n \ \text{times}}+\frac{1^2+2^2+\cdot+(n-1)^2}{n^2}\right]$

- $=\frac{1}{n}[n+\frac{(n-1) n(2 n-1)}{6  n^2}]$

- $\\{\because h=\frac{1}{n}\\}$
  
  
- $ L_n  =1+\frac{1}{6}\left(1-\frac{1}{n}\right) \frac{n}{n}\left(2-\frac{1}{n}\right)$

- $\quad$

- $ \lim _{n \rightarrow \infty} L_n  =1+\frac{1}{6} \times 2=1+\frac{1}{3}=\frac{4}{3}\\\\=\int_0^1(1+x^2) d x$

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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 3 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- **Ex:**  Area between $y=e^{-x}, y=0, x=0$ and $x=1$.

- $ \frac{1-D}{n}=h$       
- $ x_k=x_0+k h$      
- $ x_k=0+\frac{k}{n}$    
- $ x_1=\frac{1}{n}, x_2=\frac{2}{n} \ldots$        
- $ L_n=hf\left(x_1\right)+hf\left(x_2\right)+\cdots+h f\left(x_n\right)$         
- $ L_n=h\left[e^{-\frac{1}{n}}+e^{-\frac{2}{n}}+\cdots+e^{-1}\right]$

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

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- $  L_n= h[e^{-\frac{1}{n}}+e^{-\frac{2}{n}}+ \ldots +e^{-\frac{n-1}{n}}+e^{-1}]$      
- $ =h \frac{e^{-\frac{1}{n}}}{1-e^{-\frac{1}{n}}}\left(1-\left(e^{-\frac{1}{n}}\right)^n\right)$     
	
	
- $ L_n=\frac{h}{e^n-1}\left(1-e^{-1}\right)$ $\quad$   
- $ \biggl\\{\frac{1}{n}=h \\\\ n\rightarrow \infty \ \text{then} \ h\rightarrow 0  \biggl\\}$

- $ \lim _{n \rightarrow \infty}$ $ L_n=\lim _{h \rightarrow 0} L_n=1-e^{-1} \lim \left(\frac{h}{e^{h-1}}\right) \text { as } h \rightarrow 0=1$

- We can compute that $\int_0^1 e^x dx = 1-e^{-1}$

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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Example 3 </span> $\rightarrow$ Anti derivatives $\rightarrow$ First fundamental theorem of calculus </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- $ \text{Show} ,\int_0^1 {1+x^2} d x$ $= {\frac{4}{3}}= $ $\lim _{n \rightarrow \infty}$ $ L_n$

- $ \int_0^1\left(1+x^2\right) d x$ 
- =$F(x)|_{x=0} ^{x=1}$

- $ F^{\prime}(x)= 1+x^2$

- $\frac{d}{d x}\left(x+\frac{1}{3} x^3\right)=1+x^2$

- $ \Rightarrow \int_0^1 (1+x^2) d x $  
- $=x+\frac{x^3}{3}$ $|_{x=0} ^{x=1}$  
- $={1+\frac{1}{3} -0}=\frac{4}{3}$

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<footer style = "font-size: 18px">First fundamental theorem of calculus $\rightarrow$ Second fundamental theorem of calculus $\rightarrow$ <span style = "color:blue"> Example 4 </span> $\rightarrow$ Example 5 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Definite Integral L-1 </Success>
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- $ \int_0^1 e^{-x} d x  =1-e^{-1}$

- $= \lim _{n \to \infty} L_n$

- $\int_0^1 e^{-x} d x  =-\left.e^{-x}\right|_{x=0} ^{x=1}$

- $\biggl[\because \frac{d}{dx}(-e^{-x})=e^{-x}\biggl]$

- $ =-e^{-1}-\left(-e^{-0}\right)$        
- $ =-e^{-1}+1$      
- $ =1-e^{-1} .$

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<footer style = "font-size: 18px">Second fundamental theorem of calculus $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Example 5 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>