mathematics-class-12-unit-07-chapter-02-definite-integral-2-7-2zs1naeonxw

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- $\underline{{\text{Ex.}}} \quad  \int_0^1 \frac{1}{1+x^2} d x $

- $ =\left.\tan ^{-1} x\right|_{x=0} ^{x=1} $

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

- $ [ \quad \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} ] $

- $ \frac{\pi}{4}-0=\frac{\pi}{4} $

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<footer style = "font-size: 18px">Definite integral lecture-2 $\rightarrow$ Method of anti derivatives $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Definite integral $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- $\underline{{\text{Ex:}}}$

- $ I=\int_0^{\pi / 4}  {2 \tan x\left\(\sec ^2 x d x) .\right.} $

- $ {\tan x=t} $

- $ \sec ^2 x d x=d t $   
- $x=0, tan0=0=1 \\\\ x=\frac{\pi}{4}, \tan \frac{\pi}{4}=1=t $

- $ I=\int_{t=0}^1 2 t d t $

- $ =\left.2 \frac{t^2}{2}\right|_0 ^1 $

- $ =1-0=1 $

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Definite integral $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Method of substitution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- $\underline{{\text{Ex:}}}$

- $ \int_0^2 \frac{x^3 d x}{\sqrt{x^4+9}}$

- $ \sqrt{\left(x^4+9\right)}=t $

- $ \frac{1}{2}(\left(x^4+9\right)^{-1 / 2} 4 x^3 d x=d t $

- $ \frac{ x^3 d x}{\sqrt{x^4+9}}=\frac{d t}{2} $

- $ x=0 \quad t=\sqrt{9}=3 $

- $ x=2 \quad t=\sqrt{25}=5 $

- $=\int_3^5 \frac{d t}{2}=\frac{1}{2}(5-3)=1 $

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Properties of definite integrals </primary>
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- $\underline{\text{Property 2:}}$

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

- in particular $ \int_a^a f(x) d x=0 $

- $ \int_a^b f(x) d x=F(b)-F(a)=-[F(a)-F(b)] $

- $ =-\int_b^a f(x)dx $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Properties of definite integrals $\rightarrow$ <span style = "color:blue"> Properties of definite integrals </span> $\rightarrow$ Properties of definite integrals $\rightarrow$ Properties of definite integrals </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Properties of definite integrals </primary>
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- $\underline{\text{Property 4.}}$ 
 
- $ \int_a^b f(x) d x =\int_a^b f(a+b-x) d x $

- $ \int_a^b f(x) d x \quad x=a+b-t , dx = -dt $

- $ \biggl[ x = a , t = b \\\\ \quad x = b,  t = a  \biggl] $

- $ =\int_{t=b}^{t=a} f(a+b-t)(-d t) $

- $ =\int_a^b f(a+b-t) d t=\int_a^b f(a+b-x) d x $
	
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<footer style = "font-size: 18px">Properties of definite integrals $\rightarrow$ Properties of definite integrals $\rightarrow$ <span style = "color:blue"> Properties of definite integrals </span> $\rightarrow$ Properties of definite integrals $\rightarrow$ Properties of definite integrals </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Properties of definite integrals </primary>
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- $\underline{\text{Property 5.}}$ 
 
- $ \quad \int_0^a f(x) d x=\int_0^a f(a-x) d x $
 
- $ \int_0^a f(x) d x \quad x=a-t \quad x=0, t=a $

- $ [ \quad dx = -dt \quad x=a, t=0 ] $
 
- $ =\int_{t=a}^{t=0} f(a-t)-d t=\int_0^a f(a-t) d t $
 
- $ =\int_0^a f(a-x) d x $

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### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Properties of definite integrals </primary>
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- $\underline{\text{Property 6.}}$

- $ \int_0^{2a} f(x) dx = \int_0^a f(x) dx+ \int_a^{2a} f(x) dx \rightarrow (Property 3) $

- $ \int_0^{2 a} f(x) d x \quad x=2 a-t ,d x=-d t $

- $ =\int_0^0 f(2 a-t) \(-dt) $

- $ x=a, \rightarrow t=a $

- $ x=2 a \rightarrow t=0 $

- $ =\int_0^a f(2 a-k) d t =\int_0^a f(2 a-x) d x $

- $ \int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 a} f(x) d x $

- $ \int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x $

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### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Properties of definite integrals </primary>
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- $\underline{\text{Property 7.}}$

- $\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-2) d x \rightarrow 1 $

- $ \text { If }f(2 a-x)=f(x) \text { then } $

- $ \left.\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right] $

- $ \text { if } f(2 a-x)=-f(x) $

- $ \left.\int_0^{2 a} f(x) d x=0\right) $

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### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Property 8 proof </primary>
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- $ \int_{-a}^a f(x) d x  =\int_0^a f(-t) d t+\int_0^a f(t) d t $
 
- $\int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x $

- If $ f(x)$ is even, i.e., $f(-x)=f(x) $

- $ \int_{-a}^a f(x) d x=2 \int_{-a}^a f(x) d x $

- $ \text { If } f(-x)=-f(x) $

- $ \int_{-a}^a f(x) d x=0 $

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<footer style = "font-size: 18px">Properties of definite integrals $\rightarrow$ Property 8 proof  $\rightarrow$ <span style = "color:blue">Property 8 proof </span> $\rightarrow$ Property 8 proof $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- $ \int_0^4|x-2| d x $

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- $ = \int_0^2-(x-2) d x+\int_0^4(x-2) d x $

- $ = -\frac{x^2}{2}+\left.2 x\right|_0 ^2+\frac{x^2}{2}-\left.2 x\right|_2 ^4 $

- $ = -\frac{4}{2}+4-0+\frac{16}{2}-8-\frac{4}{2}+4 $

- = -2 + 4 - 0 + 8 - 8 - 2 + 4 = 4

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<footer style = "font-size: 18px">Property 8 proof $\rightarrow$ Property 8 proof $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x \quad f(2 a-x)=f(x) $

- $ I=\frac{1}{2} \times 2 \int_0^{\pi / 2} \log \sin (\pi-t) d t-\frac{\pi}{2} \log 2 $

- $ I=\int_0^{\pi / 2} \log \sin t d t-\frac{\pi}{2} \operatorname{log} $

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<footer style = "font-size: 18px">Property 8 proof $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-9 $\rightarrow$ Problem-10 </footer>

### <Success style="font-size:25px"> Definite Integral L-2</Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ I =\int_0^\pi \frac{xd x}{1+\sin x} =\int_0^\pi \frac{\pi-x}{1+\sin(\pi-x)} d x $

- $ [{\int_0^a f(x) d x}] = [{\int_0^a f(a-x) dx }] $

- $ =\int_0^\pi \frac{(\pi-x) d x}{1+\sin x} ,\quad 2 I=\int_0^\pi \frac{\pi d x}{1+\sin x} $

- $ 2 I=\pi \int_0^\pi \frac{d x}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}=\pi \int_0^\pi \frac{\sec ^2 \frac{x}{2} d x}{\left(1+\tan \frac{x}{2}\right)^2} $

- Let $\tan \frac{x}{2}=t \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \quad{2 I=2 \pi} $

- $2 I=\pi \int_0^{\infty} \frac{2 d t}{(1+t)^2}=\left.\pi \frac{(-2)}{1+t}\right|_0 ^{\infty}=-2 \pi(0-1)$

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<footer style = "font-size: 18px">Problem-9 $\rightarrow$ Problem-10 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>