mathematics-class-12-unit-07-chapter-04-definite-integral-l-4-7-wgstyjvesq0

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\quad y=b \sqrt{1-\frac{x^2}{a^2}}=\frac{b}{a} \sqrt{a^2-x^2}$
 
- $=2 \int_{a e}^a \frac{b}{a} \sqrt{a^2-x^2} d x \\  =\frac{2 b}{a} \int_{a e}^a \sqrt{a^2-x^2} d x$
  
- $ =\frac{2 b}{a}\left[\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1} \frac{x}{a}\right]_{a e}^{a}$ 
  
- $ =\frac{2 b}{a}\left[0+\frac{1}{2} a^2 \sin ^{-1} 1-\frac{1}{2} a e \sqrt{a^2-a^2    e^2}-\frac{1}{2} a^2 \sin ^{-1} e\right]$ 
  
- $  =\frac{2 b}{a}\left[\frac{\pi a^2}{4}-\frac{1}{2} a e b-\frac{1}{2} a^2  \sin^{-1} c\right]$

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<footer style = "font-size: 18px">Definite integral lecture-4 $\rightarrow$  Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Area bounded between two curves $\rightarrow$ Area bounded between two curves </footer>

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Area bounded between two curves </primary>
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- Case $2 \quad f(x) \geqslant g(x), x \in[0, c]$

- $g(x) \geqslant f(x), \quad x \in[c, b]$

- Required Area

- $\int_ a^c[f(x)-g(x)] d x$ + $\int_c^b[g(x)-f(x)] d x$

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

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\begin{aligned}& \int_{x=0}^{x=\frac{1}{2}}\left(\sqrt{2 x}-4 x^2\right) d x \end{aligned}$

- $\begin{aligned}& =\left.\left.\sqrt{2} \frac{x^{3 / 2}}{3 / 2}\right|_0 ^{\frac{1}{2}} 4 \frac{x^3}{3}\right|_0 ^{1 / 2} \end{aligned}$
- $\begin{aligned}& =\frac{2 \sqrt{2}}{3} \frac{1}{2} \frac{1}{\sqrt{2}}-\frac{4}{3} \times \frac{1}{8} \end{aligned}$
- $\begin{aligned}& =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\end{aligned}$
- $\begin{array}{r}y^2=2 x, y=4 x^2 \\ 16 x^4=2 x\left|\begin{array}{l}x=0 \\ x\left(8 x^3-x\right)=0\end{array}\right| x=\frac{1}{2}\end{array}$

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<footer style = "font-size: 18px">Area bounded between two curves $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> solution </primary>
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- $ =\left[\frac{1}{2} t \sqrt{16-t^2}+\frac{1}{2} 16 {\sin}^{-1} \frac{t}{4}\right]^0_{-4}$
 
 - $- 2\frac{x^{3 / 2}}{3 / 2}]_0^4 $
 
 - $=0+0-0-8 \sin ^{-1}(-1)-\frac{4}{3} \times(4)^{3 / 2}+0$
 - $=4 \pi-\frac{4}{3} \times 8=4 \pi-\frac{32}{3} . \\  \text { Req in area }=8 \pi-64 / 8 .$

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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"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\int_0^{\sqrt{3}}\left(2 \sqrt{4-y^2}-2\right) d y  =4$

- $\int_0^{\sqrt{3}}\left(\frac{\sqrt{4-y^2}}{2}-1\right) d y $.

- $ =4\left[\frac{1}{2} y \sqrt{4-y^2}+\frac{1}{2} 4 \sin ^{-1} \frac{y}{2}-y\right]_0^{\sqrt{3}}$

- $  =4 \int_0^{\sqrt{3}}\left({\sqrt{4-y^2}}^2-1\right) d y .$

- $=  4\left[\frac{1}{2} y \sqrt{4-y^2}+\frac{1}{2} 4 \sin ^{-1} \frac{y}{2}-y\right]_0^{\sqrt{3}}$

- $= 4\left[\frac{1}{2} \sqrt{3}+2 \sin ^{-1} \frac{\sqrt{3}}{2}-\sqrt{3}\right]$

- $=  4\left[2 \frac{\pi}{3}-\frac{1}{2} \sqrt{3}\right] \\ =  \frac{8 \pi}{3}-2 \sqrt{3}$
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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Required Area.

- $=\left|A_1\right|+A_2+\left|A_3\right|$ 
 
- $ A_1=\int_{-\pi / 2}^0 \sin x d x=-\left.\cos x\right|_{-\pi / 2} ^0=-1 $
 
- $ A_2=\int_0^{\pi / 2} \sin x d x=-\left.\cos x\right|_0 ^\pi=1-(-1)=2$ 
 
- $ A_3=\int_\pi^{3 \pi / 2} \sin x d x=-\left.\cos x\right|_\pi ^{3 \pi / 2}=-$

- Required Area $=|-1|+2+|-1|=4$.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Definite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- Ex. Find out area bounded by $|x|+|y|=1$.

- $ \int_0^1(1-x-(x-1)) d x $
 
- $ +\int_{-1}^0(1+x-(-x-2)) d x=1, \quad x-y=1,-x-y=1 .$ 
 
- $ .=\int_0^1 2(1-x) d x+\int_{-1}^0 2(1+x)$

- $d x=\left.2 \frac{(1-x)^2}{-2}\right|_0$
 
- $ ^1+\frac{(1+x)^2}{2}]_1^0$
 
- $ =-2 \times\left(-\frac{1}{2}\right)+2 \times \frac{1}{2}=1+1=2$
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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Thankyou </footer>