1) Triangle

2) Rectangle

3) Circle

etc

$A_1+A_2+A_3+A_4 \\ =$ Required Area.

$\underline{\text{Ex 1:}}$ Area bounded between a line & a curve

$y=1, y={x^{2}}$

$\underline{\text{Ex 2:}}$ Area bounded between two curves

$y=x^2$ and $y^2=x$

$\underline{\text{Ex 3:}}$ Area bounded between 3-Curves

$y=\sqrt{2 x}$

$y=\sqrt{2 x-x^2}$

$x=2$

Here $y=\sqrt{2 x-x^2}$ is equivalent to circle

$(x-1)^2+y^2=1$

$\underline{\text{Ex 4:}}$

Area bounded between 4-Curves

$y^2=4x, \quad y^2=16x$

$y=4x^2, \quad y=16x^2$

$\underline{\text{Definite Integral:-}}$

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

$a=$ lower limit

$b=$ upper limit

Methods to compute Definite Integral (A):

Limit of finite sums

Anti Derivatives

Ex. Area bounded between

$y=0$,

$y=1+x^2$,

$x=0$ and $x=1$.

$A=\int_0^1 (1+x^2) d x$

$A>L_2=R_1+R_2$

$=\frac{1}{2} \times(1+0)+\frac{1}{2} \times\left(1+\frac{1}{4}\right)$

$=\frac{1}{2}\left[1+1+\frac{1}{4}\right]\\=1+\frac{1}{8}=\frac{9}{8}=1.125$

$U_2 =\bar{R}_1+\bar{R}_2$

$=\frac{1}{2} \times\left(1+\frac{1}{4}\right)+\frac{1}{2} \\ \times(1+1)$

$=\frac{1}{2}\left(2+\frac{5}{4}\right)$

$=1+\frac{5}{8}=\frac{13}{8}=1.625$

$L_2= 1 \cdot 125$

$U_2=\text { Upper Sum }$

$L_2 =\text { Lower sum }$

$L_2 <\mathrm{A}<U_2$

Lower sum

$L_4 =\frac{1}{4}(1+0)$

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

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

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

Therefore

$L_4 =\frac{1}{4}\left[4+\frac{1+4+9}{16}\right]$

$=1+\frac{14}{{4} \times 16}=\frac{32+7}{32}\\=\frac{39}{32}=1.218$

$L_2 =1.125$

$L_2 <L_4<A$

$A<U _4 <U_2$

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

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

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

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

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

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

$U_2= 1.625$

$L_2<L_4<A<U_4\\<U_2$

$L_2<L_4<\ldots<L_n<A<U_n<\ldots<U_4<U_2$

upper sum decreases

lower sum increases

$\lim _{n \rightarrow \infty} L_n =A\\=\lim _{n \rightarrow \infty} U_n$

$A =\int_0^1\left(1+x^2\right) d x$

Ex. Let $f(x)$ be a continous function on $[a, b]$.

$f(x)>0$

$f(x)$ is increasing.

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

$x_0=a$

$x_n=b$

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

$x_k=x_0+hk$

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$

$U_n=h f\left(x_1\right)+h f\left(x_2\right)+\cdots+h f\left(x_n\right)$

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

$L_n =\sum_{k=0}^{n-1} f\left(x_k\right) h,$

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

$L_n <A<U_n$

$\underset{n \rightarrow \infty} {\lim} L_ n = \underset{n \rightarrow \infty}{\lim} \sum_{k=0}^{n-1} f(x_k) h$

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

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

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

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,

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

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

$x_k=\frac{k}{n}$

$L_n =h\left[\underbrace{1+1+\cdots+1}_{n \ \text{times}}+\frac{1^2+2^2+\cdot+(n-1)^2}{n^2}\right]$

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

$\{\because h=\frac{1}{n}\}$

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

$\quad$

Ex: Area between $y=e^{-x}, y=0, x=0$ and $x=1$.

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

$x_k=x_0+k h$

$x_k=0+\frac{k}{n}$

$x_1=\frac{1}{n}, x_2=\frac{2}{n} \ldots$

$L_n=hf\left(x_1\right)+hf\left(x_2\right)+\cdots+h f\left(x_n\right)$

$L_n=h\left[e^{-\frac{1}{n}}+e^{-\frac{2}{n}}+\cdots+e^{-1}\right]$

$L_n=\frac{h}{e^n-1}\left(1-e^{-1}\right)$ $\quad$

$\biggl\{\frac{1}{n}=h \\ n\rightarrow \infty \ \text{then} \ h\rightarrow 0 \biggl\}$

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

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

$f(x)>0$

Area Function:

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

If $f(x)$ is continuous on $[a, b]$ and area function is defined as

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

then $A^{\prime}(x)=f(x)$.

Let $f(x)$ be a continuous function on $[a, b]$ and $F(x)$ is an anti derivative of $f(x)$, i.e.,

$F^{\prime}(x)=f(x)$ then

$\int_a^b f(x) d x=\left.F(x)\right|_{x=a} ^{x=b}=F(b)-F(a)$

$\text{Show} ,\int_0^1 {1+x^2} d x$ $= {\frac{4}{3}}=$ $\lim _{n \rightarrow \infty}$ $L_n$

$\int_0^1\left(1+x^2\right) d x$

=$F(x)|_{x=0} ^{x=1}$

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

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

$\Rightarrow \int_0^1 (1+x^2) d x$

$=x+\frac{x^3}{3}$ $|_{x=0} ^{x=1}$

$={1+\frac{1}{3} -0}=\frac{4}{3}$

Ex:

$\int_0^1 e^{-x} d x =1-e^{-1}$

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

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

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

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

$=-e^{-1}+1$

$=1-e^{-1} .$