Consider a general region bounded by \(y=f(x)\) in the interval \([a,b]\) .

• Divide the length \((b-a)\) into \(n\) equal parts, each equal to \(\Delta x = \dfrac{b-a}{n}\)

\(x_1 =a+\Delta x\)

\(x_2 = a+2\,\Delta\,x\)

\(\vdots\)

\(x_{n-1}\;=a+(n-1)\,\Delta \,x\)

As  \(n\) increases these rectangles become thinner and thinner and approximation  become more and more accurate.

• When  \(x \to\infty\)  this becomes exact area.

        \(\therefore A= \lim\limits _{x\to\infty} R_n = \lim\limits _{x\to\infty} L_n \)  both limits will take the same value .

 \(A= \lim\limits _{x\to\infty} R_n\, =\lim\limits _{x\to\infty}\underbrace {\left(\dfrac{b-a}{n}\right)}_{\text {Width of each rectangle}}\left[\underbrace {f(x_1)}_{\text {Height of 1st rectangle}}+f(x_2)+.....f(x_n)\right]\)

\(=\lim\limits_{x\to\infty}(\Delta\,x)\,\sum\limits_{i=1}^n\,f(x_i) \to\Delta\,x= \dfrac{b-a}{n}, \,x_i = a+i\,\Delta\,x \ and \;\sum\) means the value of \(i\)varies from 1 to \(n \) .

\(A=\lim\limits_{x\to\infty}L_n=\lim\limits_{x\to\infty}\left( \dfrac{b-a}{n}\right)[f(a)+f(x_1)+.....f(x_{n-1})]\)

\(= \lim\limits_{x\to\infty}\,(\Delta\,x)\sum\limits^n_{i=1} f(x_{i-1})\to (a=x_0)\)

\(\Rightarrow A=\lim\limits_{x\to\infty}\,\sum\limits_{i=1}^n f(x_{i}) \,\Delta\,x = \lim\limits_{x\to\infty} \,\,\sum \limits ^n_{i=1} f(x_{i-1})\,\Delta \,x\)

\(\therefore\) Area = \(\lim\limits_{n\to \infty}\,\sum\limits_{i=1}^{n}\, \Big ( f(a+i\,\Delta x) \;\Delta x\Big) \)

Area bounded by \(y=f(x)\) and the lines \(x=a\) and \(x=b\), where \(\Delta x=\dfrac {b-a}{n}\).

• Consider the velocity time graph of a particle as shown.

• If we break the time interval [a, b] into many equal parts (say n) each measuring \(\dfrac {b-a}{n}=\Delta t\), and draw rectangles similar to area problem, we see that area of each rectangle will be the distance covered by particle in that small time interval..
• Let \(v=f(t)\), \((a\leq t \leq b)\)   \([\,f(t)>0\,]\) .
• Distance = \(f(t_0)\Delta t+f(t_1)\Delta t+\dots f(t_{n-1})\Delta t\)\(=\sum\limits_{i=1}^{n-1}\,f(t_{i-1})\Delta t\) (If we take the left end point for height of rectangle)
• Distance = \(f(t_1)\Delta t+f(t_2)\Delta t+\dots f(t_{n})\Delta t\)\(=\sum\limits_{i=1}^{n}\,f(t_{i})\Delta t\) (If we take the right end point for height of rectangle)

Instead of using Left end points \((L_n)\) or Right end points \((R_n)\) for approximating area,

we can use height of any point \(x_i^*\) which lies between \(x_{i-1}\) and \(x_i\).

\(x_1^*\to\) between \(x_0\) and \(x_1\) (or \(a\) and \(x_1\))

\(x_2^*\to\) between \(x_1\) and \(x_2\) 

\(\vdots\)

\(x_n^*\to\) between \(x_{n-1}\) and \(b\)

• Therefore, general expression for area \(A=\lim\limits_{x\to\infty}\;f(x_1^*)\,\Delta x+ f(x_2^*)\,\Delta x+\dots f(x_n^*)\,\Delta x\)

              \(=\lim\limits_{x\to\infty}\; \sum\limits_{i=1}^n\,f(x_i^*) \Delta x\)

• If we choose \(x_i^*\) to be the maximum height in each respective sub interval, we get the upper sum or the upper estimate of the area.
• If we choose \(x_i^*\) to be the minimum height in each respective sub interval we get the lower sum or the lower estimate of the area.

Tall rectangles \(\to\)upper sum.

For constant velocity motion, the distance traveled by a particle is given by 

Distance = Speed × time

D = v × t

• If 'v' varies with time, then estimate the distance by taking suitable sub intervals of time and assuming that velocity is constant in these intervals.