Informative line

### Exact Calculation Of Area By Taking Limits

Calculate Exact Area Calculation by taking limits & find total distance traveled from a velocity vs time Graph.

# Exact Area Calculation by taking n Tends to Infinity ($$n\to \infty$$) in $$R_n\, \,\text {and} \,\,L_n$$

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 $$\Delta x=\dfrac {b-a}{n}$$.

#### The expression for the area under the graph of $$'f'$$ as a limit when  $$f(x) = \sqrt{2+7\,x}, \,\,1\leq x\leq 5$$ is

A   $$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\dfrac{1}{n} \sqrt{2+i\,n}$$

B   $$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\dfrac{4}{n} \sqrt{9+\dfrac{28\,i\,}{n}}$$

C   $$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\dfrac{3}{n} \sqrt{2+\dfrac{3\,i\,}{n}}$$

D   $$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\left(\dfrac{7}{n} \sqrt{1+\dfrac{\,i\,}{n}}\right)$$

×

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

$$= \lim\limits_{n\to\infty} \,\sum\limits_{i=1}^n f(1+i\,\Delta \,x)\,\Delta x$$

$$\Rightarrow\Delta \,x = \dfrac{b-a}{n}$$

$$= \dfrac{5-1}{n}$$

$$= \dfrac{4}{n}$$

$$= \lim\limits_{n\to\infty}\,\,\sum\limits_{i=1}^n\,f\left(1+\dfrac{4\,i}{n}\right)×\dfrac{4}{n}$$

$$= \lim\limits_{n\to\infty}\,\,\sum\limits_{i=1}^n\,\dfrac{4}{n}\sqrt{2+7\left({1+\dfrac{4\,i}{n}}\right)}$$

$$= \lim\limits_{n\to\infty}\,\,\sum\limits_{i=1}^n\,\left(\dfrac{4}{n}\sqrt{9+{\dfrac{28\,i}{n}}}\right)$$

### The expression for the area under the graph of $$'f'$$ as a limit when  $$f(x) = \sqrt{2+7\,x}, \,\,1\leq x\leq 5$$ is

A

$$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\dfrac{1}{n} \sqrt{2+i\,n}$$

.

B

$$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\dfrac{4}{n} \sqrt{9+\dfrac{28\,i\,}{n}}$$

C

$$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\dfrac{3}{n} \sqrt{2+\dfrac{3\,i\,}{n}}$$

D

$$\lim\limits_{n\to\infty}\,\sum\limits_{i=1}^n\,\left(\dfrac{7}{n} \sqrt{1+\dfrac{\,i\,}{n}}\right)$$

Option B is Correct

#### The region whose area is equal to $$\lim\limits_{n\to \infty}\,\sum\limits_{i=1}^{n}\,\left ( \dfrac {4}{n}\right) \left ( 2+\dfrac {4\,i}{n}\right)^2$$ is the

A Region bounded by $$y=x^4$$ between $$x=7$$ and $$x=11$$

B Region bounded by $$y=x^2$$ between $$x=2$$ and $$x=6$$

C Region bounded by $$y=sin\,x$$ between $$x=\dfrac {\pi}{4}$$ and $$x=\dfrac {\pi}{2}$$

D Region bounded by $$y=cos\,x$$ between $$x=0$$ and $$x=\pi$$

×

Compare

$$\lim\limits_{n\to \infty}\,\sum\limits_{i=1}^{n}\,\left ( \dfrac {4}{n}\right) \left ( 2+\dfrac {4\,i}{n}\right)^2 = \lim\limits_{n\to \infty}\,(\Delta x)\,f(a+i\,\Delta x)$$

$$\Delta x=\dfrac {4}{n}$$ and $$a=2$$$$f(x_i)=x_i^2$$

$$\Delta x=\dfrac {b-a}{n}$$

$$\Rightarrow b-a=4$$

$$\Rightarrow b-2=4$$

$$\Rightarrow b=6$$

$$f(x)=x^2$$

Area is of the region bounded by $$f(x)=x^2$$ between $$x=2$$ and $$x=6$$

### The region whose area is equal to $$\lim\limits_{n\to \infty}\,\sum\limits_{i=1}^{n}\,\left ( \dfrac {4}{n}\right) \left ( 2+\dfrac {4\,i}{n}\right)^2$$ is the

A

Region bounded by $$y=x^4$$ between $$x=7$$ and $$x=11$$

.

B

Region bounded by $$y=x^2$$ between $$x=2$$ and $$x=6$$

C

Region bounded by $$y=sin\,x$$ between $$x=\dfrac {\pi}{4}$$ and $$x=\dfrac {\pi}{2}$$

D

Region bounded by $$y=cos\,x$$ between $$x=0$$ and $$x=\pi$$

Option B is Correct

#### The following table shows the speed of a truck during the first minute of its start. Find the lower estimate of the distance traveled by the truck in one min?

A 5000 m

B 330 m

C 3 m

D 78 m

×

Assume that velocity during the first 10 seconds is constant and equals 0 m/s. (Lower) Distance traveled in Ist 10 sec$$= 10 × 0 = 0\, m$$ Assume velocity during $$[10, 20]$$ second interval is constant and equals 4 m/s. Distance traveled in $$[10, 20]=10×4=40\, m$$ $$\therefore$$  Total distance $$=10×0+10×4+10×5+10×6+10×8+10×10$$

$$=10[0+4+5+6+8+10]$$

$$=10×33=330\,m$$ ### The following table shows the speed of a truck during the first minute of its start. Find the lower estimate of the distance traveled by the truck in one min? A

5000 m

.

B

330 m

C

3 m

D

78 m

Option B is Correct

# Estimating the Distance traveled by Estimating the Area of Region under the Graph of Velocity v/s Time

• 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)

#### The velocity-time graph of a particle over a period of 2 minute is as shown. Estimate the distance traveled during 2 minutes by taking $$\Delta t=20\,sec$$. (Use upper estimate)

A 720 m

B 8 m

C 502 m

D 1001 m

×

For a decreasing function $$\rightarrow$$ lower estimate $$\Rightarrow$$ (right endpoints)

Distance $$=\sum\limits_{i=1}^{n}\,f(t_{i})\Delta t$$

$$=f(20)×20+f(40)×20+ f(60)×20+ f(80)×20+ f(100)×20+ f(120)×20$$

$$=20[3+4+5+6+8+10]$$

$$=20×36=720\,m$$

Distance traveled $$= 720\, m$$

### The velocity-time graph of a particle over a period of 2 minute is as shown. Estimate the distance traveled during 2 minutes by taking $$\Delta t=20\,sec$$. (Use upper estimate) A

720 m

.

B

8 m

C

502 m

D

1001 m

Option A is Correct

# Concept of Upper sum and Lower sum

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.  #### Consider the graph of $$'f'$$. If we are approximating  the area of region under graph of f by taking $$n=6$$ ($$x=2$$ to $$x=8$$) and desire to get upper sum, then

A $$x_5^*=6.2$$

B $$x_1^*=5.8$$

C $$x_2^*=9$$

D $$x_5^*=4$$

×

$$x_5^*$$ lies in $$[6, 7]\to$$fifth interval

It is the point at which height is greatest.

$$x_5^*=6.2=$$ greatest height point in $$[6,7]$$

### Consider the graph of $$'f'$$. If we are approximating  the area of region under graph of f by taking $$n=6$$ ($$x=2$$ to $$x=8$$) and desire to get upper sum, then A

$$x_5^*=6.2$$

.

B

$$x_1^*=5.8$$

C

$$x_2^*=9$$

D

$$x_5^*=4$$

Option A is Correct

# Calculation of Distance Traveled

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.

#### The following table shows the speed of a truck during the first minute of its start. Find the upper estimate of the distance traveled by the truck in one minute.

A 10 m

B 460 m

C 1000 m

D 5 km

×

Assume that velocity during the first 10 seconds is constant and equals 4 m/s (Upper) Distance traveled in Ist 10 sec $$= 10 × 4 = 40\, m$$ Assume velocity during $$[10, 20]$$ second interval is constant and equals 5 m/s (Upper) Distance traveled in $$[10, 20]=10×5=50\, m$$ $$\therefore$$  Total distance $$=10×4+10×5+10×6+10×8+10×10+10×13$$

$$=10[4+5+6+8+10+13]$$

$$=10×46=460\,m$$ ### The following table shows the speed of a truck during the first minute of its start. Find the upper estimate of the distance traveled by the truck in one minute. A

10 m

.

B

460 m

C

1000 m

D

5 km

Option B is Correct