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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 \(\Delta x=\dfrac {b-a}{n}\).

Illustration Questions

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

Illustration Questions

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

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.

Illustration Questions

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

image
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.

Illustration Questions

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)

image

Distance traveled in Ist 10 sec \( = 10 × 4 = 40\, m\)

image

Assume velocity during \([10, 20]\) second interval is constant and equals 5 m/s (Upper)

image

Distance traveled in \([10, 20]=10×5=50\, m\)

image

\(\therefore\)  Total distance \(=10×4+10×5+10×6+10×8+10×10+10×13\)

\(=10[4+5+6+8+10+13]\)

\(=10×46=460\,m\)

image

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.

image
A

10 m

.

B

460 m

C

1000 m

D

5 km

Option B is Correct

Illustration Questions

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)

image

Distance traveled in Ist 10 sec\( = 10 × 0 = 0\, m\)

image

Assume velocity during \([10, 20]\) second interval is constant and equals 4 m/s.

image

Distance traveled in \([10, 20]=10×4=40\, m\)

image

\(\therefore\)  Total distance \(=10×0+10×4+10×5+10×6+10×8+10×10\)

\(=10[0+4+5+6+8+10]\)

\(=10×33=330\,m\)

image

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?

image
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)

Illustration Questions

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)

image
A

720 m

.

B

8 m

C

502 m

D

1001 m

Option A is Correct

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