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Errors In Approximate Integration And Tabular Form

Find the approximation value of Integral in which function is given in tabular form method. Practice error bound for midpoint & trapezoidal rule & simpson rule error.

Error Bound For Midpoint Rule:

  • Let \(E_M\) be the error while we calculate the value of integral using Midpoint Rule.
  • Midpoint Rule says that,

 In this method of approximation, we write

  • \(\int\limits_a^bf(x)dx\simeq M_{n}=\Delta x \Bigg[f(\overline x_1)+f(\overline x_2) + .........f(\overline x_n)\Bigg]\)

    where, \(\Delta\,x=\dfrac{b-a} {n}\)and  \(\overline x_i=\dfrac{x_{i–1}+x_i}{2}\rightarrow\) midpoint of \([x_{i–1},x_i]\)

    Interval [a,b] is divided into n equal parts each measuring \(\dfrac{b-a}{n}\)=\(\Delta u\) and then height  \(f(\overline x)\) is taken of the midpoint of each interval. 

 

  • Let  \(|f"(x)|\leq k\) for \(a\leq x\leq b\) then,

    \(|E_M| \leq \dfrac{k(b-a)^3}{24\,n^2}\)

  •   The R.H.S value in this inequality is the maximum error that can creep in 
  • In general, the actual error is substantiality less than maximum error.

  • Let  \(|f"(x)|\leq k\) for \(a\leq x\leq b\) then,

    \(|E_M| \leq \dfrac{k(b-a)^3}{24\,n^2}\)

  •   The R.H.S value in this inequality is the maximum error that can creep in 
  • In general, the actual error is substantiality less than maximum error.

Illustration Questions

Estimate the error while calculating M8 for the integral \(I = \int\limits ^4_1 \dfrac{1}{\sqrt x} dx\) .

A \((E_M) \leq.2652\)

B \((E_M) \leq 0.01318\)

C \((E_M) \leq 0.00526\)

D \((E_M) \geq 1.326\)

×

For the Midpoint Rule 

\(|E_M| = Error \leq \dfrac{k(b-a)^3}{24\,n^2}\)    where,  \(|f"(x)|\leq k\) for \(a\leq x\leq b\)

For the integral \(I = \int\limits^b_a f(x) dx\) and \(\Delta x = \dfrac{b-a}{n}\)

 

In this case,

\(f(x) = \dfrac{1}{\sqrt x} = x^{-{1}/{2}}\)

\(= f'(x) = -\dfrac{1}{2} \, x ^{-{3}/{2}}\)

\(\Rightarrow f'' (x) = -\dfrac{1}{2} × -\dfrac{3}{2}× x^{-{5}/{2}} = \dfrac{3}{4}\,x^{-{5}/{2}}\)

a= 1, b=4  

 The maximum value of \(f"(x)\)  in [1,4] is \(\dfrac{3}{4} \)

\(\Rightarrow k= \dfrac{3}{4}\)

\(\therefore \, |E_M| \leq \,\dfrac{3}{4} \,\dfrac{(4-1)^3}{24× 64} \\= \dfrac{3}{4}× \dfrac{27}{24× 64}\\ = 0.01318\)

\(\therefore\, |E_M| \leq \, 0.01318\)

Estimate the error while calculating M8 for the integral \(I = \int\limits ^4_1 \dfrac{1}{\sqrt x} dx\) .

A

\((E_M) \leq.2652\)

.

B

\((E_M) \leq 0.01318\)

C

\((E_M) \leq 0.00526\)

D

\((E_M) \geq 1.326\)

Option B is Correct

Error Bounds For Trapezoidal Rule 

  • Let ET be the error while we calculate the value of integral using Trapezoidal  Rule.
  • Trapezoidal  Rule says that,  

In this rule of approximation ,

\(\int\limits_a^b\,f(x)\,dx \simeq T_n = \dfrac{\Delta x}{2} \Bigg[f(x_0) + 2f(x_1) + 2f(x_2) + .......+ 2f(x_{n–1}) + f(x_n)\Bigg]\)

where, \(\Delta x = \dfrac {b-a}{n},x_i=a+i\Delta x\) 

 

Each shaded portion is a trapezoidal and we add all the trapezium areas.

For n=5,

Shaded area = \(=\dfrac{1}{2}\bigg(f(x_0) + f(x_1)\bigg) \Delta x + \dfrac {1}{2} \bigg(f(x_1) + f(x_2)\bigg) \Delta x + ..........+ \dfrac {1}{2} \bigg(f(x_n) + f(b)\bigg)\Delta x\)

\(=\dfrac{\Delta x}{2}\Bigg[\underbrace {f(x_0)}_{f(a)} + 2f(x_1) + ..............+ 2f(x_4) + \underbrace {f(x_5)}_{f(b)}\Bigg]\)

  • Let \(|f''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b\,\,\text{then}\) 

\(|E_T| \leq \dfrac{k(b-a)^3}{12\,n^2}\)  

 

  • The R.H.S. value in the inequality  is the maximum value of error  that can creep in .

  • In general the actual error is substantially  less than the maximum error. 

Illustration Questions

Estimate the error ,while calculating T8 for the integral \(I = \int\limits ^4_1 \dfrac{1}{x^2} dx\).

A \(|E_T| \leq 0.0156\)

B \(|E_T| \leq 0.2109\)

C \(|E_T| \leq 0.258\)

D \(|E_T| \leq 5.16\)

×

For the Trapezoidal Rule,

\(|E_T| = Error \leq \dfrac{k(b-a)^3}{12\,n^2}\)

where, \(|f''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b\)

For the integral , \(I=\int\limits^b_a f(x) dx\)  and \(\Delta x = \dfrac{b-a}{n}\)

 

In this case, \(f(x) = \dfrac{1}{x^2} \)     

\(= f'(x) = -2x^{-3}\)

\(\Rightarrow f''(x) = 6x^{-4} = \dfrac{6}{x^4}\)

a=1 , b=4  

\(\rightarrow\) maximum value of \(f''(x) \) in [1,4] is 6.

\(\Rightarrow k = 6\)

n=8,

\(\therefore |E_T| \leq \dfrac{6× (4-1)^3}{12× 64} \\= \dfrac{6× 3^3}{12× 64}\\ = 0.2109\)

\(\therefore |E_T| \leq 0.2109\)

Estimate the error ,while calculating T8 for the integral \(I = \int\limits ^4_1 \dfrac{1}{x^2} dx\).

A

\(|E_T| \leq 0.0156\)

.

B

\(|E_T| \leq 0.2109\)

C

\(|E_T| \leq 0.258\)

D

\(|E_T| \leq 5.16\)

Option B is Correct

Error Bounds For Simpsons Rule

  • Let ES be the error while calculating the value of integral using  Simpson Rule.
  • Simpson Rule says that, 

Simpsons rule for approximating integrals make use of parabolas instead of straight line for approximating curves we divide the intervals [a,b] into n equal parts,

each part \(=\Delta x=\dfrac{b-a}{n}\) where n is even, then for each pair of intervals, we approximate y = f(x) >0 by a parabola as shown in the figures. 

We take the interval \([x_0, x_2]\) where, \(x_0=-h, x_1=0, x_2=h\)the equation of parabola through P0, P1 & P2 is of the form y = Ax2+Bx +C  (where A,B,C are constants)

The area = \(\int\limits_{-h}^h(Ax^2 + Bx + C)dx=\dfrac {h} {3}(2Ah^2 + 6C)\)      .............(1)

Now the parabola passes through \(P_0(-h,y_0), P_1(0,y_1), P_2 (h,y_2)\)

\(\therefore\)\(y_0=Ah^2 - Bh + C\\y_1=C\\y_2=Ah^2 + Bh + C\)           (\(\therefore y_0 + 4y_1 + y_2=2Ah^2+6C\))

put this in 1 we get,

Area of parabola = \(\dfrac{h}{3}(y_0+4y_1+y_2)\)

Similarly, area under the parabola \(P_2, P_3,P_4\) will be \(\dfrac{h}{3}(y_2 + 4{y_3} + y_4)\)

\(\therefore\)Sum of all areas = \(\dfrac{h}{3}\Bigg[y_0+4y_1+y_0\Bigg] + \dfrac{h}{3}\Bigg[y_2+4y_3+y_4\Bigg] + \dfrac{h}{3}\Bigg[y_4+4y_5+y_0\Bigg] + .....+ \dfrac{h}{3}\Bigg[y_{n-2}+4y_{n-1}+y_n\Bigg]\)\(\dfrac{h}{3}\Bigg[y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + .............+2y_n-2 + 4y_{n-1} + y_n\Bigg]\)

\(\simeq\int\limits_a^bf(x)dx\)

\(\therefore \,\int\limits_a^bf(x)dx =S_n=\dfrac{\Delta x}{3}\Bigg[f(x_0) + 4f(x_1) + 2f(x_2) + ........+ 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)\Bigg]\)

where, n is even and \(\Delta x=\dfrac{b-a}{n}\)   

  • Let \(|f'' (x)| \leq k\), ( \(f''''(x)\) is the fourth derivation  of \(f\) w.r.t. \(x\))

    for \(a\leq x\leq b\) then,    

    \(|E_S| \leq \dfrac{k(b-a)^5}{180\,n^4}\)    

  • The R.H.S. value in this inequality is the maximum error that can creep in. 

  • In general the actual error is substantially less than maximum error.

 

Illustration Questions

Estimate the  error while calculating  S8 for the integral  \(I = \int\limits^2_1 \dfrac{1}{x^2} dx\)

A \(|E_S| \leq 0.000012\)

B \(|E_S| \leq 0.000163\)

C \(|E_S| \leq 0.23\)

D \(|E_S| \leq 1.1136\)

×

For the Simpsons Rule 

\(|E_S| = Error \leq \dfrac{k(b-a)^5}{180\, n^4}\)

where, \(|f''''(x)| \leq k\,\, \text{for}\,\,a\leq x\leq b\) 

For the integral,  \(I = \int\limits^b_a f(x) dx\) and  \(\Delta x= \dfrac{b-a}{n}\)

 

In this case,

 \(f(x) = \dfrac{1}{x^2}\) , \(\Rightarrow f'(x) = -2x^{-3}\) 

\(\Rightarrow f''(x) = 6\,x^{-4} \)

\(\Rightarrow f'''(x) = -24\,x^{-5}\)

\(\Rightarrow f'''' (x) = 120\, x^{-6}\)

a=1 , b=2  

 The maximum value of  \(f^4(x) = 120\).

\(\Rightarrow k=120\)

\(n=8\),

\(\therefore |E_S| \leq \dfrac{k(b-a)^5}{180\,n^4} \\= \dfrac{120× (2-1)^5}{180× 8^4} \\= 0.000163\)

\(= |E_S| \leq 0.000163\)

Estimate the  error while calculating  S8 for the integral  \(I = \int\limits^2_1 \dfrac{1}{x^2} dx\)

A

\(|E_S| \leq 0.000012\)

.

B

\(|E_S| \leq 0.000163\)

C

\(|E_S| \leq 0.23\)

D

\(|E_S| \leq 1.1136\)

Option B is Correct

Approximation of Integral in which Function is given in Tabular Form :

  • Sometimes to find the approximate value of \(I = \int\limits ^b_a f(x) dx\) we are not given the formula for \('f'\) but a table which mention some values of \(x \) and \(f(x)\) for those corresponding values.
  • We can use any of the three rules  Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
  • We have to choose \(n\)  according to the data given . 

Illustration Questions

Use Midpoint Rule and given data to estimate the value of  \(I = \int \limits^6_2 f(x) dx\). \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9      

A 102.58

B 1.63

C 37.30

D 15.2

×

Midpoint rule says that,

\(\int\limits_a^bf(x)dx\cong\,\Delta\,x\,\Bigg[f(\overline x_1)+f(\overline x_2) + .........f(\overline x_n)\Bigg]\)

                \(=M_{n}\)

where, \(\Delta\,x=\dfrac{b-a} {n}\) and  \({\over{x_i}}=\dfrac{1}{2}(x_{i–1}+x_i)\)

 

In this problem,

we take n= 4 ,a = 2 , b= 6,

\(\Delta x = \dfrac{6-2}{4} = 1\) , \(\bar x_1 = \dfrac{2+3}{2} = \dfrac{5}{2}\)\(\bar x_2 = \dfrac{3+4}{2} = \dfrac{7}{2}\),\(\bar x_3 = \dfrac{4+5}{2} = \dfrac{9}{2}\)\(\bar x_4 = \dfrac{5+6}{2} = \dfrac{11}{2}\)

 

\(\therefore \int\limits^6_2 f(x) dx \simeq M_4 = \Delta x\Bigg[ f(\bar x_1)+f(\bar x_2)+f(\bar x_3)+f(\bar x_4)\Bigg]\)

Now, read the values of function from the table given 

\(=1\Bigg[ f(2.5)+f(3.5)+f(4.5)+f(5.5)\Bigg]\)

\(= 1 \Bigg[6.3+8.1+10.6+12.3\Bigg]\)

\(= 37.3\)

Use Midpoint Rule and given data to estimate the value of  \(I = \int \limits^6_2 f(x) dx\). \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9      

A

102.58

.

B

1.63

C

37.30

D

15.2

Option C is Correct

Approximation of Integral in Which Function is Given in Tabular Form 

  • Sometimes to find the approximate value of \(I = \int\limits ^b_a f(x) dx\) we are given the formula for \('f'\) but a table which mentation same values of \(x \) and \(f(x)\) for those corresponding values.
  • We can use any of the three rules Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
  • We have to choose \(n\)  according to the data given . 

Illustration Questions

Use Trapezoidal Rule and give data to estimate the value of \(I= \int\limits^6_2 f(x) dx\) \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9      

A 15.82

B 36.85

C 1.93

D 105.62

×

By trapezoidal rule of approximation,

\(\int\limits_a^b\,f(x) \simeq \dfrac{\Delta x}{2} \Bigg[f(x_0) + 2f(x_1) + 2f(x_2) + .......+ 2f(x_{n–1}) + f(x_n)\Bigg]\)

\(=T_n\)

Where,  \(\Delta x = \dfrac {b-a}{n},x_i=a+i\Delta x\) 

In this case,

we take \(n= 4\) , a=2 , b=6

\(\Delta x= \dfrac{6-2}{4} = 1,\,x_0 = 2,\,x_1= 3, x_2= 4,x_3= 5,x_4=b= 6 \)

\(\therefore \, \int\limits^6_2 f(x) dx \simeq T_4 = \dfrac{\Delta x}{2}\Bigg[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)\Bigg]\)

\(= \dfrac{1}{2} \Bigg[f(2)+2f(3)+2f(4)+2f(5)+f(6)\Bigg]\)

(Read the values from table)

\(= \dfrac{1}{2} \Bigg[5.6+2× 7.2+2× 8.9+2× 11.2+13.5\Bigg]\)

\(= \dfrac{1}{2} \Bigg[73.7\Bigg] = 36.85\)

 

Use Trapezoidal Rule and give data to estimate the value of \(I= \int\limits^6_2 f(x) dx\) \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9      

A

15.82

.

B

36.85

C

1.93

D

105.62

Option B is Correct

Approximation of Integral in which Function is given in Tabular Form 

  • Sometimes to find the approximate value of \(I = \int\limits ^b_a f(x) dx\) we are given the formula for \('f'\) but a table which mentation same values of \(x \) and \(f(x)\) for those corresponding values.
  • We can use any of the three rules Midpoint Rule, Trapezoidal  Rule  or Simpson Rule to estimate the value of this integral.
  • We have to choose \(x\)  according to the data given . 

Illustration Questions

Use Simpson Rule and give data to estimate the value of \(I= \int\limits^6_2 f(x) dx\) \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9    

A 59.62

B 36.83

C 0.02

D 511.23

×

Simpsons Rule says that ,

\(\int\limits_a^bf(x)dx \simeq S_n=\dfrac{\Delta x}{3} \Bigg[f(x_0) + 4f(x_1) + 2f(x_2) + .......2f(x_{n-2})+4f(x_{n-1}) + f(x_n) \Bigg]\)

Where, \(\Delta x=\dfrac{b-a}{n}\) and n is even, \(x_i=a+i\Delta x\)

 

In this case,

we take n= 4 , a=2 , b=6,

\(\Delta x= \dfrac{6-2}{4}=1,\, x_0= 2, \,x_1= 3\,,\,x_2= 4\,,\,x_3= 5\,,\,x_4= b=6\)

\(\therefore \, \int\limits^6_2 f(x) dx \simeq S_4 = \dfrac{\Delta x}{3}\Bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)\Bigg]\)

\(= \dfrac{1}{3} \Bigg[f(2)+4\,f(3)+2\,f(4)+4\,f(5)+f(6)\Bigg]\)

(Read the values from table)

\(= \dfrac{1}{3} \Bigg[5.6+4× 7.2+2× 8.9+4× 11.2+13.5\Bigg]\)

\(= \dfrac{1}{3} \Bigg[5.6+ 28.8+17.8+44.8+13.5\Bigg]\)

\(= \dfrac{1}{3} \Bigg[110.5\Bigg]\\ = 36.83\)

 

Use Simpson Rule and give data to estimate the value of \(I= \int\limits^6_2 f(x) dx\) \(x\) \(f(x)\) \(x\) \(f(x)\) 2.0 5.6 4.5 10.6 2.5 6.3 5.0 11.2 3.0 7.2 5.5 12.3 3.5 8.1 6.0 13.5 4.0 8.9    

A

59.62

.

B

36.83

C

0.02

D

511.23

Option B is Correct

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