Question

The force F on a beam of length 5m is given by

𝐹= ∫⁵₀ 𝑒𝑥/1+ √𝑥

Find the value of F using the trapezium rule, Simpsons rule and the mid ordinate rule with a suitable number of intervals.

Answer 1

$F= \int_0^5 ex +\sqrt{x}$

Trapezium rule:

$\displaystyle{\int^{x_n}_{x_0}}f(x)dx=\frac{1}{2}h[y_0+y_n+2(y_1+y_2+...y_n)]$

Value of e = 2.718

Now:

$y_0=f(0)=0$

$y_1=f(1)=2.718$

$y_2=f(2)=6.85$

$y_3=f(3)=9.88$

$y_4=f(4)=12.87$

$y_5=f(5)=15.82$

Let the value of h = 1 ,then:

$\int_0^5ex+ \sqrt{x} =\dfrac{1}{2} [0+15.82+2(2.718+6.85+9.88+12.87)]$

$\int_0^5ex+ \sqrt{x} =40.228$

Simpson’s rule:

$\displaystyle{\int^{x_n}_{x_0}}f(x)dx=\frac{1}{3}h[y_0+4y_1+2y_2+4y_3+2y_4+...+4y_{n-1}+y_n]$

Let h = 0.5 ,as n is even here so n = 10

$y_0=f(0)= 0$

$y_1= f(0.5)= 2.066$

$y_2= f(1)=2.718$

$y_3= f(1.5)= 5.30$

$y_4= f(2)=$ $6.85$

$y_5= f(2.5)=$$8.37$

$y_6= f(3)= 9.88$

$y_7= f(3.5)= 11.38$

$y_8= f(4)= 12.87$

$y_9= f(4.5)= 14.35$

$y_{10}= f(5)= 15.82$

Putting the values in above formula:

$\int_0^5 ex+\sqrt{x}= \dfrac{0.5}{3}[0+4(2.066)+2(2.718)+4(5.30)+2(6.85)$

$4(8.317)+2(9.88)+4(11.38)+2(12.87)+4(14.35)+15.82$

$= 41.018$

Mid ordinate rule:

for n=5, h=1

$\int_0^5 ex+ \sqrt{x} = h[y(0.5)+y(1.5)+y(2.5)+y(3.5)+y(4.5)]$

$=$ $1[2.066+5.30+8.37+11.38+14.35]$

$=$ $41.466$