Answer to Question #199573 in Quantitative Methods for VPM

Question

4. The equation for the velocity of a car is given as

Y = √𝑥 + 1𝑥

Use the trapezium rule, Simpson’s rule and the mid ordinate rule ( with a suitable number of intervals ) to find the distance travelled between 1 and 6 seconds.

Compare your answers to the value given by the definite integral and comment on the accuracy of the numerical methods.

Answer 1

$\displaystyle{\int^{6}_{1}}x\sqrt{x+1}dx=\frac{(x+1)^{3/2}(3x-2)}{15}|^6_1=$

$=\frac{(6+1)^{3/2}(3\cdot6-2)}{15}-\frac{(1+1)^{3/2}(3-2)}{15}=39.13$

Trapezium rule:

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

Let $h=\Delta x=1\ sec$ , then:

$y_0=f(1)=\sqrt{2}=1.41$

$y_1=f(2)=2\sqrt{3}=3.46$

$y_2=f(3)=6$

$y_3=f(4)=4\sqrt{5}=8.94$

$y_4=f(5)=5\sqrt{6}=12.25$

$y_5=f(6)=6\sqrt{7}=15.87$

$\displaystyle{\int^{6}_{1}}x\sqrt{x+1}dx=\frac{1}{2}[1.41+15.87+2(3.46+6+8.94+12.25)]=39.29$

Accuracy:

$\varepsilon=39.29-39.13=0.16$

Simpson’s rule:

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

In Simpson's Rule, n must be even.

Let n=10, then:

$h=0.5$

$y_0=f(1)=1.41$

$y_1=f(1.5)=1.5\sqrt{2.5}=2.37$

$y_2=f(2)=3.46$

$y_3=f(2.5)=2.5\sqrt{3.5}=4.68$

$y_4=f(3)=6$

$y_5=f(3.5)=3.5\sqrt{4.5}=7.42$

$y_6=f(4)=8.94$

$y_7=f(4.5)=4.5\sqrt{5.5}=10.55$

$y_8=f(5)=12.25$

$y_9=f(5.5)=5.5\sqrt{6.5}=14.02$

$y_{10}=f(6)=15.87$

$\displaystyle{\int^{6}_{1}}x\sqrt{x+1}dx=\frac{0.5}{3}(1.41+4\cdot2.37+2\cdot3.46+4\cdot4.68+2\cdot6+$

$+4\cdot7.42+2\cdot8.94+4\cdot10.55+2\cdot12.25+4\cdot14.02+15.87)=39.12$

Accuracy:

$\varepsilon=39.13-39.12=0.01$

Mid ordinate rule:

for n=5, h=1

$\displaystyle{\int^{6}_{1}}x\sqrt{x+1}dx=h(y(1.5)+y(2.5)+y(3.5)+y(4.5)+y(5.5))=$

$=2.27+4.68+7.42+10.55+14.02=38.94$

Accuracy:

$\varepsilon=39.13-38.94=0.16$

So, in this case, the best accuracy is for Simpson’s rule.