Answer to Question #199573 in Quantitative Methods for VPM

Question #199573

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

Y = √π‘₯ + 1π‘₯

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.


1
Expert's answer
2021-05-28T10:05:50-0400

∫16xx+1dx=(x+1)3/2(3xβˆ’2)15∣16=\displaystyle{\int^{6}_{1}}x\sqrt{x+1}dx=\frac{(x+1)^{3/2}(3x-2)}{15}|^6_1=

=(6+1)3/2(3β‹…6βˆ’2)15βˆ’(1+1)3/2(3βˆ’2)15=39.13=\frac{(6+1)^{3/2}(3\cdot6-2)}{15}-\frac{(1+1)^{3/2}(3-2)}{15}=39.13


Trapezium rule:

∫x0xnf(x)dx=12h[y0+yn+2(y1+y2+...yn)]\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=Ξ”x=1 sech=\Delta x=1\ sec , then:

y0=f(1)=2=1.41y_0=f(1)=\sqrt{2}=1.41

y1=f(2)=23=3.46y_1=f(2)=2\sqrt{3}=3.46

y2=f(3)=6y_2=f(3)=6

y3=f(4)=45=8.94y_3=f(4)=4\sqrt{5}=8.94

y4=f(5)=56=12.25y_4=f(5)=5\sqrt{6}=12.25

y5=f(6)=67=15.87y_5=f(6)=6\sqrt{7}=15.87

∫16xx+1dx=12[1.41+15.87+2(3.46+6+8.94+12.25)]=39.29\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:

Ξ΅=39.29βˆ’39.13=0.16\varepsilon=39.29-39.13=0.16


Simpson’s rule:

∫x0xnf(x)dx=13h[y0+4y1+2y2+4y3+2y4+...+4ynβˆ’1+yn]\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=0.5h=0.5

y0=f(1)=1.41y_0=f(1)=1.41

y1=f(1.5)=1.52.5=2.37y_1=f(1.5)=1.5\sqrt{2.5}=2.37

y2=f(2)=3.46y_2=f(2)=3.46

y3=f(2.5)=2.53.5=4.68y_3=f(2.5)=2.5\sqrt{3.5}=4.68

y4=f(3)=6y_4=f(3)=6

y5=f(3.5)=3.54.5=7.42y_5=f(3.5)=3.5\sqrt{4.5}=7.42

y6=f(4)=8.94y_6=f(4)=8.94

y7=f(4.5)=4.55.5=10.55y_7=f(4.5)=4.5\sqrt{5.5}=10.55

y8=f(5)=12.25y_8=f(5)=12.25

y9=f(5.5)=5.56.5=14.02y_9=f(5.5)=5.5\sqrt{6.5}=14.02

y10=f(6)=15.87y_{10}=f(6)=15.87

∫16xx+1dx=0.53(1.41+4β‹…2.37+2β‹…3.46+4β‹…4.68+2β‹…6+\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+

+4β‹…7.42+2β‹…8.94+4β‹…10.55+2β‹…12.25+4β‹…14.02+15.87)=39.12+4\cdot7.42+2\cdot8.94+4\cdot10.55+2\cdot12.25+4\cdot14.02+15.87)=39.12

Accuracy:

Ξ΅=39.13βˆ’39.12=0.01\varepsilon=39.13-39.12=0.01


Mid ordinate rule:

for n=5, h=1

∫16xx+1dx=h(y(1.5)+y(2.5)+y(3.5)+y(4.5)+y(5.5))=\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=2.27+4.68+7.42+10.55+14.02=38.94

Accuracy:

Ξ΅=39.13βˆ’38.94=0.16\varepsilon=39.13-38.94=0.16


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


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