The function plotted in the graph for Exercise 2a is given below. Find the exact
ID: 2851496 • Letter: T
Question
The function plotted in the graph for Exercise 2a is given below. Find the exact area for the same interval by integrating the function, and compare the results to Exercise 2a by finding their percent difference. F(x) = e^x - 5x^2 +10x A = 4 Integrity U (e^x - 5x^2 + 10dx) By now By now you should be convinced that the Trapezoidal Rule is a reasonable approximation to an exact integral. Next week's lab is a study of Newton s Second. will be acquiring data that must be integrated. Since the data will fit no known function the integrals must be done . The integration will be no more difficult than Exercise 2 of this assignment, but you should have a good understanding of the Trapezoidal rule before going to lab to avoid wasting valuable lab time.Explanation / Answer
[0 to 4] ex -5x2 +10x dx
= [0 to 4] ex -5x2+1/(2 +1) +10(x2/2)
= [0 to 4] ex -5x3/3 +5x2
= e4 -5(4)3/3 +5(4)2 - (e0 -5(0)3/3 +5(0)2)
= 26.93
Using trapezoidal rule:
let n = 8
==> x = (b -a)/n = (4 -0)/8 = 1/2 = 0.5
==> [0 to 4] ex -5x2 +10x dx = (0.5/2)[ f(0) + 2f(0.5) + 2f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + 2f(3) + 2f(3.5) + f(4) ]
= (0.25)[ (e0 -5(0)2 +10(0)) + 2[e0.5 -5(0.5)2 +10(0.5)] + 2[e1 -5(1)2 +10(1)] + 2[e1.5-5(1.5)2 +10(1.5)] + 2[e2 -5(2)2 +10(2)] + 2[e2.5 -5(2.5)2 +10(2.5)] + 2[e3 -5(3)2 +10(3)] + 2[e3.5 -5(3.5)2 +10(3.5)]+ e4 -5(4)2 +10(4) ]
= (0.25)[ 1 + 10.797 + 15.4366 + 16.463378 + 14.778 + 11.86499 + 10.171074 + 13.730904+ 14.5982 ]
= 0.25[108.840]
= 27.21
Hence percentage difference = (27.21 - 26.93)/100 = 0.0028 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.