solve my stupid questions(1)!!! LOL?

Question

Q1. (a) Use the Trapezoidal Rule with n =10 to approximate the integral e^-x^2 dx.
(b) Use the Trapezoidal Rule with n = 8 to approximate the integral
cos(x^2)dx.

Answer 1

Trapezoidal rule says INTEGRAL FROM a to b of f(x)dx can be approximated by:

[(b-a)/(2n)] * [ f(x0) + 2f(x1) + 2f(x2) + … + 2f(x n-1) + f(xn) ]


A. Let f(x) = e^(x^2), a=0, b=1, n=10:

f(x0) = f(0.0) = 1.000; 1f(x0) = 1.000
f(x1) = f(0.1) = 1.010; 2f(x1) = 2.020100334
f(x2) = f(0.2) = 1.041; 2f(x2) = 2.081621548
f(x3) = f(0.3) = 1.094; 2f(x3) = 2.188348567
f(x4) = f(0.4) = 1.174; 2f(x4) = 2.347021742
f(x5) = f(0.5) = 1.284; 2f(x5) = 2.568050833
f(x6) = f(0.6) = 1.433; 2f(x6) = 2.866658829
f(x7) = f(0.7) = 1.632; 2f(x7) = 3.26463244
f(x8) = f(0.8) = 1.896; 2f(x8) = 3.792961759
f(x9) = f(0.9) = 2.248; 2f(x9) = 4.495815973
f(x10) = f(1.0) = 2.718; 1f(x10) = 2.718

Sum of last column:29.343

(b-a)/(2n) =(1 - 0)/(2*10) = 0.05

[(b-a)/(2n)] * [ f(x0) + 2f(x1) + 2f(x2) + … + 2f(x n-1) + f(xn) ] = .05 * 29.343 = 1.46715

Answer 2

You do not state the limits of integration.

It is too cumbersome to solve this just saying the limits are from a to b. All I can say is set h=(b-a)/n = (b-a)/10.

Then find x0 through x10 where x0 = a and xn = b and all intervening x values are h apart.
Let y= e^-x^2
Then find y0 through y10

Then apply the trapezoidal rule:
int a to b e^-x^2= h/2y0 +2y1 +2y2 + ,,, +y10)

Answer 3

Good question. I'm still searching for another of my kind. I don't want to go the whole asexual route but may have to do it in order to protect my seed. It's tough being a velociraptor.

Answer 4

hahaha, they are not funny or that easy, and I don't feel like taking my Calculus 2 book out and refreshing