Question

I calculated 0.776 but that was not the right answer so I need help. Suppose the function g(x) = x^2 · 2^x on the interval [-1, 1]. Use the midpoints of four rectangles of equal width to approximate the integral from -1 to 1 of g(x) dx

Answer 1

show your work, please

Answer 2

I keep getting a different answer every time I try it. I just got 0 as my answer now:
delta x= (1 - -1)/4
delta x= 2/4
delta x = 0.5
(0.5) * ( f((-1+-0.5)/2)) + f((-0.5+0)/2)) + f((0+0.5)/2)) + f((0.5+1)/2)) )
(0.5) * ( f(-0.75) + f(-0.25) + f(0.25) + f(0.75) )
=0

Answer 3

I don't know if I am even doing it right.

Answer 4

I really don't understand what are you doing. what is f(x)?

Answer 5

The function. I guess it should be g(x) instead. Sorry

Answer 6

Also if you take the integral of the function from -1 to 1 it is also 0

Answer 7

ok, I am not sure whether I can help or not, just check your work
if you said f(x) is g(x) , then the first term of your expression is (0.5)* (g(-1 -0.5) /2) let check it
= 0.5*(g(-1.5)/2 and g(-1.5) = (-1.5)^2 * 2*(-1.5) =0.795495
then *0.5 / 2 = 0.1988

If you take the integral, it not =0

Answer 8

oh, I messed up somewhere. But anyway, it is not 0 http://www.wolframalpha.com/input/?i=int%28x^2*2^x%29+dx%28from-1to1%29

Answer 9

so 0.198 would be correct.

Answer 10

because I wrote 0.776 last time but they said that I was wrong.

Answer 11

the wolfram said it is 0.765 , not 0.776

Answer 12

oh okay. That is probably what I did wrong.