0 to 1
1 to 2 all of width 1
2 to 3
y(0) = 0
y(1) = 1
y(2) = 8
y(3) = 27
(1/2)(1+0) = .5
(1/2)(1+8) = 4.5
(1/2)(8+27) = 35/2 = 17.5
sum = 22.5
now do it perffecctly :)
y^4/4 at 3 minus at 0
81 /3 = 27
Find the absolute value of the resulting error if the value of integral from 0 to 3 (x^3 dx) is estimated with 3 trapezoids of equal width.
a) 2.25
b) 5.25
c) 15.75
d) 47.25
12 answers
Thank you Mr. Damon but 27 is not in my options. How would I have to do to find the answer.
Or in other words none of the above. Check my arithmetic.
I already see but the answer is supposed to be one of the 4 options. I don't know what is happening
I do not think the list in your text is correct, sorry.
e) 4.5 ???
it asked for the error by the way which is 27 - 22.5 = 4.5
lmao Damon the options are 2.25, 5.25, 15.75, and 47.25. the problem isn’t wrong you are.
Chill guys. Damon has a small error in his calculations. The actual integral is 81/4, not 81/3. So it's 22.5 - 20.25 = 2.25.
Damon chill bro
yep
Thanks, this problem gets me a lot in calculus