Trapezoidal Rule Homework: Integrate 0 to 2 x^3

[mxthuy95]

Homework Statement

Determine and evaluate a definite integral for which (1/40( (0)^3 + 2(.05)^3 +2(.1)^3 +... 2(1.95)^3 + (2)^3 )) is a trapezoidal approximation. Which is greater, the integral or trapezoidal approximation why

Homework Equations

The Attempt at a Solution

So i figure that the the original equation is x^3 and the limits are 0 to 2 so i got this integral

2
∫ X^3 = 4.0025 using trapezoidal rule with n=20 on my Riemann sum program. I said that this is
0

greater than the actual integral because the graph is increasing and concave up. I check with my

teacher and it was wrong. So where did i go wrong.

Thank you.

 

[Simon Bridge]

I said that [the trapezium approx for $\int_0^2x^3dx$] is greater than the actual integral because the graph is increasing and concave up.

That's what I'd have said too.
Have you correctly identified the function being integrated?
Have you use the correct reasoning?

(1/40( (0)^3 + 2(.05)^3 +2(.1)^3 +... 2(1.95)^3 + (2)^3 ))

... that would be: $$\frac{1}{40}\left (
0^3+2(0.05)^3+2(0.1)^3+\cdots + 2(1.95)^3+(2.0)^3
\right )$$

Compare with the trapezoidal rule:
$$\int_a^b f(x)dx \approx \frac{b-a}{2N}\left ( f(x_1)+2f(x_2)+\cdots +2f(x_{N-1})+f(x_N) \right )$$... I'm having trouble faulting this.
Perhaps the teacher means something else?

 

[TheoMcCloskey]

" ... with n=20 ..."

n = 40, not 20. That's the only thing I can see.'wrong'.

 

[Simon Bridge]

It could be that simple.