If
$\mathrm{f}\left(x\right) = 2\cos\left(x\right)$
$ 0 \leq x \leq 3\pi/4$
evaluate the Riemann sum with $n = 6$, taking the sample points to be left endpoints. ( Round your answer to six decimal places ).
I've tried this several times now by using $\delta x = \pi/8$ and doing:
$$ \frac{\pi}{8}\left[2\cos\left(0\right) + 2\cos\left(\frac{\pi}{8}\right) + 2\cos\left(\frac{2\pi}{8}\right) +\cdots + 2\cos\left(\frac{5\pi}{8}\right)\right] $$
and the answer I keep coming up with is $4.711374$ but my online homework is telling me this is wrong. Can anybody help ?.
Edit: I tried subtracting a $\pi/8$ from each value inside the cosine functions and got $3.926437$ can anyone tell me if this is correct ?.
Your setup is right.
You might make use of the fact that $\cos(\frac{\pi}{2})=0$ and that $$ \cos \left(\frac{5\pi}{8} \right) = -\cos \left(\frac{3\pi}{8} \right) $$ to simplify your sum to $$ \frac{\pi}{8} \left( 2\cos (0) + 2\cos \left( \frac{\pi}{8} \right) +2 \cos \left(\frac{2\pi}{8} \right) \right) $$ so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!