# Consider $I = \int^{14}_{8} e^{-x^4} dx$. Match each riemann sum.

$I = \int^{14}_{8} e^{-x^4} dx$

We have $L_{1000}, L_{10}, R_{1000}$.

We have $I= 0.335, 0.368, 0.367$.

Match each sum with each $I$ value.

I know that the graph is decreasing and approaches $0$ on the interval $[8,14]$, so $R_n \leq \text{ Actual Area I} \leq L_n$.

I know $R_{1000} = 0.335$, as it is the number most away from the actual area $I$.

I think $L_{10} = 0.367$ and $L_{1000} = 0.368$ since $1000$ rectangles is much more accurate then $10$ rectangles, and thus it will be more larger then the actual area $I$.

Is this correct? It's the first time I'm doing these types of questions so I'm not $100%$ sure.

(not real approximations)

• It looks increasing to me... – Simply Beautiful Art Jan 11 '17 at 23:08
• Sorry I forgot a negative. – K Split X Jan 11 '17 at 23:09
• The question is now fixed – K Split X Jan 11 '17 at 23:10
• Why do you claim that $R_{1000}$ is a worse approximation than $L_{10}$? – Anonymous Jan 11 '17 at 23:12
• Since the function is decreasing, if you take the $\text{right}$ sum, then you will have $x$ values that lead to smaller $f(x)$ values, since the function is decreasing. So the area covered by these rectangles isnt as much. If you took the left sum, you take x values where $f(x)$ is larger then the right $f(x)$ values – K Split X Jan 11 '17 at 23:27

Since the function is decreasing.

$L_n \ge \int_a^b f(x) dx \ge R_n$

The right sum is less than the true value. The left sum is greater than the true value.

As $n$ gets larger $L_n$ gets closer to the true value.

Except then $L_{n}\ge L_{n+1}$
• But I did say that $L_{1000} > L_{10}$? – K Split X Jan 11 '17 at 23:31
• $L_{10}> L_{1000}$ – Doug M Jan 11 '17 at 23:32
• $L_{n} > L_{n+1}$ because each left sum is overstated, and making the partition finer makes the estimate more accurate. – Doug M Jan 11 '17 at 23:33
• So I have got my $L$ valued switched up? – K Split X Jan 11 '17 at 23:39