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

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)

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    It looks increasing to me... – 2017-01-11
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    Sorry I forgot a negative. – 2017-01-11
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    The question is now fixed – 2017-01-11
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    Why do you claim that $R_{1000}$ is a worse approximation than $L_{10}$? – 2017-01-11
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    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 – 2017-01-11
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    those numbers look waaaaaayyyy too big. – 2017-01-11
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    They are not real – 2017-01-11
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    Which is odd, if they give numbers, the numbers should at least make sense, even if only in respect to each other. $R_{1000}$ and $L_{1000}$ should be much closer together than $L_{1000}$ and $L_{10}$ but that's not the case with the numbers given. Not a criticism of you, but rather of the source of the question. – 2017-01-11
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    Its to test our knowledge about how much we understand sums. Really, the only thing we needed is to know whether the function is increasing/decreasing, and approximations. I understand your point tho, and I agree with you – 2017-01-11

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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.

Your logic is good there.

Except then $L_{n}\ge L_{n+1}$

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    But I did say that $L_{1000} > L_{10}$? – 2017-01-11
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    $L_{10}> L_{1000}$ – 2017-01-11
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    Why do you say that? – 2017-01-11
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    $L_{n} > L_{n+1}$ because each left sum is overstated, and making the partition finer makes the estimate more accurate. – 2017-01-11
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    So I have got my $L$ valued switched up? – 2017-01-11
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    From the point of view of the logic, yes. However, it is worth noting that e^{-14^4} is minuscule. So, all of these estimates are way off, from the actual value of the integral. – 2017-01-11