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I'm given the following problem:

Calculate the net area between $f(x) = x^3+8$ and the $x$ axis on the interval $[-1,1]$.

I do so by finding the Riemann sum, then taking a limit. I've audited this class before, so I check these answers by taking the definite integral over the same interval. For some reason, the definite integral and my Riemann sum don't seem to agree, this time.

Here's my solution:

Find the exact net area between $f(x) = x^3 + 8$ and the $x$-axis on $[-1,1]$ by finding the Riemann sum and then taking a limit.

  1. $\Delta x = \frac{-1 - (-1)}{n} = \frac{2}{n}$

  2. $x_k^* = a + k \Delta x = -1 + \frac{2k}{n}$

  3. $f(x_k^*) = (-1 + \frac{2k}{n})^3 + 8$

  4. $\Sigma_{k=1}^{n} ((-1 + \frac{2k}{n})^3 + 8)(\frac{2}{n})$

  5. Now, we just simplify. After distributing and substituting in summation formulas, we have: $\frac{16}{n^4}(\frac{n^2(n+1)^2}{4}) - \frac{24}{n^4}(\frac{n(n+1)(2n+1)}{6}) + 12n(\frac{n(n+1)}{2}) + 14n^2$ (verified by simply plugging into WolframAlpha the above expression).

When I take the limit as $n\to \infty$ of this expression, however, I get $\infty$. The definite integral is $16$. What did I do wrong here?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Is the interval $[-1,1]$ or $[1,6]$? $\endgroup$ Commented Sep 15, 2020 at 19:31
  • $\begingroup$ Your last two terms are not correct. $\endgroup$ Commented Sep 15, 2020 at 19:32
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    $\begingroup$ @J.W.Tanner Sorry, that's $[-1,1]$. I'll fix that. $\endgroup$
    – 10GeV
    Commented Sep 15, 2020 at 19:33

2 Answers 2

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You must have made a mistake in step 5. It should simplify a lot, to $$16 + \frac2n,$$ and obviously the limit will be $16$ as $n\to\infty$.

The reason it reduces to just $16 + 2/n$ is that most terms will cancel out in pairs: for every $(-1+2k/n)^3$, you'll have a $(1-2k/n)^3$. The only element without such a matching pair is the far right endpoint, $1$.

Extending the reasoning, you can get the area without integration or even Riemann sums. Observe that the area of $g(x)=x^3$ on the given interval is zero, by symmetry. The area under the line $h(x)=8$ is $16$. Your function is $g+h$.

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  • $\begingroup$ Very neat evaluation! $\endgroup$ Commented Sep 15, 2020 at 19:33
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When I expand the expression in step 4, I get $$ \sum_{k=1}^n \bigg( \frac{16 k^3}{n^4}-\frac{24 k^2}{n^3}+\frac{12 k}{n^2}+\frac{14}{n}\bigg). $$ After using summation formulas, this becomes $$ \frac{4 (n+1)^2}{n^2}-\frac{4 (n+1) (2 n+1)}{n^2}+\frac{6 (n+1)}{n}+14. $$

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