# Where is my arithmetic messed up?

I can't find my dumb mistake.

I'm figuring the definite integral from first principles of $2x+3$ with limits $x=1$ to $x=4$. No big deal! But for some reason I can't find where my arithmetic went screwy. (Maybe because it's 2:46am @_@).

so

$\delta x=\frac{3}{n}$ and $x_i^*=\frac{3i}{n}$

where $x_i^*$ is the right end point of each rectangle under the curve.

So the sum of the areas of the $n$ rectangles is

$\Sigma_{i=1}^n f(\delta xi)\delta x$

$=\Sigma_{i=1}^n f(\frac{3i}{n})\frac{3}{n}$

$=\Sigma_{i=1}^n (2(\frac{3i}{n})+3)\frac{3}{n}$

$=\frac{3}{n}\Sigma_{i=1}^n (2(\frac{3i}{n})+3)$

$=\frac{3}{n}\Sigma_{i=1}^n ((\frac{6i}{n})+3)$

$=\frac{3}{n} (\frac{6}{n}\Sigma_{i=1}^ni+ 3\Sigma_{i=1}^n1)$

$=\frac{3}{n} (\frac{6}{n}\frac{n(n+1)}{2}+ 3n)$

$=\frac{18}{n}\frac{(n+1)}{2}+ 9$

$=\frac{9(n+1)}{n}+ 9$

$\lim_{n\to\infty} \frac{9(n+1)}{n}+ 9 = 18$

But the correct answer is 24.

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Hint: $$\int_a^bf(x)dx=\lim_{n\to\infty}\sum_1^nf(a+\frac{b-a}{n}i)\frac{b-a}{n}$$

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!!!!!! You know, I had a half-thought that I was sliding the graph to the left so to avoid the $a$ term, but I only half-calculated that. Thanks! – Korgan Rivera Aug 27 '12 at 19:42
@KorganRivera: All of us here want to help each other in Maths. :) – Babak S. Aug 28 '12 at 11:10
Too long without an upvote on this nice answer! +1 – amWhy Mar 14 '13 at 2:02

You want $$f\left(1+\frac{3i}{n}\right).$$

The $+3$ in the third line (and later) will change to $+5$.

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The problem lies at the very beginning. You took

$$x^{*}_i = \frac{3i}{n}$$.

But this is wrong. Remember, you are looking for the right endpoints of the partitioning of $[1, 4]$. So at $i = n$, this should equal 4... but it doesn't. And at $i = 1$, this should equal $1 + \frac{3}{n}$... but it doesn't. You want

$$x^{*}_i = 1 + \frac{3i}{n}$$.

Try it now. It should work.

(See, the right endpoint of the first subinterval will be $1 + \delta x$, as the left endpoint is 1 and the width of the subinterval is $\delta x$. The right endpoint of the second subinterval will then be $1 + 2 \delta x$, and so forth, up to the last subinterval, with right endpoint $1 + n \delta x = 4$.)

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