Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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

share|improve this answer
    
!!!!!! 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. :) –  B. 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 $+5$.

share|improve this answer

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 $1 + n \delta x = 4$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.