0
$\begingroup$

How to calculate $\int\limits^{3}_{-1}(4x+3x^2)dx$ using definition $\int^\limits{b}_{a} f(x) = \lim\limits_{\Delta x\rightarrow 0} \sum\limits_{i=0}^{n-1} f(\xi_i)\Delta x_i$?

  • 0
    Can you show us some of the steps you have used? Or any progress? – 2017-02-21
  • 0
    @Hushus46 I stuck in defining of $\xi_i$ and $\Delta x_i$, then it's clear for me how to calculate. – 2017-02-21
  • 0
    $$\Delta x_i = \frac{b-a}{n} = \frac{3-(-1)}{n}=\frac{4}{n}$$ $$\xi_i = i\Delta x = \frac{4i}{n}$$ – 2017-02-21
  • 0
    you're welcome, just to clarify if you didn't know this is the integral as a Left Riemann Sum (which is why $n-1$ is the limit in the summation). If you have any questions about any of the variables, don't hesitate to ask – 2017-02-21
  • 0
    @Hushus46 I saw somewhere, that instead of defining $\xi_i = i\Delta x$, we define it as $a + i \Delta x$. What is the difference? – 2017-02-21
  • 0
    Oh, snap I forgot to add that in. Usually when I have problems like this $a = 0 $ so i forget about it, but yes you have to add $a$. The reason is because this variable $\xi_i$ are the intervals for infinitely small width that you are adding up. So you start adding from the left bound all the way to the right bound. So your first term in the sum would be $f(a)$, then $f(a+\Delta x)$ then $f(a+2\Delta x)$ and so on. I can't edit my comment before anymore but you are correct it is $f(a + i \Delta x)$. My mistake, sorry. – 2017-02-21
  • 0
    Your answer should evaluate to 44, incase you need to check whether you calculated it correctly. – 2017-02-21

1 Answers 1

0

I think solution can be useful for other users, so I'll write it step-by-step:

$\int\limits^{3}_{-1}(4x+3x^2)dx$.

Let's write the definition: $\int^\limits{b}_{a} f(x) = \lim\limits_{\Delta x\rightarrow 0} \sum\limits_{i=0}^{n-1} f(\xi_i)\Delta x_i$, where $$\Delta x_i = \frac{b-a}{n} = \frac{3-(-1)}{n}=\frac{4}{n}$$ $$\xi_i = a + i \Delta x = -1 + \frac{4}{n}$$

Now we can calculate $f(\xi_i)$: $$f(\xi_i) = 4\xi_i+3\xi^2_i = 4(-1+\frac{16i}{n})+3(-1+\frac{16i}{n})^2=$$ $$= \frac{48i^2}{n^2}-\frac{8i}{n}-1$$

Let's write the sum: $$\sum\limits_{i=0}^{n-1}(\frac{48i^2}{n^2}-\frac{8i}{n}-1)\frac{4}{n}= \frac{4}{n} (\frac{48}{n^2} \sum\limits_{i=0}^{n-1}i^2 - \frac{8}{n} \sum\limits_{i=0}^{n-1} i - \sum\limits_{i=0}^{n-1}1) =$$

$$=|\sum\limits_{i=0}^{n-1}i^2=\frac{n(n-1)(2n-1)}{6}, \sum\limits_{i=0}^{n-1}i=\frac{n(n-1)}{2}|=$$

$$=\frac{4}{n}(\frac{8(n-1)(2n-1)}{n}-4(n-1)-n)= 44 + \frac{-80n+32}{n^2}$$

Finally, limit is: $$\lim\limits_{n \rightarrow \infty}(44 + \frac{-80n+32}{n^2}) = 44 + 0 = 44$$

And integral: $$\int\limits^{3}_{-1}(4x+3x^2)dx = 44$$