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For the integral: $\int_{-1}^{5} \left( x^{2} -4 \right) dx$

My calculations:

$\begin{align*}\Delta x &= \frac6n\\\\ x_i &= -1 + \frac{6i}n\\\\ f(x_i) &= 1 + \frac{36i^2}{n^2} -4\\\\ A&=72 \end{align*}$

I'm unsure if this is correct as it is my first attempt at doing this type of problem.

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    @DavidMitra Thank you for pointing that out, it seems I accidentally removed it when editing the latex. Added the$-4$back to the function2012-05-07

2 Answers 2

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If your integrand is $x^2-4$, then $\displaystyle f(x_i) = x_i^2 - 4 = \left( - 1 + \frac{6i}{n} \right)^2 - 4 = -3 -\frac{12i}{n} + \frac{36 i^2}{n^2}$,

where your $\displaystyle \Delta x = \frac6n$. Your integral then becomes : $ \begin{align} \int_{-1}^{5} \left(x^2 -4 \right) dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \Delta x \right) = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) \end{align} $ Hence, all we need is to evaluate $\displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right)$ and take the limit as $n \rightarrow \infty$.

$ \begin{align} \displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = 6 \times \displaystyle \sum_{i=0}^{n-1} \left( \frac{-3}n - \frac{12i}{n^2} + \frac{36i^2}{n^3} \right)\\ & = 6 \times \left(-3 - \frac{12 n(n-1)/2}{n^2} + \frac{36 n(n-1)(2n-1)/6}{n^3} \right)\\ & = 6 \times \left( -3 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \end{align} $ where we made use of the following summations. $ \sum_{i=0}^{n-1} 1 = n $ $ \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2} $ and $ \sum_{i=0}^{n-1} i^2 = \frac{n(n-1)(2n-1)}{6} $

Now taking the limit as $n \rightarrow \infty$, we get $ \begin{align} \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = \lim_{n \rightarrow \infty} 6 \times \left( -3 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \\ & = 6 \times \left( -3 - 6 + 12 \right) = 6 \times 3 = 18 \end{align} $ Hence, $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) = 18 \end{align} $


EDIT:

Note that if $f(x) = x^2$ and $\displaystyle x_i = - 1 + \frac{6i}{n}$, then $\displaystyle f(x_i) = x_i^2 = \left( - 1 + \frac{6i}{n} \right)^2 = 1 -\frac{12i}{n} + \frac{36 i^2}{n^2}$ where your $\displaystyle \Delta x = \frac6n$. Your integral then becomes : $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \Delta x \right) = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) \end{align} $ Hence, all we need is to evaluate $\displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right)$ and take the limit as $n \rightarrow \infty$.

$ \begin{align} \displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = 6 \times \displaystyle \sum_{i=0}^{n-1} \left( \frac1n - \frac{12i}{n^2} + \frac{36i^2}{n^3} \right)\\ & = 6 \times \left(1 - \frac{12 n(n-1)/2}{n^2} + \frac{36 n(n-1)(2n-1)/6}{n^3} \right)\\ & = 6 \times \left( 1 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \end{align} $ where we made use of the following summations. $ \sum_{i=0}^{n-1} 1 = n $ $ \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2} $ and $ \sum_{i=0}^{n-1} i^2 = \frac{n(n-1)(2n-1)}{6} $

Now taking the limit as $n \rightarrow \infty$, we get $ \begin{align} \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = \lim_{n \rightarrow \infty} 6 \times \left( 1 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \\ & = 6 \times \left( 1 - 6 + 12 \right) = 6 \times 7 = 42 \end{align} $ Hence, $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) = 42 \end{align} $

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    When you put the final answer, should it also include the -4? Also, are there shorter methods to getting these answers rather than using Riemanns Sum?2012-05-07
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Your $\Delta x$ and $x_i$ are fine, but you've gone astray after that: $f(x_i)=x_i^2=\left(-1+\frac{6i}n\right)^2=1-\frac{12i}n+\frac{36i^2}{n^2}\;.$ Your Riemann sum is then

$\begin{align*} \sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\frac6n&=\frac6n\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\\ &=\frac6n\left(\sum_{i=1}^n1-\frac{12}n\sum_{i=1}^ni+\frac{36}{n^2}\sum_{i=1}^ni^2\right)\\ &=\frac6n\left(n-\frac{12}n\cdot\frac{n(n+1)}2+\frac{36}{n^2}\cdot\frac{n(n+1)(2n+1}6\right)\\ &=\frac6n\left(n-6(n+1)+\frac{6(n+1)(2n+1)}n\right)\\ &=\frac6n\cdot\frac{-5n^2-6n+12n^2+18n+6}n\\ &=6\left(7+\frac{12}n+\frac6{n^2}\right)\;, \end{align*}$

which converges to $42$ as $n\to\infty$.

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    Thank you for the reply and the explanation. It was easy to follow and made it easy to understand. (I did it the same way you did, however I just didn't square my value $c$orrectly)2012-05-07