How do i answer this?  int_2^4 \ (2x)/(x^2+1)  via a Riemann sum.

Nov 20, 2017

 A = lim_(n rarr oo) \ sum_(i=1)^n 2/n \ (2((2n+2i)/n)) / (( (2n+2i)^2/n^2 + 1)

$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} \frac{4 \left(2 n + 2 i\right)}{{\left(2 n + 2 i\right)}^{2} + {n}^{2}}$

Explanation:

We seek:

${\int}_{2}^{4} \setminus \frac{2 x}{{x}^{2} + 1}$

via a Riemann sum. So, we have:

$A = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} f \left(a + i \Delta x\right) \Delta x$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} f \left(2 + \frac{\left(4 - 2\right) i}{n}\right) \frac{4 - 2}{n}$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} f \left(2 + \frac{2 i}{n}\right) \frac{2}{n}$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} \frac{2}{n} \setminus f \left(\frac{2 n + 2 i}{n}\right)$

Then using the definition of $f \left(x\right) = \frac{2 x}{{x}^{2} + 1}$ this becomes:

$A = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} \frac{2}{n} \setminus \frac{2 \left(\frac{2 n + 2 i}{n}\right)}{{\left(\frac{2 n + 2 i}{n}\right)}^{2} + 1}$

for a 1 mark question this answer is probably sufficient, but it could be simplified further if required:

 A = lim_(n rarr oo) \ sum_(i=1)^n 2/n \ (2((2n+2i)/n)) / (( (2n+2i)^2/n^2 + n^2/n^2)

$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} \frac{4 \left(\frac{2 n + 2 i}{n} ^ 2\right)}{\frac{{\left(2 n + 2 i\right)}^{2} + {n}^{2}}{n} ^ 2}$

$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \setminus {\sum}_{i = 1}^{n} \frac{4 \left(2 n + 2 i\right)}{{\left(2 n + 2 i\right)}^{2} + {n}^{2}}$