Question

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

Answer 1

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

`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n (4(2n+2i)) / ( (2n+2i)^2+n^2)`

Explanation:

We seek:

`int_2^4 \ (2x)/(x^2+1)`

via a Riemann sum. So, we have:

`A = lim_(n rarr oo) \ sum_(i=1)^n f(x_i) Delta x`
`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n f(a+iDelta x) Delta x`
`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n f(2+((4-2)i)/n) (4-2)/n`
`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n f(2+(2i)/n) (2)/n`
`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n 2/n \ f( (2n+2i )/n )`

Then using the definition of `f(x)=(2x)/(x^2+1)` this becomes:

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

`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n (4((2n+2i)/n^2)) / (( (2n+2i)^2+n^2)/n^2)`

`\ \ \ = lim_(n rarr oo) \ sum_(i=1)^n (4(2n+2i)) / ( (2n+2i)^2+n^2)`