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

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1 Answer
Nov 20, 2017

Answer:

# 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) #