The following sum √(7+(6/n)^2)*(6/n)+√(7+(612n)^2)*(6/n)+…+√(7+(6n/n)^2)*(6/n) is a right Riemann sum with n subintervals of equal length for the definite integral ∫b and 5 f(x)dx where b= f(x)= ?

1 Answer
Aug 28, 2015

Depending on what is meant by ∫b and 5 f(x)dx, we get either #b=-1# and #f(x) = sqrt(7+(x+1)^2)#, OR we get #b = 11# and #f(x) = sqrt(7+(x-5)^2)#

Explanation:

If I have read correctly, we have:

#sqrt(7+(6/n)^2) * (6/n) + sqrt(7+(12/n)^2) * (6/n)+ * * * + sqrt(7+((6n)/n)^2)*(6/n)#

Is a right Riemann sum with #n# subintervals for #int_b^5 f(x) dx#

We are asked to find #b# and #f(x)#

(I'm not at all certain that I have identified the integral from where to where. If not, you will need to make some changes in this answer.)

So it is #f(b+ Deltax) * Deltax + f(b+ 2Deltax) * Deltax + * * * + f(b+ nDeltax) * Deltax#

We must, therefore have #Deltax = 6/n#

Since #Delta x = (5-b)/n = 6/n# we must have #5- b =6#, so #b = -1#

Now, #sqrt(7+(6/n)^2) = f(b+ Deltax) = f(-1+6/n)#

Which will be true if #f(x) = sqrt(7+(x+1)^2)#

This #f(x)# works for the other two given terms, so it must be the correct #f(x)#.

(I wonder if it should be #int_5^b f(x) dx#? If so, then #b = 11# and
we would have #f(5+6/n) = sqrt(7+(6/n)^2)#,
so we would get #f(x) = sqrt(7+(x-5)^2)#.)