As you can see in the image I made the example with n=6. In general, if the interval is divided into n equal parts we have that te right value of x in the kth rectangle (Rk) is
Rk = 2+ (1/2)k
So, f(Rk) = f(2+(1/2)k) = 2(2+(1/2)k)+1 = 4+k+1 = 5+k.
By "<span>f(x = 2x + 1" you likely meant "f(x) = 2x + 1."
The given interval is [2,5], and so the width of each interval is (5-2)/n, or (3/n). Since we are using right end points, the x value at the right endpoint of the kth rectangle is 2+k(3/n). Examples: if k=1, the value at the right endpoint is 2+1(3/n); if k=2, 2+2(3/n), and so on. If k=n, then the value at the right endpoint is 2+n(3/n), or 2+3=5, which agrees with the given interval [2,5].
Once again, the given function is f(x) = 2x + 1. At x = 2+k(3/n), the value of the function is 2[2+k(3/n)] + 1 (Answer). Check: If k=n, x = 2 + n(3/n) = 2+3 + 5, and f(5)=2(5) + 1 = 11.</span>
