Solving Limits and Riemann Sums: Tips from Nemo

[Nemo1]

Hi Community,

I have the following question:
View attachment 5613

I have done basic solving of limits and also of Riemann sums but never had to do them in the same question.

Would I be correct in saying that I need to solve for the Riemann sum first then take the limit of the integral?

Cheers Nemo

 

[Prove It]

Hi Community,

I have the following question:I have done basic solving of limits and also of Riemann sums but never had to do them in the same question.

Would I be correct in saying that I need to solve for the Riemann sum first then take the limit of the integral?

Cheers Nemo

No, this entire expression, which is a Riemann Sum, represents a definite integral. You need to figure out what the definite integral is, and then solve the definite integral. You aren't trying to solve this by taking the sum.

 

[Nemo1]

So after, quite a few days of working thru this one. I think I have it.

\(\displaystyle \lim_{{n}\to{\infty}}\sum_{k=0}^{n-1}\sqrt{4+\frac{5k}{n}}\cdot\frac{5}{n}\)

Using the formula:

\(\displaystyle \int_{a}^{b} f(x)\,dx = \lim_{{n}\to{\infty}}\sum_{i=1}^{n}f(xi)\Delta x\)

Using:

\(\displaystyle \Delta x = \frac{b-a}{n} = \frac{5}{n}\)

\(\displaystyle xi = 4+\frac{5k}{n}\)

\(\displaystyle a = 4\)

\(\displaystyle b = 9$ $from$ $b - 4 = 5$ $solve$ $for$ $b = 9\)

To then get:

\(\displaystyle \lim_{{n}\to{\infty}}\sum_{k=0}^{n-1}\sqrt{{xi}}$ $\Delta x =\)\(\displaystyle \int_{4}^{9}\sqrt{{xi}} \,dx\)

\(\displaystyle \int \sqrt{x} \,dx\) $=$ \(\displaystyle \int {x^{\frac{1}{2}}} \,dx\)- Power Rule - \(\displaystyle \int x^a$ $dx = \frac{x^{a+1}}{a+1}$ $Where$ $a\ne 1\)

To then get:

\(\displaystyle x^{\frac{1}{2}}\,dx$ $=$ $\frac{x^{\frac{1}{2}+1}}{{\frac{1}{2}}+1}$ $=$ $\frac{2x^{\frac{3}{2}}}{3}+c\)

Using the FTOC.

\(\displaystyle F(b)=\frac{2\cdot9^{\frac{3}{2}}}{3}=18\) - \(\displaystyle F(a)=\frac{2\cdot4^{\frac{3}{2}}}{3}=\frac{16}{3}\)

To get:

\(\displaystyle \frac{38}{3}$ $\approx12.6667\)

Its been a bit of a long journey of learning but I got there.

Cheers Nemo.