# Thread: Limit definition of definite integrals

1. ## Limit definition of definite integrals

In a practice quiz I'm working on for my calc class I'm asked to evaluate the given definite integral from 3 to 7 of (x^2-3x+5)dx using the limit definition of a definite integral. I know how to express an integral as a limit using the limit definition, and I know how to evaluate definite integrals usint areas, but I don't know how to EVALUATE a definite integral using the limit definition. What does this even mean exactly and how could I go about doing it?

2. You may define the partition $\displaystyle x_0<x_1<\ldots<x_{n-1}<x_n$, where $\displaystyle x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right)$ so when $\displaystyle n\to\infty, \Delta x_k \to 0$.

Finally, $\displaystyle \int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5$ and calculate that sum isn't a problem knowing $\displaystyle \sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$.

3. thanks guys I'll check that link out for sure

4. Originally Posted by Abu-Khalil
You may define the partition $\displaystyle x_0<x_1<\ldots<x_{n-1}<x_n$, where $\displaystyle x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right)$ so when $\displaystyle n\to\infty, \Delta x_k \to 0$.

Finally, $\displaystyle \int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5$ and calculate that sum isn't a problem knowing $\displaystyle \sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$.
I'm sorry that's a bit too jumbled for me I don't really understand any of what you said. What is x subscript k even representing here?

5. I keep getting wrong answers. I just found the answer by using the integral function on my calculator but I can't seem to get it. Could someone walk me through this step by step?