Limit definition of definite integrals

• Feb 6th 2009, 02:34 PM
fattydq
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?
• Feb 6th 2009, 02:42 PM
Mush
• Feb 6th 2009, 02:45 PM
Abu-Khalil
You may define the partition $x_0, where $x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right)$ so when $n\to\infty, \Delta x_k \to 0$.

Finally, $\int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5$ and calculate that sum isn't a problem knowing $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$.
• Feb 6th 2009, 02:47 PM
fattydq
thanks guys I'll check that link out for sure
• Feb 6th 2009, 02:50 PM
fattydq
Quote:

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

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

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?
• Feb 8th 2009, 03:49 PM
fattydq
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?