If $f$ is integrable on $[a,b]$, then $$ \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x_i $$ where $\Delta x = (b-a)/n$ and $x_i = a + i\Delta x$. Use this definition of the integral to evaluate $\int_2^4 (1-2x)dx$.
- We have $\int_2^4 (1-2x)dx = \lim_{n \to \infty} \sum_{i=1}^n ???$
- Evaluating the sum gives $\int_2^4 (1-2x)dx = \lim_{n \to \infty} ???$
- Evaluating the limit gives $\int_2^4 (1-2x)dx = ???$
So I'm able to solve the first part of this question correctly with getting $(1-2(2+\frac{2i}{n}))(\frac{2}{n})$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.
It's not the same as evaluating the limit so this is confusing me a bit.
It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.
Thank you!
Without checking the correctness of your representation, you have $$ \frac{2}{n} \left[1 - 2\left(2+\frac{2i}{n}\right)\right] = \frac{2}{n} \left[-3 - \frac{4i}{n}\right] = -\frac{6}{n} - \frac{8i}{n} $$ Therefore, $$ \sum_{i=0}^{n-1} \frac{2}{n} \left[1 - 2\left(2+\frac{2i}{n}\right)\right] = -\sum_{i=0}^{n-1} \frac{6}{n}-\sum_{i=0}^{n-1} \frac{8i}{n} = -6-\frac{8}{n}\sum_{i=0}^{n-1} i $$ Can you finish this?