Question

Turn the Riemann Sum (-1+k(3/n))^4 3/n into a definite integral

Answer 1

I have the limits, I know they go from -1 to 2, but I'm not sure how to find f(x)

Answer 2

\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n}f(a+i \Delta x) \cdot \Delta x=\int\limits_{a}^{b}f(x) dx \text{ where } \Delta x=\frac{b-a}{n}\]

Answer 3

so you say a is -1 and bis 2
so delta x is (2-(-1))/n=3/n

so you have \[\lim_{n \rightarrow \infty} \sum_{i=1}^{n}(a+i \Delta x)^4 \cdot \Delta x \]
which is the exact form on the left hand side of the equality above
you just have to identify f(x) now

Answer 4

Maybe I wrote it wrong, but I believe that delta x is 3/n . So would f(x) be 3x^4?

Answer 5

yes delta x is 3/n

Answer 6

so I replaced the 3/n 's with delta x

Answer 7

you mean f(x)=x^4

Answer 8

Deltax is 3/n not 1/n

Answer 9

Okay, so what is inside the parentheses is x in f(x)?