Exact area using limits and Riemann sum

Question

the i need to find the exact area under tha curve of the function $f(x)=4+3x-x^2$ on the interval $[-1,3]$ using limits and a Riemann Sum. I have nothing started, because I am confused on where to start, and where I would need to go from there, I need help with the entire problem. I hope someone can help me! Thanks

Answer 1

Some ideas for you to understand, prove and develop:

$$f(x)=-x^2+3x+4$$

and since this is a continuous and thus integrable function in any finite interval, we can choose now partitions of the interval $\,[-1,3]\;$ and choose conveniently the points within each subinterval in each partition, say:

$$\forall\,n\in\Bbb N\;;\;\;P_n:=\left\{x_0:=-1\,,\,x_1:=-1+\frac4n\,,\,x_2:=-1+\frac8n\,,\ldots\,,\,x_n=-1+\frac{4n}n=3\right\}$$

and we choose conveniently in each subinterval

$$c_k=\left(-1+\frac{4k}n\right)\in [x_{k-1},x_k]\;,\;k=1,2,\ldots,n$$

so we get the Riemann sum

$$\sum_{k=1}^nf(c_k)(x_k-x_{k-1})=\frac4n\sum_{k=1}^n\left[-\left(-1+\frac{4k}n\right)^2+3\left(-1+\frac{4k}n\right)+4\right]=$$

$$=\frac4n\sum_{k=1}^n\left[-\frac{16k^2}{n^2}+\frac{20k}n\right]=\frac{16}{n^3}\sum_{k=1}^n\left(-4k^2+5nk\right)=$$

$$\frac{16}{n^3}\left(-4\frac{n(n+1)(2n+1)}6+5n\frac{n(n+1)}2\right)=$$

$$=-\frac{32}3\frac{n(n+1)(2n+1)}{n^3}+40\frac{n^2(n+1)}{n^3}\xrightarrow[n\to\infty]{}\ldots?$$