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Use Definition 2 to find an expression for the area under the graph of $ f $ as a limit. Do not evaluate the limit.

$ f(x) = \dfrac{2x}{x^2 + 1}, \hspace{5mm} 1 \le x \le 3 $

$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{2(1+^2 \frac i {n})}{(1+^2 \frac i{n})^{2}+1} \cdot \frac{2}{n}$

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Dishant P.

September 30, 2020

hui

So the general theme when we start talking about Integral is the idea of area under the curve on definition two states that the area is equal to the limit as an approaches infinity of Delta X, the change in X times The summation from I equals one toe end of f of x abi. So in this case, we know that Delta X is equal to B minus a our ending point in R minus our initial point divided by the number of kind of rectangles we have or whatever we're using to split up the area. So what this problem that we have is f of X being equal to two X over X squared, plus one. So we want to set up our limit. Um, we know that a equals one and B equals three. So right off the bat, we confined Delta X to be equal to three minus one over n. So it'll be too over end. Then, um, with this information, we see that the area will be equal to the limit as an approaches infinity of two over em times the sum from I equals one toe end of two times our X value and we know they're X value is one plus Chu I over end over X squared. So are once again or X values me one plus two i over and squared plus one Andi. The way that we can get that is knowing that we're evaluating X being equal to one plus I times two over end. Once we plug that into ffx, we'll be able to do everything that we see right here for our final answer.

California Baptist University