Consider $\int_{4}^{9} \frac{1}{\sqrt x} \ dx$
1) Find a value of $n$ where $|L_n - R_n| \leq 0.1$
2) Use that value of $n$ to find $L_n$ and $R_n$
3) If you were to take the average of the Left-hand sum and the Right-hand sum, will your approximation be larger or smaller than the integral? $$\\$$
*$L_n$ is left-hand sum
*$R_n$ is right-hand sum
$$\\$$ I have no clue on how to go about this problem. I was think I might have to use the error formulas but I'm not certain.
Your help would be greatly appreciated. :)
Notice that the left hand sum is $\displaystyle \frac{b-a}{n}\sum_{k=0}^{n-1}f\left(a+\displaystyle \frac{k}{n}(b-a)\right)$
Similarly, the right hand sum is $\displaystyle \frac{b-a}{n}\sum_{k=1}^{n}f\left(a+\displaystyle \frac{k}{n}(b-a)\right).$
They're almost the same. The difference is that the rectangle at the beginning of the left hand sum is "replaced" with the rectangle at the end of the right hand sum.
So we have the difference of areas is $\displaystyle L_n-R_n=\frac{b-a}{n}\left(f(a)-f(b)\right)$.
We need that $|L_n-R_n| \leq 0.1$. We have that $\displaystyle f(a)-f(b)=\frac{1}{6}$, and that $b-a=5$.
Plugging right in gives us $\displaystyle \frac{5}{6}\left(\frac{1}{n}\right) \leq \frac{1}{10}$. (We just drop absolute value because $n$ is positive).
Now we isolate $n$ to get $\displaystyle \frac{1}{n} \leq \frac{3}{25}$.
Flipping the inequality gives us $\displaystyle n \geq \frac{25}{3}$.
Because we can't have half a rectangle, we say that $\boxed{n \geq 9}$.
$L_n = \displaystyle \frac{b-a}{n}\sum_{k=0}^{n-1}f\left(a+\displaystyle \frac{k}{n}(b-a)\right)=\displaystyle \frac{b-a}{n}f(a) + \frac{b-a}{n}\sum_{k=1}^{n-1}f\left(a+\displaystyle \frac{k}{n}(b-a)\right)$
$R_n = \displaystyle \frac{b-a}{n}\sum_{k=1}^{n}f\left(a+\displaystyle \frac{k}{n}(b-a)\right)=\displaystyle \frac{b-a}{n}\sum_{k=1}^{n-1}f\left(a+\displaystyle \frac{k}{n}(b-a)\right)+\frac{b-a}{n}f(b)$
That's how $\displaystyle L_n-R_n = \frac{b-a}{n}(f(a)-f(b))$ comes about.