The Area under the curve $\sqrt{x}$ from $1\le x\le 4$ using no integrals but using riemann sums

Question

I tried to solve this problem by using summation with about $1000$ rectangles and then kept adding more until about $100000$ rectangles. I got closer and closer to $\frac{14}{3}$ and my closest answer was $4.66668166$ . Is it safe to assume that if I keep going infinitely the answer will be exactly $\frac {14}{3}$?

Answer 1

Is it safe to assume that if I keep going infinitely the answer will be exactly 14/3?

Strictly speaking, no, it is not. The answer could be some number really close to $\frac{14}{3}$, and the only way to tell would be to actually do the exact integral. At no point will a finite Riemann sum be able to tell you that the answer is definitely $\frac{14}3$.

However, in this case, with a very simple function and very simple bounds... Odds are definitely in your favour. Also, it is entirely correct to use your calculations to conclude that $$ \int_1^4\sqrt x\,dx\approx\frac{14}3 $$

Answer 2

You can actually exactly evaluate the area, without integration. Observe that required area is $$A=A_1-A_2-A_3$$ where $A_1$ is the area of the rectangle having vertices $(0,0), (0,2), (4,0), (4,2)$, hence, $$A_1=4\times 2=8$$ $A_2$ is the area of the small rectangle having vertices $(0,0), (0,1), (1,0), (1,1)$, hence, $$A_2=1\times 1=1$$ Now, $A_3$ is the area of graph of $y=\sqrt{x}$ with $y$-axis from $y=1$ to $y=2$. This area is actually area of curve $y=x^2$ from $x=1$ to $x=2$. This area can be evaluated as limit of a sum. $$A_3=\lim_{n\to\infty}\frac{1}{n}\sum_{r=n+1}^{2n}\Big(\frac{r}{n}\Big)^2$$ $$=\lim_{n\to\infty}\frac{1}{n^3}(\sum_{r=1}^{2n} r-\sum_{r=1}^{n} r)$$ $$=\lim_{n\to\infty}\frac{1}{n^3}\Big(\frac{2n(2n+1)(4n+1)}{6}-\frac{n(n+1)(2n+1)}{6}\Big)$$ $$\frac{2\times 2\times 4}{6}-\frac{1\times 1\times 2}{6}$$ $$=\frac{7}{3}$$ So, required area is $$A=8-1-\frac{7}{3}=\frac{14}{3}$$

Answer 3

The area of the $A$-region can be computed as the difference between the area of a $4\times 2$ rectangle and the area of the $B$-region. Since the inverse function of $y=\sqrt{x}$ is $x=y^2$ the area of the $B$-region is given by

$$ \lim_{n\to +\infty}\frac{2}{n}\sum_{k=1}^{n}\left(\frac{2k}{n}\right)^2 = 8 \lim_{n\to +\infty}\frac{n(n+1)(2n+1)}{6n^3}=\frac{8}{3}$$ and the area of the $A$-region equals $8-\frac{8}{3}=\frac{16}{3}$. Similarly the area you want to compute is given by $$ 3+\left[3-\left(\int_{1}^{2}y^2\,dy-1\right)\right] = 7-\frac{8-1}{3}=\color{red}{\frac{14}{3}}.$$

Answer 4

It is not necessary to choose equispaced division points. You can choose the division points as $$x_k=\left(1+\frac{k}{n}\right)^2$$ and use the Riemann sum $$S_n=\sum_{k=1}^{n}f(x_k)(x_k-x_{k-1})$$ where $f(x) =\sqrt{x} $.

Then $$S_n=\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\left\{\left(1+\frac{k}{n}\right)^2-\left(1+\frac{k-1}{n}\right)^2\right\}$$ which can be simplified as $$S_n=\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\left(2+\frac{2k-1}{n}\right)=\frac{1}{n}\sum_{k=1}^{n}\left(2+\frac{4k-1}{n}+\frac{2k^2-k}{n^2}\right)$$ and this further simplifies to $$S_n=2+\frac{2n(n+1)-n}{n^2}+\frac{n(n+1)(2n+1)} {3n^3}-\frac{n(n+1)}{2n^3}$$ and clearly the above sum tends to $$2+2+\frac{2}{3}=\frac{14}{3}$$ as $n\to\infty $.