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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 $\frac {14}{3}$?

weareallin
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    Do you know how to apply limits to a summation? My guess is that's what you are supposed to do. – Calvin Lin Oct 04 '19 at 04:08
  • One of the definitions of the definite integral is the sum of all rectangles as the width of rectangles approaches zero (or if the number of rectangles approaches infinity). If you know the antiderivative, then you will find that you are approaching it correctly. – Andrew Chin Oct 04 '19 at 04:10

4 Answers4

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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 $$ \int_1^4\sqrt x\,dx\approx\frac{14}3 $$

Arthur
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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 $$A=8-1-\frac{7}{3}=\frac{14}{3}$$

Martund
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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 $$ 3+\left[3-\left(\int_{1}^{2}y^2\,dy-1\right)\right] = 7-\frac{8-1}{3}=\color{red}{\frac{14}{3}}.$$

Jack D'Aurizio
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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 $.

Paramanand Singh
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