Evaluating ${\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x}$ using $\int f^{-1}(x)dx=x f^{-1}(x)-F(f^{-1}(x))+C$

Question

While studying, I countered problem $\text{#}2$ here. $\text{(UCHICAGO REU 2019 - MATH GRE PREP: WEEK 2)}$.

I saw this also. But wanted to know if my way is also correct (regardless of the time consuming).

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$\color{red}{\text{Problem: Which of the following is closest to the value of this integral}}$

$$\color{red}{\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x?}$$

$\color{red}{\text{(A) }1 \space\space\space\space\space\space\space\space\space\space\text{(B) }1.2 \space\space\space\space\space\space\space\space\space\space \color{blue}{\boxed{{\text{(D) }1.6}}}\space\space\space\space\space\space\space\space\space\space\text{(D) }2\space\space\space\space\space\space\space\space\space\space\text{(E) }\text{The integral doesn’t converge.} }$

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My Attempt:

First, I found the inverse of $f^{-1}(x)=\sqrt{1+\frac{1}{3x}}$ as a separate problem. As usual, I replaced $x$ with $y$ and $y$ with $x$, then I solved for $y$. I got $f(x)=\frac{1}{3(x^2-1)}$, which (using partial fractions), $f(x)=\frac{1}{6}\bigg(\frac{1}{x-1}-\frac{1}{x+1}\bigg)$.

Second, I evaluated $F(x)=\int f(x)\text{ d}x=\frac{1}{6}\int \big(\frac{1}{x-1}-\frac{1}{x+1}\big)\text{ d}x=\frac{1}{6}\log\bigg(\frac{x-1}{x+1}\bigg)$ [constant of integration is not added in this step, because we have definite integral, actually].

Next, I evaluated $F(f^{-1}(x))=\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)$.

Finally, I used the formula: $$\int f^{-1}(x)dx=x f^{-1}(x)-F(f^{-1}(x))+C$$

So, $\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x=x\sqrt{1+\frac{1}{3x}}-\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)\bigg|_{x=0}^{x=1}$

Plugging the upper limit of integration, $x=1$, we get $\frac{2\sqrt{3}}{3}-\frac{1}{6}\log(7-4\sqrt{3})$

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My problems:

• Even though this might not be an efficient way to save the time in GRE MATH SUBJECT TEST, I am still confused if my way was correct to solve such integral.

• If the way is correct, how can we evaluate it at the lower limit of integration, that is:

$$\lim_{x \rightarrow {0^+}}\bigg(x\sqrt{1+\frac{1}{3x}}-\frac{1}{6}\log\bigg(\frac{\sqrt{1+\frac{1}{3x}}-1}{\sqrt{1+\frac{1}{3x}}+1}\bigg)\bigg)$$

This should be $0$ as my calculator says. (By the way, calculators are prohibited in the test). I thought about applying L'Hospital's rule, but I did not success. I am not sure if this limit problem is harder than the original integral problem.

• How can we estimate $\frac{2\sqrt{3}}{3}-\frac{1}{6}\log(7-4\sqrt{3})$ to at least one decimal place? (No high accuracy is required as the given choices are far enough).

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Your help would be appreciated. THANKS!

Answer 1

Hm let's see. If we're only focused on approximating the integral, analytically solving it seems inefficient because in the end we do not seek such a thing, and as you can see, even if you have the closed form answer, it's useless if you cannot convert this into a decimal!

Instead let's use integral approximation methods. Our function decreases rapidly from infinity at $0$, before smoothing out to an asymptote. Hence, on the interval $(0,1]$, a good approximation method would be simpson's rule.

Since $0$ is a vertical asymptote, we'll use a left endpoint of $0.01$ for a close enough approximation. We'll use the 1/3 and 4-2-...-2-4 version of the rule.

Partition the interval into $4$ subintervals, to wit, $$[0.01, 0.25],\quad[0.25, 0.5],\quad[0.5, 0.75],\quad [0.75, 1]$$

First, establish $$f(x)=\sqrt{1+\frac1{3x}}$$ We have $f(0.01)\approx 5.86$, $4f(0.25)\approx 6.11$, $2f(0.5)\approx 2.58$, $4f(0.75)\approx 4.81$, $f(1)\approx 1.15$.

Lastly, $\frac{\Delta x}3\approx 0.0833$.

In the end, our integral is approximately $$0.0833(5.86+6.11+2.58+4.81+1.15)\approx 1.7$$ This is closest to $\boxed{\text{D}}$.

Arguably, this margin of error is still pretty large. It would be up to you during the test to determine the number of subintervals/accuracy you want to compute this to. imo $4$ subintervals is a pretty good approximation, but if you want to be more sure you could to $6$ or $8$. Depends on how much time you have to spare i guess.

Answer 2

I will concentrate only on the last two questions.

For the limit, you could simplify your expression. In the first term, move $x$ under the square root. For the expression in the log, multiply both numerator and denominator by $\sqrt{3x}$. Then you get $$\lim_{x\to 0^+}\left(\sqrt{x^2+\frac x3}-\frac16\log\frac{\sqrt{3x+1}-\sqrt{3x}}{\sqrt{3x+1}+\sqrt{3x}}\right)$$ Then you just plug in $x=0$ and you get $$\sqrt 0-\frac16\log\frac 11=0$$

For the last question, you can easily calculate $\sqrt3\approx 1.73$. Just keep two digits after decimal point. Then $$\frac{2\sqrt 3}3\approx 1.15$$For the logarithm, it would be easier if you rewrite it using $-\log x=\log\frac1x$. Therefore your expression is $$1.15+\frac16\log(13.92)$$ The log is between $2$ and $3$, so your result will be between $1.15+0.33=1.48$ and $1.15+0.5=1.65$.

Answer 3

A Hopefully Simpler Estimate $$ \begin{align} \int_0^1\sqrt{1+\frac1{3x}}\,\mathrm{d}x &=\int_0^1\frac1{\sqrt{3x}}\sqrt{1+3x\Rule{0em}{.75em}{0em}}\ \mathrm{d}x\tag{1a}\\ &=\int_0^1\frac1{\sqrt{3x}}\left[1+x,1+\frac32x\right]_\#\,\mathrm{d}x\tag{1b}\\ &=\left[\frac2{\sqrt3}+\frac2{3\sqrt3},\frac2{\sqrt3}+\frac1{\sqrt3}\right]_\#\tag{1c}\\ &=\left[\frac89\sqrt3,\sqrt3\right]_\#\tag{1d}\\[3pt] &=\left[1.52,1.74\right]_\#\vphantom{\frac89}\tag{1e} \end{align} $$ where $[a,b]_\#$ represents a number in $[a,b]$.

Explanation:
$\text{(1a):}$ $\sqrt{1+\frac{1}{3x}}=\frac1{\sqrt{3x}}\sqrt{1+3x\Rule{0em}{.75em}{0em}}$
$\text{(1b):}$ $1+x\le\sqrt{1+3x\Rule{0em}{.75em}{0em}}$ since $\sqrt{x}$ is concave on $[0,1]$
$\phantom{\text{(1b):}}\sqrt{1+3x\Rule{0em}{.75em}{0em}}\le1+\frac32x$ by Bernoulli's Inequality
$\text{(1c):}$ integrate each bound
$\text{(1d):}$ simplify each bound
$\text{(1e):}$ $1.52=\frac89\cdot1.71\le\frac89\sqrt3$ and $\sqrt3\le1.74$

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Comment on the Answer in the Question

The computation in the question is correct. Here is essentially what is given in the question, but using integration by parts to simplify it a bit. $$ \begin{align} \int_0^1\sqrt{1+\frac1{3x}}\mathrm{d}x &=-\int_{2/\sqrt3}^\infty u\,\mathrm{d}\frac1{3\!\left(u^2-1\right)}\tag{2a}\\ &=\frac2{\sqrt3}+\int_{2/\sqrt3}^\infty\frac1{3\!\left(u^2-1\right)}\,\mathrm{d}u\tag{2b}\\ &=\frac2{\sqrt3}+\frac16\int_{2/\sqrt{3}}^\infty\left(\frac1{u-1}-\frac1{u+1}\right)\mathrm{d}u\tag{2c}\\ &=\frac2{\sqrt3}+\frac16\log\left(\frac{2/\sqrt3+1}{2/\sqrt3-1}\right)\tag{2d}\\ &=\frac2{\sqrt3}+\frac13\log\left(2+\sqrt3\right)\tag{2e}\\ \end{align} $$ Explanation:
$\text{(2a):}$ substitute $x=\frac1{3\left(u^2-1\right)}$
$\text{(2b):}$ integrate by parts
$\text{(2c):}$ partial fractions
$\text{(2d):}$ integrate
$\text{(2e):}$ simplify

To evaluate the answer numerically, we can use $\frac87\le\frac2{\sqrt3}\le\frac76$ to approximate $$ \frac2{\sqrt3}\approx\frac{15}{13}\tag3 $$ Since $\frac43-\frac{15^2}{13^2}=\frac1{507}$, we get that $$ 0\lt\frac2{\sqrt3}-\frac{15}{13}=\frac{\frac43-\frac{15^2}{13^2}}{\frac2{\sqrt3}+\frac{15}{13}}\lt\frac{\frac1{507}}{\frac{15}{13}+\frac{15}{13}}=\frac1{1170}\tag4 $$ Thus, $\frac{15}{13}\approx\frac2{\sqrt3}$ to $3$ decimal places.

We can also use the series $$ \log(u)=\sum_{k=0}^\infty\frac2{2k+1}\left(\frac{u-1}{u+1}\right)^{2k+1}\tag5 $$ where for $u=2+\sqrt3$, $$ \frac{u-1}{u+1}=\frac1{\sqrt3}\tag6 $$ Therefore, using the first $3$ terms to get $2$ decimal places $$ \begin{align} \log\left(2+\sqrt3\right) &=\frac2{\sqrt3}\left(\vphantom{\sum_{k=3}^\infty}\right.1+\frac19+\frac1{45}+\overbrace{\color{#090}{\sum_{k=3}^\infty}\color{#C00}{\frac1{2k+1}}\color{#090}{\frac1{3^k}}}^{\le\color{#C00}{\frac17}\color{#090}{\frac{1/27}{1-1/3}}=\frac1{126}}\left.\vphantom{\sum_{k=3}^\infty}\right)\tag{7a}\\ &\approx\frac{15}{13}\cdot\frac{17}{15}\tag{7b}\\[3pt] &=\frac{17}{13}\tag{7c} \end{align} $$ Thus, to $2$ decimal places, $$ \begin{align} \frac2{\sqrt3}+\frac13\log\left(2+\sqrt3\right) &\approx\frac{62}{39}\tag{8a}\\ &\approx1.59\tag{8b} \end{align} $$