Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$.

Question

I'm trying to solve this:

Which of the following is the closest to the value of this integral?

$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$

(A) 1

(B) 1.2

(C) 1.6

(D) 2

(E) The integral doesn't converge.

I've found a lower bound by manually calculating $\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.

From GRE problem sets by UChicago

Answer 1

Starting from

$$\int_0^1\sqrt{1+{1\over3x}}\,dx=2\int_0^1\sqrt{t^2+{1\over3}}\,dt$$

(from the subsitution $x=t^2$) as in Yves Daoust's answer, integration by parts gives

$$\int_0^1\sqrt{t^2+{1\over3}}\,dt=t\sqrt{t^2+{1\over3}}\Big|_0^1-\int_0^1{t^2\over\sqrt{t^2+{1\over3}}}\,dt={2\over\sqrt3}-\int_0^1{t^2+{1\over3}-{1\over3}\over\sqrt{t^2+{1\over3}}}\,dt$$

hence

$$2\int_0^1\sqrt{t^2+{1\over3}}\,dt={2\over\sqrt3}+{1\over3}\int_0^1{dt\over\sqrt{t^2+{1\over3}}}={2\over\sqrt3}+{1\over\sqrt3}\int_0^1{dt\over\sqrt{3t^2+1}}$$

Since $1\le\sqrt{3t^2+1}\le2$ for $0\le t\le1$, we have

$${1\over2}\le\int_0^1{dt\over\sqrt{3t^2+1}}\le1$$

Thus

$${2\over\sqrt3}+{1\over2\sqrt3}\le2\int_0^1\sqrt{t^2+{1\over3}}\,dt\le{2\over\sqrt3}+{1\over\sqrt3}$$

Now

$${2\over\sqrt3}+{1\over2\sqrt3}={5\sqrt3\over6}=\sqrt{75\over36}\gt\sqrt2\gt1.4$$

and

$${2\over\sqrt3}+{1\over\sqrt3}=\sqrt3\lt\sqrt{3.24}=1.8$$

Consquently

$$1.4\lt\int_0^1\sqrt{1+{1\over3x}}\,dx\lt1.8$$

and thus (C) is the correct answer.

Answer 2

$$\int_0^1\sqrt{1+\dfrac1{3x}}dx=2\int_0^1\sqrt{t^2+\dfrac13}dt$$ proves convergence.

Then

$$\frac1{\sqrt 3}\le\sqrt{t^2+\frac13}\le t+\frac1{\sqrt3}$$ implies

$$\frac2{\sqrt 3}\approx 1.155\le I\le1+\frac2{\sqrt 3}\approx2.155$$

A tighter upper bound is obtained by noting that the function is convex and

$$\sqrt{t^2+\frac13}\le \frac1{\sqrt3}+t\left(\sqrt{\frac 43}-\frac1{\sqrt3}\right),$$ giving $$I\le\sqrt3\approx1.732$$ A tighter lower bound could be found by considering the tangents at both endpoints up to their intersection, but we can already conclude C.

The exact value is $$1.5936865\cdots$$ The bounds can be computed by hand, by squaring to avoid square roots.

Answer 3

We know that,

${\displaystyle\int}\sqrt{\dfrac{1}{3x}+1}\,\mathrm{d}x$=$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{3}}}}{\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$

Substitute $u=\sqrt{\dfrac{1}{x}+3}$ and $\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\frac{1}{x}+3}x^2}$,i.e $\mathrm{d}x=-2\sqrt{\dfrac{1}{x}+3}x^2\,\mathrm{d}u$

${\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$=$-\class{steps-node}{\cssId{steps-node-2}{2}}{\displaystyle\int}\dfrac{u^2}{\left(u^2-3\right)^2}\,\mathrm{d}u$ $={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-5}{u^2-3}}}{\left(u^2-3\right)^2}+\dfrac{\class{steps-node}{\cssId{steps-node-6}{3}}}{\left(u^2-3\right)^2}\right)\mathrm{d}u$

$={\displaystyle\int}\dfrac{1}{u^2-3}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-7}{3}}{\displaystyle\int}\dfrac{1}{\left(u^2-3\right)^2}\,\mathrm{d}u$

Perform partial fraction decomposition:

$={\displaystyle\int}\left(\dfrac{1}{2\sqrt{3}\left(u-\sqrt{3}\right)}-\dfrac{1}{2\sqrt{3}\left(u+\sqrt{3}\right)}\right)\mathrm{d}u$ + ${\displaystyle\int}\dfrac{1}{u+\sqrt{3}}\,\mathrm{d}u$

On solving this further we get

$\dfrac{\sqrt{\frac{1}{3x}+1}}{3\left(\frac{1}{3x}+1\right)-3}+\dfrac{\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)}{6}-\dfrac{\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$

I.e,

$\dfrac{6\sqrt{\frac{1}{3x}+1}x+\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)-\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$

$\boldsymbol{\int\limits^{1}_{0}{f(x)}\,\mathrm{d}x =}$=${1\over6}[4 \sqrt(3) + \ln(7 + 4 \sqrt(3)]$

Approximation: $1.593686504020857$,

i.e $1.6$.

The answer is option $(C)$

Answer 4

Let's try to use integration by parts to $I = \int\limits_0^1 \sqrt{1 + \frac{1}{3x}}dx$. First, transform integral into $\frac{2}{\sqrt3}\int\limits_0^1\frac{\sqrt{1 + 3x}}{2\sqrt{x}}$. Now $u = \sqrt{3x+1}$ and $dv = \frac{dx}{2\sqrt{x}}$ and what we get after IBP is $$\frac{2}{\sqrt3}\sqrt{x(3x+1)}|_0^1 - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx$$. We have $$\frac{5}{2\sqrt3} = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x + x}}dx < I <\frac{4}{\sqrt3} - \sqrt3\int\limits_0^1 \sqrt{\frac{x}{3 + 1}}dx = \frac{4}{\sqrt3} - \frac{\sqrt3}{2} \frac{2}{3}x\sqrt{x}|_0^1 = \sqrt3$$ $\frac{5}{2\sqrt3} \approx 1.44$ and $\sqrt3 \approx 1.73$, so the answer is (C).

If one doesn't know the value of $\sqrt3$, we can check that $1.7^2 < 3 < 1.8^2$ and then $3 < 1.75^2$. Therefore, $\sqrt3 < 1.75$. From it we have $\frac{5}{2\sqrt3} > \frac{5}{2\cdot1.75} > 1.42$ and, for integral, $1.42 < I < 1.75$.

Answer 5

Since $\sqrt{t^2+1/3}$ is a convex function on $[0,1]$, you may simply use the Hermite-Hadamard inequality to derive that

$$ \sqrt{2+\frac{1}{3}}\leq 2\int_{0}^{1}\sqrt{t^2+1/3}\,dt \leq \sqrt{3} $$ so $(C)$ is the correct option.

Answer 6

Although it is an old post, I would like to show my attempt and people answers. However, there I was not trying to answer it as a timed test, but I was checking if that is a valid way of answering it. In fact, it is.

See My Post Here

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Here is a recent attempt, it may be useful for someone who sees the problem after a long time since posted.

If $\int_{a}^{b}f(x)\text{ d}x$ converges, then it can be estimated by the following:

$$\large\boxed{\int_{a}^{b}f(x)\text{ d}x \approx \frac{b-a}{4}\bigg(f(a)+2f\bigg(\frac{a+b}{2}\bigg)+f(b)\bigg)}$$

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

Put $x=t^2 \implies \text{d}x=2t\text{ d}t$. As $x$ ranges from $0$ to $1$, $t$ ranges from $0$ to $1$.

$$\int_{0}^{1}\sqrt{1+\frac{1}{3x}}\text{ d}x \xrightarrow{x \rightarrow t^2} \int_{0}^{1}\sqrt{4t^2+\frac{4}{3}}\text{ d}t \space\space\space \text{[This proves convergence]}$$

Using the boxed formula above, we get:

$$\int_{0}^{1}\sqrt{4t^2+\frac{4}{3}}\text{ d}t \approx \frac{1-0}{4}\bigg(\sqrt{\frac{4}{3}}+2\sqrt{\frac{7}{3}}+\sqrt{\frac{16}{3}}\bigg)$$

$$=\frac{1}{4\sqrt{3}}\bigg(2+2\sqrt{7}+4\bigg)=\frac{6+2\sqrt{7}}{4\sqrt{3}}=\frac{3+\sqrt{7}}{2\sqrt{3}}$$

But $\sqrt{7} \approx 2.6$ and $\sqrt{3} \approx 1.7$ implying that

$$\frac{3+\sqrt{7}}{2\sqrt{3}} \approx \frac{3+2.6}{2(1.7)}=\frac{5.6}{3.4}=\frac{28}{17} \approx 1.6$$