About $I(a,b)=\int_{a}^{b}\sqrt{1+x+x^2+x^3+x^4}\text{ d}x$

Question

The following is an MCQ question, one should answer it without a calculator, within $3$ minutes.

---

Consider the expression

$$I(a,b)=\int_{a}^{b}\sqrt{1+x+x^2+x^3+x^4}\text{ d}x.$$

Which of the following is true?

I. $I(2,3)+I(3,4)=I(4,5)$

II. $I(2,3)+I(3,4)<I(4,5)$

III. $16<I(4,5)<25$

$\text{(A) }$I only

$\text{(B) }$II only

$\text{(C) }$III only

$\text{(D) }$I and III

$\text{(E) }$II and III

---

My Attempt (which, I think, is not a feasible way since it is not that accurate and also time consuming):

$\sqrt{1+x+x^2+x^3+x^4}\bigg|_{x=2}=\sqrt{1+2+2^2+2^3+2^4}=\sqrt{1+2+4+8+16}=\sqrt{31}\approx 6$

and

$\sqrt{1+x+x^2+x^3+x^4}\bigg|_{x=5}=\sqrt{1+5+5^2+5^3+5^4}=\sqrt{1+5+25+125+625}=\sqrt{781}\approx 28$

Now considering the integrand, approximately, as the straight line joining the points $(2,6)$ and $(5,28)$. The equation of the line can be found by first determining the slope, $m$:

$$m=\frac{28-6}{5-2}=\frac{22}{3}\approx 7$$

The equation of the line is therefore

$$y-6=7(x-2) \implies y=7x-8$$

now $I(a,b) \approx \int_{a}^{b} (7x-8)\text{ d}x=\frac{7}{2}x^2-8x \bigg|_{x=a}^{x=b} = \dots$

and $I(2,3) \approx \int_{2}^{3} (7x-8)\text{ d}x=9.5$

and $I(3,4) \approx \int_{3}^{4} (7x-8)\text{ d}x=16.5$

and $I(4,5) \approx \int_{4}^{5} (7x-8)\text{ d}x=23.5$

From these approximations, we can say, with low confidence, that $\text{(C)}$ is the correct option, which is not. In fact $\text{(E)}$ is the correct option. Noting that we can do better approximations, but that will cost time.

---

Attempt $\text{#}2$ (Not sure if it is a correct way):

In the interval $[2,5]$, function $f(x)=\sqrt{1+x+x^2+x^3+x^4}$ is always (positive), is always (increasing), and is always (concaving up). In that case, the inequalities will be the same when we do not consider the radical. So simply we evaluate:

$$\int_{2}^{3}(1+x+x^2+x^3+x^4)\text{ d}x$$

which gives $(x+x^2/2+x^3/3+x^4/4+x^5/5) \bigg|_{x=2}^{x=3}\approx 68$

Similarly, we evaluate:

$$\int_{3}^{4}(1+x+x^2+x^3+x^4)\text{ d}x \text{ and } \int_{4}^{5}(1+x+x^2+x^3+x^4)\text{ d}x$$

which give $\approx 217$ and $\approx 538$, respectively.

Clearly $68+217 < 538$. Hence II is true.

AND

$16^2 = 256 < 538 < 625 = 25^2$. Hence III is true.

Well, this attempt may also require time, but still it would be nice to know if it is a correct way. So, the approach (which might be wrong) is to observe that; between the limits of integration, the integrand is positive, increasing, and concaving up, the we deal with it without the radical. If this is wrong, please support by giving a counter-example.

---

Your help would be appreciated. Thanks!

Answer 1

First of all we will prove that for any $\,x\geqslant\sqrt2\,$ it results that

$\left(\dfrac12x+x^2\right)^{\!2}\!<1+x+x^2+x^3+x^4<\left(x+x^2\right)^2\,.\qquad\color{blue}{(1)}$

Proof :

For any $\,x\in\Bbb R\,$ it results that

$\begin{align}\left(\!\dfrac12x\!+\!x^2\!\right)^{\!2}\!\!&=\!\dfrac14x^2\!+\!x^3\!+\!x^4\!<\!\left(\!1\!+\!\dfrac12x\!\right)^{\!2}\!\!+\!\dfrac12x^2\!+\!\dfrac14x^2\!+\!x^3\!+\!x^4=\\[3pt]&=1\!+\!x\!+\!x^2\!+\!x^3\!+\!x^4\;.\end{align}$

Moreover, for any $\,x\geqslant\sqrt2\,$ it results that

$\begin{align}1\!+\!x\!+\!x^2\!+\!x^3\!+\!x^4&<2x\!+\!x^2\!+\!x^3\!+\!x^4\leqslant x^2\!\!\cdot\!x\!+\!x^2\!+\!x^3\!+\!x^4=\\[3pt]&=x^2\!+\!2x^3\!+\!x^4=\left(x\!+\!x^2\right)^2\;.\end{align}$

---

Now we will prove that for any $\,x\geqslant2\,$ it results that

$\dfrac38<\sqrt{1+x+x^2+x^3+x^4}-\dfrac12x-x^2 <\dfrac35\;.\quad\color{blue}{(2)}$

Proof :

For any $\,x\geqslant\sqrt2\,$ it results that

$\sqrt{1+x+x^2+x^3+x^4}-\dfrac12x-x^2=$

$=\dfrac{1+x+x^2+x^3+x^4-\left(\dfrac12x+x^2\right)^{\!2}}{\sqrt{1+x+x^2+x^3+x^4}+\dfrac12x+x^2}=$

$=\dfrac{1+x+\dfrac34x^2}{\sqrt{1+x+x^2+x^3+x^4}+\dfrac12x+x^2}\underset{\overbrace{\;\text{ by using }(1)\;}}{>}$

$>\dfrac{1+x+\dfrac34x^2}{\sqrt{\left(x+x^2\right)^2}+\dfrac12x+x^2}=\dfrac{1+x+\dfrac34x^2}{\dfrac32x+2x^2}>$

$>\dfrac{\dfrac9{16}x+\dfrac34x^2}{\dfrac32x+2x^2}=\dfrac{\dfrac38\left(\dfrac32x+2x^2\right)}{\dfrac32x+2x^2}=\dfrac38\;.$

Moreover, for any $\,x\geqslant2\,$ it results that

$\sqrt{1+x+x^2+x^3+x^4}-\dfrac12x-x^2=$

$=\dfrac{1+x+\dfrac34x^2}{\sqrt{1+x+x^2+x^3+x^4}+\dfrac12x+x^2}\underset{\overbrace{\;\text{ by using }(1)\;}}{<}$

$<\dfrac{1+x+\dfrac34x^2}{\sqrt{\left(\dfrac12x+x^2\right)^2}+\dfrac12x+x^2}=\dfrac{1+x+\dfrac34x^2}{x+2x^2}\leqslant$

$\leqslant\dfrac{1+x+\dfrac34x^2+\dfrac14\left(x^2-4\right)+\dfrac15x(x-2)}{x+2x^2}=$

$=\dfrac{\dfrac35x+\dfrac65x^2}{x+2x^2}=\dfrac{\dfrac35\left(x+2x^2\right)}{x+2x^2}=\dfrac35\;.$

---

From $\,(2)\,$ it follows that

$\sqrt{1\!+\!x\!+\!x^2\!+\!x^3\!+\!x^4}<\dfrac35\!+\!\dfrac12x\!+\!x^2\;,\;\;$ for any $\,x\geqslant2\;.\quad\color{blue}{(3)}$

$\sqrt{1\!+\!x\!+\!x^2\!+\!x^3\!+\!x^4}>\dfrac38\!+\!\dfrac12x\!+\!x^2 \;,\;\;$ for any $\,x\geqslant2\;.\quad\color{blue}{(4)}$

---

Now we will estimate the values of the definite integrals $\,I(2,3)+I(3,4)\,$ and $\,I(4,5)\,.$

$\begin{align}I(2,3)+I(3,4)&=\displaystyle\!\!\int_2^4\!\!\sqrt{1+x+x^2+x^3+x^4}\,\mathrm dx\!\!\underset{\overbrace{\;\text{ by using }(3)\;}}{<}\\[3pt]&<\int_2^4\!\left(\dfrac35+\dfrac12x+x^2\right)\mathrm dx=\\[3pt]&=\left[\dfrac35x+\dfrac14x^2+\dfrac13x^3\right]_2^4=\\[3pt]&=\dfrac{12}5+4+\dfrac{64}3-\dfrac65-1-\dfrac83=\dfrac{343}{15}\;.\end{align}$

$\begin{align}I(4,5)&=\displaystyle\!\!\int_4^5\!\!\sqrt{1+x+x^2+x^3+x^4}\,\mathrm dx\!\!\underset{\overbrace{\;\text{ by using }(4)\;}}{>}\\[3pt]&>\int_4^5\!\left(\dfrac38+\dfrac12x+x^2\right)\mathrm dx=\\[3pt]&=\left[\dfrac38x+\dfrac14x^2+\dfrac13x^3\right]_4^5=\\[3pt]&=\dfrac{15}8+\dfrac{25}4+\dfrac{125}3-\dfrac32-4-\dfrac{64}3=\dfrac{551}{24}\;.\end{align}$

Since $\;\dfrac{343}{15}<\dfrac{344}{15}=\dfrac{344\!\cdot\!8}{15\!\cdot\!8}=\dfrac{2752}{120}<\dfrac{2755}{120}=\dfrac{551}{24}$

it follows that

$I(2,3)+I(3,4)<\dfrac{343}{15}<\dfrac{551}{24}<I(4,5)\;.$

Hence, the statement II is true.

Moreover ,

$\begin{align}I(4,5)&=\displaystyle\!\!\int_4^5\!\!\sqrt{1+x+x^2+x^3+x^4}\,\mathrm dx\!\!\underset{\overbrace{\;\text{ by using }(3)\;}}{<}\\[3pt]&<\int_4^5\!\left(\dfrac35+\dfrac12x+x^2\right)\mathrm dx=\\[3pt]&=\left[\dfrac35x+\dfrac14x^2+\dfrac13x^3\right]_4^5=\\[3pt]&=3+\dfrac{25}4+\dfrac{125}3-\dfrac{12}5-4-\dfrac{64}3=\dfrac{1391}{60}\;.\end{align}$

Consequently ,

$16<22<\dfrac{551}{24}<I(4,5)<\dfrac{1391}{60}<24<25\,.$

Hence the statement III is true too.

Answer 2

The trick is that $\sqrt{1+x+x^2+x^3+x^4} = x^2 \sqrt{\frac1{x^4}+\frac1{x^3}+\frac1{x^2}+\frac1x +1}$ which is approximately $x^2$.

For the first two statements we need to compare $I(2,4)$ with $I(4,5)$ and we see that with this approximation that $I(4,5)-I(2,4) = \frac{61}3-\frac{56}3 = \frac53$ which we can see is almost $2$. This gives us II. as true. For the final question we noted that $\frac{56}3 = 18 \frac13$ so III. is true as well.

Answer 3

For III, $\int_{4}^{5}\sqrt{x^4}\text{ d}x<\int_{4}^{5}\sqrt{1+x+x^2+x^3+x^4}\text{ d}x<\int_{4}^{5}\sqrt{(x^2+\frac{x}{2}+1)^2}\text{ d}x$ It's a very good approximation because the first two largest terms are conserved as is, on solving you should get a range for III somewhere between $(19,24)$.

Answer 4

Three minutes is a short time. What we need is some simple, plausible reasoning and a bit of luck. Here is an attempt.

Using the finite geometric series formula we have \begin{align*} \int_{a}^{b}\sqrt{1+x+x^2+x^3+x^4+x^5}\,dx=\int_{a}^b\sqrt{\frac{x^5-1}{x-1}}\,dx \end{align*}

We consider $f(x)=\sqrt{\frac{x^5-1}{x-1}}$ and calculate for $x\in\{2,3,4,5\}$ \begin{align*} \color{blue}{f(2)}&=\sqrt{31}\color{blue}{\approx 5.5}\tag{$5<\sqrt{31}<6$}\\ \color{blue}{f(3)}&=\sqrt{121}\color{blue}{=11}\\ \color{blue}{f(4)}&\approx\sqrt{\frac{1}{3}2^{10}}=\sqrt{\frac{1024}{3}}\color{blue}{\approx 18}\tag{$18^2=324$}\\ \color{blue}{f(5)}&\approx\sqrt{\frac{1}{4}5^5}\approx\frac{1}{2}25\sqrt{5}\color{blue}{\approx 28}\tag{$\sqrt{5}\approx 2.2$} \end{align*}

Now we do some linear approximation \begin{align*} \color{blue}{I(n,n+1)}&\approx f(n)+\frac{1}{2}\left(f(n+1)-f(n)\right)\color{blue}{=\frac{1}{2}\left(f(n+1)+f(n)\right)}\\ \\ I(2,3)&\approx\frac{1}{2}\left(11+5.5\right)=8.25\\ I(3,4)&\approx\frac{1}{2}\left(18+11\right)=14.5\\ I(4,5)&\approx\frac{1}{2}\left(28+18\right)=23\\ \end{align*}

We obtain \begin{align*} \color{blue}{I(2,3)+I(3,4)}&\approx 8.25+14.5=22.75\color{blue}{< 23\approx I(4,5)}\tag{II.}\\ \color{blue}{16}&\color{blue}{<23\approx I(4,5)<25}\tag{III.} \end{align*} and conclude $\color{blue}{\mathrm{(E)}}$ is the solution.

Answer 5

Remark: The following can not be done during three minutes.

First, III is true.

We have $$I(4, 5) \ge \int_4^5 x^2\,\mathrm{d} x = \frac{61}{3} > 16,$$ and $$I(4, 5) \le \int_4^5 (x^2 + x/2 + 1)\,\mathrm{d} x = \frac{283}{12} < 25$$ where we use $$(x^2 + x/2 + 1)^2 = x^4 + x^3 + (2 + 1/4)x^2 + x + 1 \ge 1 + x + x^2 + x^3 + x^4.$$

---

Second, II is true.

Note that $f(x) := \sqrt{1 + x + x^2 + x^3 + x^4}$ is convex on $x \ge 0$. We have $I(2, 3) \le \frac12[f(2) + f(3)]$ and $I(3, 4) \le \frac12[f(3) + f(4)]$. Thus, we have $$I(2, 3) + I(3, 4) \le \frac12 f(2) + f(3) + \frac12 f(4) = \frac12\sqrt{31} + 11 + \frac12\sqrt{341}.$$

To find a lower bound of $I(4, 5)$, note that \begin{align*} 1 + x + x^2 + x^3 + x^4 &= (x^2 + x/2)^2 + \frac34 x^2 + x + 1\\ &= (x^2 + x/2)^2 + \frac34 (x^2 + x/2) + \frac58 x + 1\\ &= (x^2 + x/2 + 3/8)^2 + \frac58 x + \frac{55}{64}. \end{align*} We hope to find a constant $c_1 > 3/8$ such that $$ (x^2 + x/2 + 3/8)^2 + \frac58 x + \frac{55}{64} \ge (x^2 + x/2 + c_1)^2, \quad \forall x \in [4, 5]$$ that is $$g(x) := \frac58 x + \frac{55}{64} - (c_1 - 3/8)(2x^2 + x + 3/8 + c_1)\ge 0, \quad \forall x \in [4, 5]. \tag{1}$$ Since $g(x)$ is concave. (1) is equivalent to $g(4) \ge 0, g(5) \ge 0$. We have $c_1 \le -\frac{55}{2} + \sqrt{781}$. We choose $c_1 = 4/9$. We have $$\sqrt{1 + x + x^2 + x^3 + x^4} \ge x^2 + x/2 + 4/9, \quad \forall x \in [4, 5].$$

We have $$I(4, 5) \ge \int_4^5 (x^2 + x/2 + 4/9)\,\mathrm{d} x = 829/36.$$

Thus, $$I(4, 5) - I(2, 3) - I(3, 4) \ge 829/36 - \left(\frac12\sqrt{31} + 11 + \frac12\sqrt{341}\right) > 0.$$

We are done.

Answer 6

This can be done in 3 minutes if you know the dirty method :

All body knows that $f(x)=x^2$ is convex and the inverse function of $f$ is concave so we have :

$$\sqrt{\frac{x^5-1}{x-1}}\geq \sqrt{f'(a^2)\left(\sqrt{\frac{x^5-1}{x-1}}-a^2\right)+f(a^2)}$$

A lower bound can be find with $f^{-1}(x)=\sqrt{x}$