Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $$ \begin{eqnarray*} F(x) &=& \displaystyle\int_{0}^{x} \dfrac{1}{\sqrt{1 + t^2}} dt = \log(x + \sqrt{x^2 + 1})\\ G(x) &=& \displaystyle\int_{1}^{x} \dfrac{1}{t} dt = \log(x) \end{eqnarray*} $$ Prove that $F(x) \geq G(x) $ for all $x \geq 1$.

My initial approach is to integrate them, but my teacher said there is a direct way using the Fundamental Theorem of Calculus, and I have no idea how could I approach this problem? My thought is to use the lower sum and upper sum, but that doesn't seem promising. So could anyone share me some ideas?

Edit
Please ignore the $\log(x + \sqrt{x^2 +1})$ part because it was the result after I integrated $F(x)$.

share|improve this question
2  
If the "given" is really given, then what's problem? We clearly have $$\forall\,x>1\,\,,\,\,x+\sqrt {x^2+1}>x\,\,\Longrightarrow \log(x+\sqrt{x^2+1})>\log x...$$ –  DonAntonio Dec 5 '12 at 4:12
    
@DonAntonio: Sorry for the confusion. The $\log(x + \sqrt{x^2 + 1})$ is after I integrated $F(x)$. So I have to show that $F(x) \geq G(x)$ without referring to $\log(x + \sqrt{x^2 + 1})$ and $\log(x)$. –  Chan Dec 5 '12 at 4:18
add comment

3 Answers

up vote 2 down vote accepted

He was likely referring to the Second Fundamental Theorem of Calculus, which states $$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$. You differentiate both sides to see that $$\frac{d}{dx}\int_0^x\frac{dt}{\sqrt{1+t^2}}=\frac{1}{\sqrt{1+x^2}}\geq\frac{d}{dx}\int_1^x\frac{dt}{t}=\frac{1}{t}$$ Then, you will just have to check the case $x=1$, where you can see that $$\int_0^1\frac{dt}{\sqrt{1+t^2}}>\int_1^1\frac{dt}{t}$$ since $\frac{1}{\sqrt{1+x^2}}$ is positive-definite and the right-hand side is obviously $0$.

Then, we have that since $F(1)>G(1)$ and $\frac{dF}{dx}\geq\frac{dG}{dx}$, $$F(x)\geq G(x)$$ for all $x\geq 1$.

share|improve this answer
1  
But it's not true that $1/(\sqrt{x^2+1}) \ge 1/x$. That is true if and only if $x\ge \sqrt{x^2+1} > x$ which is false for $x\ge 1$... –  Potato Dec 5 '12 at 4:53
add comment

Show that $F(1)\geq G(1)$ and that $F'(x)>G'(x)$ if $x>1$. Having those facts in hand, it shouldn't be too tough.

share|improve this answer
1  
I don't think your derivative inequality is true. See my comment on the other answer. –  Potato Dec 5 '12 at 4:55
add comment

Substituting $t=u-1$ in the expression for $F(x)$, we get \begin{eqnarray*} F(x)&=&\int_1^{x+1}\frac{1}{\sqrt{u^2-2(u-1)}}\,du\\ &\geq& \int_1^x\frac{1}{u}\,du\\ &=&G(x) \end{eqnarray*}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.