Constantly (well, not really) 0 integral

Since we aren’t given any values for $a$ or $b,$ let’s see if we can find what the integral is equal to.

Let $u=\ln x.$ Then $du=\frac{1}{x}dx.$ Proceeding with the $u\text{-substitution,}$ the integral becomes this. $\int_{\ln a}^{\ln b}u\text{ }du$ This is equal to $\left.\frac{1}{2}u^2\right]_{\ln a}^{\ln b}$ $=$ $\frac{(\ln a)^2-(\ln b)^2}{2}=0$

A little rearranging gives $(\ln a)^2=(\ln b)^2,$ so $\ln a=\pm\ln b\Rightarrow a=b^{\pm1}$ We have two possible relations that could be $f(a).$ These are $f(a)=a$ and $f(a)=\frac{1}{a}.$ However, since $f(a)$ is non-linear, we can eliminate the possibility that $f(a)=a$ and know that $f(a)=\frac{1}{a}.$

Now we need to find the minimum value of a new relation $g(a)=a+f(a)=a+\frac{1}{a}.$ Deriving, we get $g’(a)=1-\frac{1}{a^2}$ This equals $0$ when $a=\pm1,$ but $-1$ is not in the domain of $f(a)$ and is therefore not in the domain of $g(a)$ either, so we can disregard it. Applying the First Derivative Test, we find that $a=1$ is the $a\text{-coordinate}$ of the local minimum of $g(a),$ but since $g(a)$ does not change direction again and is continuous on its domain, $g(a)=2$ is also the absolute minimum.

Therefore, the point $(a,b)$ we are looking for is $(1,1)$ so $a+b=\boxed{2}$

If we put $\ln x = t$ , we get $\frac{dx}{x} = dt$ and the integral as $1/2((\ln b)^2 - (\ln a)^2$ ) = $\frac{1}{2} \ln (\frac{b}{a}) \ln (ab)$ .

For it to be $0$ , either $b = \frac{1}{a}$ or $b=a$ . I assume $b = \frac{1}{a}$ , otherwise there would be no min. value.

Hence, $a + b = a + \frac{1}{a} \geq \boxed{2}$ for $a > 0$

First we integrate. We find that the indefinite integral is 1/2 times ln(x) squared. Let's start just by setting this equal to zero. Ln of 1 is zero (anything to the zero is one), therefore if a and b both were one, then the integral will come down to zero. Of course, we must remember: this is the same thing as asking what the limit of the size of delta x is. Of course, this will be true for any numbers that are equal, because it's a tiny sliver of space that is actually not quite zero, but we'll just say that it is. That was the easy part. Now, this is going to require some differential equations and a bit of topology. We are going to do one of the Millenium Challenges. We are going to add 1 to 1. I know, I know, it's impossible. Nonetheless! The first thing we do is add: 1+1. We get 2. Done.