Prove value of definite integral is greater than b

1. The problem statement, all variables and given/known data
Prove ##\int_{0}^{b} \frac{e^x}{1+x} dx>b## for every ##b>0##.

3. The attempt at a solution
Honestly, no idea. I suppose I cannot just calculate the definite integral and estimate. I think it needs to be proven in a more rigorious way.

 

Hmm... if ##f(x)>g(x)##, then it follows that ##\int_{a}^{b} f(x) dx## > ##\int_{a}^{b} g(x)dx##?

I need to show that ## \frac{e^x}{1+x}>1##?

 

The first hint you gave was very helpful.

Let ##0≤x≤b##,

then ## \frac{e^x}{1+x}≥1##⇔## e^x≥1+x## which is true since we know the expansion ##e^x=1+x+\frac{x^2}{2}+...≥1+x##

Therefore ##\int_{0}^{b} \frac{e^x}{1+x} dx## ≥ ##\int_{0}^{b} 1 dx=b##

and ##\int_{0}^{b} \frac{e^x}{1+x} dx## = ##\int_{0}^{b} 1 dx=b## iff ## x=0##.

I think I am almost there...is there a mistake in the problem statement because ##\int_{0}^{b} \frac{e^x}{1+x} dx## can have value of b when ##x=0##? I think it wouldn't contradict anything, would it?

if ##x=0##, then ##f(x)=\frac{e^x}{1+x}## ruduces to constant function so maybe that's why we assume ##x≠0##?

 

Is that obvious or need to be proven?

 

I am asking about it now because I am checking and reviewing my work. This piece of homework is due to tomorrow. It is very obvious to me that ##\int_0^b 1 dx = b##, but we haven't proven that in class and I'm wondering if I am supposed to prove that as well or if I can just say to my teacher that "oh, that's obvious." And it's better ask late than never, right?