# Prove value of definite integral is greater than b

## Homework Statement

Prove ##\int_{0}^{b} \frac{e^x}{1+x} dx>b## for every ##b>0##.

## 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.

Last edited:

blue_leaf77
Homework Helper
Consider the fact that ##\int_0^b 1 dx = b##.

Consider the fact that ##\int_0^b 1 dx = b##.
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##?

blue_leaf77
Homework Helper
I need to show that ## \frac{e^x}{1+x}>1##?
Yes, but only within the interval ##[0,b]##.

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##?

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
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##?

No. When ##x = 0## we have ##f(x) = 1##, but ##f## is NOT the constant function!

Anyway, on ##[0,\infty)## you have ##f(x) \geq 1##, with strict inequality when ##x > 0##.

blue_leaf77
Homework Helper
Thanks to Ray I know that the OP edited his last post after the last time I checked it.
##\int_{0}^{b} \frac{e^x}{1+x} dx## = ##\int_{0}^{b} 1 dx=b## iff ## x=0##.
This line is unnecessary. Moreover those integrals are not a function of ##x## (if you want you can define it as a function of ##b##, though), thus evaluating it at certain value of ##x## makes no sense. It's like you want to evaluate a function ##f(y)## at ##x=0##.
is there a mistake in the problem statement because ∫b0ex1+xdx∫0bex1+xdx\int_{0}^{b} \frac{e^x}{1+x} dx can have value of b when x=0x=0x=0? I think it wouldn't contradict anything, would it?
$$\int_{0}^{b} \frac{e^x}{1+x} dx = b$$
is true if when ##b=0##, but the problem states that the condition for ##b## is that it is bigger than zero, so there is no contradiction.

Consider the fact that ##\int_0^b 1 dx = b##.
Is that obvious or need to be proven?

blue_leaf77
Homework Helper
Is that obvious or need to be proven?