Prove value of definite integral is greater than b

[lep11]

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.

 

[blue_leaf77]

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

 

[lep11]

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]

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

Yes, but only within the interval $[0,b]$.

 

[lep11]

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 $x≠0$?

 

[Ray Vickson]

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]

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.

 

[lep11]

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

Is that obvious or need to be proven?

 

[blue_leaf77]

Is that obvious or need to be proven?

It's obvious but why do you ask about it now?

 

[lep11]

It's obvious but why do you ask about it now?

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?

 

[blue_leaf77]

Then in that case, prove it. It should require only one line.