Prove value of definite integral is greater than b

  • Thread starter lep11
  • Start date
  • #1
380
7

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:

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
784
Consider the fact that ##\int_0^b 1 dx = b##.
 
  • #3
380
7
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##?
 
  • #4
blue_leaf77
Science Advisor
Homework Helper
2,629
784
I need to show that ## \frac{e^x}{1+x}>1##?
Yes, but only within the interval ##[0,b]##.
 
  • #5
380
7
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:
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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##.
 
  • #7
blue_leaf77
Science Advisor
Homework Helper
2,629
784
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.
 
  • #8
380
7
Consider the fact that ##\int_0^b 1 dx = b##.
Is that obvious or need to be proven?
 
  • #9
blue_leaf77
Science Advisor
Homework Helper
2,629
784
Is that obvious or need to be proven?
It's obvious but why do you ask about it now?
 
  • #10
380
7
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?
 
Last edited:
  • #11
blue_leaf77
Science Advisor
Homework Helper
2,629
784
Then in that case, prove it. It should require only one line.
 

Related Threads on Prove value of definite integral is greater than b

Replies
2
Views
7K
Replies
6
Views
3K
Replies
3
Views
5K
Replies
1
Views
6K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
4
Views
2K
Top