# Prove value of definite integral is greater than b

• lep11
In summary: I need to show that ## \frac{e^x}{1+x}>1##?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.

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

blue_leaf77 said:
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##?

lep11 said:
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:
lep11 said:
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##.

Thanks to Ray I know that the OP edited his last post after the last time I checked it.
lep11 said:
##\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##.
lep11 said:
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.

blue_leaf77 said:
Consider the fact that ##\int_0^b 1 dx = b##.
Is that obvious or need to be proven?

lep11 said:
Is that obvious or need to be proven?
It's obvious but why do you ask about it now?

blue_leaf77 said:
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:
Then in that case, prove it. It should require only one line.

## 1. What is a definite integral and how is it calculated?

A definite integral is a mathematical concept that represents the area under a curve on a graph. It is calculated by dividing the area into smaller rectangles and summing their areas, using a limit to approach an accurate value. This process is known as Riemann integration.

## 2. How can we prove that the value of a definite integral is greater than a given number, b?

To prove that the value of a definite integral is greater than b, we can use the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b], then for any value y between f(a) and f(b), there exists a value c between a and b such that f(c) = y. By setting y equal to b and applying this theorem to the function representing the definite integral, we can show that there exists a value c between the upper and lower limits of integration where the definite integral is equal to b. Therefore, the value of the definite integral must be greater than b.

## 3. Can a definite integral ever be negative?

Yes, a definite integral can be negative. This occurs when the function being integrated has values below the x-axis, which results in negative areas being added to the overall sum. In this case, the value of the definite integral represents the net area between the function and the x-axis.

## 4. How does the value of a definite integral change when the limits of integration are switched?

The value of a definite integral remains the same when the limits of integration are switched. This is because the definite integral represents the net area between the function and the x-axis, regardless of the direction of integration. Switching the limits simply changes the sign of the integral, but the magnitude remains the same.

## 5. Can the value of a definite integral be infinite?

Yes, the value of a definite integral can be infinite. This occurs when the function being integrated has unbounded values, meaning that it approaches infinity at one or both of the limits of integration. In this case, the definite integral represents the area under an unbounded curve, which is infinite.

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