Calculus II question, prove ∫ ≤ 14/3 without evaluating the integral

[dillon131222]

Homework Statement

Show that ∫(0,3) √(x+e^-x)dx ≤ 14/3 (hint: do not attempt to evaluate the integral)when looking at the integral from 0 to 3 on the graph I can see that this is true, but not sure how to go about showing this without evaluating the integral, any help as to how to go about this would be great.

thanks in advance

 

[SqueeSpleen]

I don't know how exactly they want you to solve this, but you can find functions that are greater than your's in the integration interval.
For example:
√(x+e^-x) ≤ √3 in (0,2) and √(x+e^-x) ≤ √4 in (2,3).
So ∫(0,3) √(x+e^-x)dx ≤ 2√3+2
This bounding is awful but I think it's useful as example.

 

[TheForce]

Find the maximum value of the function on the interval and multiply that by the interval length, tat is the maximum possible value for the integral (the upper bound). Try that and let us know how it goes.

 

[LCKurtz]

I think you will have good luck by overestimating the $e^{-x}$ and evaluating the resulting integral.

 

[dillon131222]

Find the maximum value of the function on the interval and multiply that by the interval length, tat is the maximum possible value for the integral (the upper bound). Try that and let us know how it goes.

so find the maximum value of √(x+e^-x) and multiply by the interval length of 3?

the function is increasing from [0,3] so max is at position 3, √(3+e^-3 ≈ 1.7464, 1.7464x3 ≈ 5.2391, which is greater than 14/3 so guessing I'm doing something wrong for your suggestion?


gonna attempt the other suggestions now

 

[dillon131222]

I don't know how exactly they want you to solve this, but you can find functions that are greater than your's in the integration interval.
For example:
√(x+e^-x) ≤ √3 in (0,2) and √(x+e^-x) ≤ √4 in (2,3).
So ∫(0,3) √(x+e^-x)dx ≤ 2√3+2
This bounding is awful but I think it's useful as example.

need to show that the integral is ≤ 14/3, so don't see how showing that its ≤ 2√3+2 which = 5.46 helps(?) or am i misunderstanding?

I think you will have good luck by overestimating the e^-x and evaluating the resulting integral.

not sure what you mean by overestimating e^-x

really not sure what's wanted for this question, not very good at this yet :/ I am just starting to understand the basics of integrals right now

 

[jbunniii]

need to show that the integral is ≤ 14/3, so don't see how showing that its ≤ 2√3+2 which = 5.46 helps(?) or am i misunderstanding?

You have found a rather crude bound by drawing a horizontal line at $y = \sqrt{3 + e^{-3}}$. But the function value is much smaller at $x = 0$: $\sqrt{0 + e^0} = 1$. Try taking advantage of that fact, for example, by drawing a line from the point $(x=0, y=1)$ to $(x = 3, y=\sqrt{3 + e^{-3}})$. Does your function stay below that line? If so, that will give you a smaller bound.

 

[STEMucator]

You could always compare your integral to something else since it's not integrable by elementary means, but this question is odd as is.

I was thinking that you should bound your integral with an easier integral to evaluate which is bigger, but still less than or equal to 14/3.

So since the bigger integral would evaluate easily to something nice and the value was less than 14/3, it would imply your integral was also less than 14/3.

 

[LCKurtz]

need to show that the integral is ≤ 14/3...
not sure what you mean by overestimating e^-x

What is the largest value $e^{-x}$ attains on [0,3]? Overestimate the integral by replacing $e^{-x}$ with that largest value. It will give you a simpler integral you can work.

 

[dillon131222]

You have found a rather crude bound by drawing a horizontal line at $y = \sqrt{3 + e^{-3}}$. But the function value is much smaller at $x = 0$: $\sqrt{0 + e^0} = 1$. Try taking advantage of that fact, for example, by drawing a line from the point $(x=0, y=1)$ to $(x = 3, y=\sqrt{3 + e^{-3}})$. Does your function stay below that line? If so, that will give you a smaller bound.

dont understand how to find a bound, isn't the integral bound from 0 to 3?

 

[dillon131222]

What is the largest value $e^{-x}$ attains on [0,3]? Overestimate the integral by replacing $e^{-x}$ with that largest value. It will give you a simpler integral you can work.

I see, that worked perfectly and much simpler, thank you so much :D