# Thread: Estimate The Value Of Integral

1. ## Estimate The Value Of Integral

Hi Everyone,

I am having trouble getting started on this problem.

Book Problem 47

Given that $\,3\, \leq\, f(x)\, \leq\, 6\,$ for $\, -8\, \leq\, x\, \leq\, 8,\,$ estimate the value of $\, \int_{-8}^{8}\, f(x)\, dx$

I honestly have no idea where to go with this one. Any help would be appreciated.

2. Well, you know that the integral can be represented by the area between the graph of the function and the x-axis. So, then, to find the upper bound of the integral, you have to find a function that satisfies the domain and range with the most area underneath it. How would you go about that? And similarly, for the lower bound, can you find the function with the least area underneath it? I recommend sketching several possible functions which have the given domain and range to see if you can find a pattern.

3. Originally Posted by ksdhart
Well, you know that the integral can be represented by the area between the graph of the function and the x-axis.
That is not always true.
Consider $\displaystyle\int_0^\pi {\cos (t)dt} = 0$. But the bounded area is two.

4. Originally Posted by chrisalau32
Book Problem 47

Given that $\,3\, \leq\, f(x)\, \leq\, 6\,$ for $\, -8\, \leq\, x\, \leq\, 8,\,$ estimate the value of $\, \int_{-8}^{8}\, f(x)\, dx$

I honestly have no idea where to go with this one.
What is the lowest value that f(x) can be? If f(x) is this lowest value over the whole length (from x = -8 to x = 8), what would be the area?

What is the highest value that f(x) can be? If f(x) is this highest value over the whole length, what would be the area?

So what are the bounds on the area?

5. Originally Posted by stapel
What is the lowest value that f(x) can be? If f(x) is this lowest value over the whole length (from x = -8 to x = 8), what would be the area?

What is the highest value that f(x) can be? If f(x) is this highest value over the whole length, what would be the area?

So what are the bounds on the area?

The lowest value f(x) could be is 3 and the highest that it could be is 6 but I don't see how that helps me with this problem. Not saying you are wrong, just saying I don't really understand.

6. Originally Posted by chrisalau32
The lowest value f(x) could be is 3 and the highest that it could be is 6 but I don't see how that helps me with this problem. Not saying you are wrong, just saying I don't really understand.
I don't know why this question irritates me so, but is does. You should have been given this theorem.

If each of $g~\&~h$ is an intergable function on $[a,b]$ and if $\left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {g(x) \le h(x)} \right]$ then $\displaystyle\int_a^b {g(x)dx} \le \int_a^b {h(x)dx}$.

Thus your answer is $\displaystyle 3\cdot 16\le\int_{-8}^8 {f(x)dx} \le 6\cdot 16$

7. I find that to solve problems - I need to sketch the information.

If you do that - the theorem that pka is talking about (and the solution) would basically jump out.

Knowing that the maximum value of f(x) = 6 and the minimum value is 3 then

$\displaystyle{\int_{-8}^{8}f(x)|_{min}dx \ \le \ \int_{-8}^{8}f(x)dx \le \ \int_{-8}^{8}f(x)|_{max}dx }$

= $\ \displaystyle{\int_{-8}^{8}3 \ dx \ \le \ \int_{-8}^{8}f(x)dx \le \ \int_{-8}^{8}6 \ dx }$

8. Originally Posted by chrisalau32
The lowest value f(x) could be is 3 and the highest that it could be is 6 but I don't see how that helps me with this problem. Not saying you are wrong, just saying I don't really understand.
Okay. If the function is always at the lowest value, then what shape is the area? How wide is the area? What is the height of the area? What then is the area?

If the function is always at the highest value, then what shape is the area? How wide is the area? What is the height of the area? When then is the area?

Since the function is always somewhere between the low and high values (including "or equal to" those values), what then are the bounds on the area?