Question

Please Help I am stuck on this question!
Given that 2 is less than or equal to f(x)less than or equal to 3
for -9 less than or equal to x less than or equal to 5 use property 8 to estimate the value of integral from 5 to -9 f(x)dx.
I will write this out as an equation below as well.

Answer 1

Given that \[2\le f(x)\le3 for -9\le x \le5 \] use property 8 to estimate the value of \[\int\limits_{-9}^{5}f(x)dx\]

Answer 2

I know that this is what property 8 says but I am still lost as to how to use it.
If\[m \le f(x)\le M for a \le x \le b, then, m(b-a)\le \int\limits_{a}^{b}f(x)dx \le M(b-a)\]

Answer 3

This is an application of the Mean Value Theorem whereby "m" is your minimum for f(x) and "M" is your maximum and b-a is 5 - (-9) = 14. Plug those values in for b-a, m, and M and you will "squeeze" the integral into that range.

Answer 4

Could you help show me a little further how this works? I just need to visualize this a little more to fully understand. Thanks.

Answer 5

Okay, first, m is 2 and M is 3, for your computations. Now, I have a bit of typing to do, so please bear with me. Here goes...

Answer 6

Got it know sorry that was a really dumb question. so it is between 28 and 42

Answer 7

From the Mean Value Theorem, we have\[f(x) = \frac{ F(b) - F(a) }{ b-a }\]where F'(x) = f(x). Now,\[(b - a)timesf(x) = F(b) - F(a) = \int\limits_{a}^{b}f(x)dx\]so,\[f(x)=\frac{ \int\limits_{a}^{b}f(x)dx }{ b-a}\]so,\[m \le \frac{ \int\limits_{a}^{b}f(x)dx }{ b-a } \le M\]and now just multiplyall 3 terms by b-a

Answer 8

Thanks!

Answer 9

uw!

Answer 10

Yes, between 28 and 42, and don't worry about the question. It's a great one and easy to get lost in.