Please help me with this integration problem

C

Csax

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Feb 27, 2020
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I can't figure out what to do when f(x) is inside of f(x). I can't use u-substitution since I have no f'(x) or can I just use the fundamental theorem of calculus and plug in the numbers?
 
well... I think this is a bit of a crap problem. I wouldn't be surprised if there are multiple functions f that satisfy the first two equations.

Eyeballing it, it should be fairly clear that \(\displaystyle f(x)=2\) is one function that satisfies the first two equations.

We can plug that in to the 3rd equation to obtain

\(\displaystyle \displaystyle \int \limits_0^5 2~dx = 10\)
 
I agree with Romsek saying that there are multiple functions satisfying the first two equation.

There are many many functions whose area from x=0 to x=3 is 6 and many many functions whose area from x=3 to x=5 is 4. We can then make a piecewise function that lives up to both integrals.

Not only do I think that this is a crappy problem I do not think it can be solved.
 
Romsek said:
well... I think this is a bit of a crap problem. I wouldn't be surprised if there are multiple functions f that satisfy the first two equations.

Eyeballing it, it should be fairly clear that \(\displaystyle f(x)=2\) is one function that satisfies the first two equations.

We can plug that in to the 3rd equation to obtain

\(\displaystyle \displaystyle \int \limits_0^5 2~dx = 10\)
Click to expand...

does it matter that it is a calculator problem?
 
Romsek said:
well... I think this is a bit of a crap problem. I wouldn't be surprised if there are multiple functions f that satisfy the first two equations.

Eyeballing it, it should be fairly clear that \(\displaystyle f(x)=2\) is one function that satisfies the first two equations.

We can plug that in to the 3rd equation to obtain

\(\displaystyle \displaystyle \int \limits_0^5 2~dx = 10\)
Click to expand...
How do we know that f(x) = 2?
 
Csax said:
does it matter that it is a calculator problem?
Click to expand...
Can you explain that? How could you possibly use a calculator to solve this? Maybe you need to show us the entire problem, including instructions.

But I strongly suspect it is a typo, and they meant something like this: [MATH]\int_0^5 (f(3)+2f(x))dx[/MATH]. That makes a lot of sense, tests something important, but still doesn't seem like a calculator problem.
 
Dr.Peterson said:
Can you explain that? How could you possibly use a calculator to solve this? Maybe you need to show us the entire problem, including instructions.

But I strongly suspect it is a typo, and they meant something like this: [MATH]\int_0^5 (f(3)+2f(x))dx[/MATH]. That makes a lot of sense, tests something important, but still doesn't seem like a calculator problem.
Click to expand...
Wouldn't we need to know what f(3) equals to do the problem you stated?
 
Yes, I just said "something like"! I didn't try hard enough to make it make as much sense as I claimed it did.

Any better ideas?
 
Maybe you are correct with your assumption and the final answer is in terms of f(3).
 
so sorry that I did not include the multiple choice. Here are the answers....and the final answer is D. 35Screen Shot 2020-02-27 at 7.59.13 PM.png
 
That is what they say the answer is....I can't figure it out.
I thought that maybe f(x)=2 and then I get f(3 + 2f(x)) turns into f(7) but it is asking from 0 to 5.
so confused but what if it is a constant at 7 and from 0 to 5, which would then be 35....I feel like I am making stuff up though.....arrrrggghhhh
 
Csax said:
That is what they say the answer is....I can't figure it out.
I thought that maybe f(x)=2 and then I get f(3 + 2f(x)) turns into f(7) but it is asking from 0 to 5.
so confused but what if it is a constant at 7 and from 0 to 5, which would then be 35....I feel like I am making stuff up though.....arrrrggghhhh
Click to expand...
If f(x)=2, a constant, then f(anything) =2. The point I am making is you should not bother to compute 3 + 2f(x), since f(3 + 2f(x)) = 2, not 7.

Using f(x)=2, there is no chance that f(3 + 2f(x)) will equal 7, as you were (slightly) saying.

There is an expert on this forum (certainly not me) who said that this problem is flawed. I would not question that response.
 
Now I am convinced they meant this: [MATH]\int_0^5 (3+2f(x))dx[/MATH]. This one does make sense, and does give the claimed answer!

As has been pointed out, there are infinitely many possible functions f; the requirement is only that the average over [0,3] and the average over [3,5] are both 2. But the given facts are sufficient to evaluate this integral without thinking about what f is.

Flawed? Clearly. Fixable? Yes!
 
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