Calculus Integration question

Calculus Integration question

Postby Guest » Thu Feb 27, 2020 2:18 am

How do you do this?
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Re: Calculus Integration question

Postby HallsofIvy » Mon Mar 02, 2020 6:09 pm

I am thinking that I saw this on another board and the answer was given there is no one answer- you can choose different f that match the two conditions but give different answers for the question.

For example, if f(x)= 2 for all x then [tex]\int_0^3 f(x)= 6[/tex] and [tex]\int_3^5 f(x)dx= 4[/tex] as required and, in this case, [tex]\int_0^5 f(3+ 2f(x))dx= \int_0^5 2 dx= 10[/tex].

But if f(x)= (4/3)x for [tex]0\le x\le 3[/tex] and [tex]f(x)= (1/2)x[/tex] for [tex]3\le x[/tex] then is is still true that [tex]\int_0^3 f(x)dx 6[/tex] and that [tex]\int_3^5 f(x)dx= 4[/tex]. But now [tex]\int_0^5 f(3+ 2f(x)) dx= 65/3[/tex].

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