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## Main Question or Discussion Point

I have been attempting to make the leap from AP Calc to more theory-based calculus by trying to solve problems. Unfortunately, there are no solutions to the book I'm using so I need some guidance. Here's the problem:

Let f: [0,1] -> (0,1) be continuous. Show that the equation

[tex] 2x - \int^{x}_{0}\ f(t)dt = 1 [/tex]

has one and only one solution in the interval [0,1].

My first instinct was to try the endpoints to support the claim. If x = 0, we find that 0 = 1, which is evidently not a solution. Trying x = 1, we get

[tex] \int^{1}_{0}\ f(t)dt = 1 [/tex]

which appears to be valid.

Noticing the form of the integral, I decided to differentiate both sides with respect to x, getting

[tex] 2 - f(x) = 0 \Rightarrow f(x) = 2 [/tex]

But [tex] f(x) = 2 [/tex] implies that the original equation becomes 2x - 2x = 0 = 1, which is clearly not a solution. Since we can choose any x in the interval (0,1) to reach this false conclusion, there is only one solution and it occurs at x = 1.

I'm pretty confident that the beginning of my proof justifies that x = 0 cannot be a solution and x = 1 is a solution. But I feel that my reasoning as to why there is no solution on (0,1) is faulty. But then again, my reasoning/method of attack could be entirely faulty.

Let f: [0,1] -> (0,1) be continuous. Show that the equation

[tex] 2x - \int^{x}_{0}\ f(t)dt = 1 [/tex]

has one and only one solution in the interval [0,1].

My first instinct was to try the endpoints to support the claim. If x = 0, we find that 0 = 1, which is evidently not a solution. Trying x = 1, we get

[tex] \int^{1}_{0}\ f(t)dt = 1 [/tex]

which appears to be valid.

Noticing the form of the integral, I decided to differentiate both sides with respect to x, getting

[tex] 2 - f(x) = 0 \Rightarrow f(x) = 2 [/tex]

But [tex] f(x) = 2 [/tex] implies that the original equation becomes 2x - 2x = 0 = 1, which is clearly not a solution. Since we can choose any x in the interval (0,1) to reach this false conclusion, there is only one solution and it occurs at x = 1.

I'm pretty confident that the beginning of my proof justifies that x = 0 cannot be a solution and x = 1 is a solution. But I feel that my reasoning as to why there is no solution on (0,1) is faulty. But then again, my reasoning/method of attack could be entirely faulty.