If the function f has a continuous derivative on [0,c], the the integral(o to

Cof f'(x)dx=

a.f(

c.-f(0)

b.absolute value (f(

c.- f(0))

Cf(
C

d.f'(x)=c

e.f"(

c.-f"(0) My work: so the the answer to the integral is f(x) and when find the answer from o t0 c, it is f(

c.-f(0).

is that the right answer? i'm confused because is there anything i have to do with the point [0,

Cor is that unneccessary info.

problem #2: let f be a polynomial function with degree greater than 2. if a does not equal b and f(

a.=f(

b.=1, which of the following must be true for atleast one value of x between a and b?

I)f(x)=0 II)f'(x)=0 III)f"(x)=0 you can choose more than one choice in the choices mentioned of I, II, III

i'm having trouble coming up with the equation and choosing a and b

The correct answer to (#1) is f(

c.-f(0). That is because the definite integral of a function (

f. is the difference between the indefinite integral (

f. evaluated at the two limits of integration.

The correct answer to (#2) is f'(x) = 0. Imagine all possible continuous curves you can draw from a to b, going though f = 1 at both points. The curve MUST have zero slope somewhere. There is no requirement that f or f'' be zero at intermediate points. you don't need an equation to prove this. You just need to invoke the Mean Value Theorem

For the first problem,
The definite integral from 0 to c is
[f(c) - f(0)] / (c - 0)
since the function is continuous in the interval 0 to c.

Since f(a) = f(b) = 1, then by the Mean Value Theorem, there is a point somewhere in the middle two values of x that f'(x) = 0 since the curve has a degree of more than 2.

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