<p>If I'm given the derivative of the function f', finding f(x) would only be the "y"-coordinate of the function f, correct?</p>
<p>And if I'm given two indices from zero to c, and I'm given f, by integrating f and using the first fundamental theorem of calculus, it gives me the area under the graph of function f?</p>
<p>Thanks!</p>
<p>Derivative of F' would be F''...I don't know what you're saying</p>
<p>If I'm given the derivative of the function f', finding f(x) would only be the "y"-coordinate of the function f, correct?</p>
<p>By this, I mean integrating f' would give me f, and THAT f(x) would be the y-coordinate of the function f.</p>
<p>This is a completely different question:</p>
<p>If I'm given f and I take the definite integral, that gives me the area under the function f (in that fixed interval)?</p>
<p>Another question:</p>
<p>Function continuity implies differentiability but function differentiability does not necessarily imply continuity, correct?</p>
<p>Your first part is right except it would be the integrated form with a point say x at (x, integrate(f'(x)) + c), so you would not have the same exact graph, but for all purposes it would be the same. On your second question
you get the area under the curve from say x1 to x2, you would integrate the function and then go (f(x2) - f(x1)) where f(x) means the integrated function.
I think.</p>
<p>I'll try to word my question a little simpler.</p>
<p>I have the function F. If I find the definite integral of F from 1 to 2, that's giving me the area under the curve from (1,2), correct?</p>
<p>However, finding the antiderivative of function f' and finding f(x) would not give me an area but the point (x, f(x)) on the graph of the antiderivative f, right?</p>
<p>It's difficult to understand my question without seeing any diagrams of it!</p>
<p>The integral of f'(x) over the inteval "I' is the function f(x) over the same interval I. It is an equation expression and not just the number for the function f evaluated at a particular point in the interval.</p>
<p>Differentiability implies continuity, not the other way around. Example: absolute value function which is continuous everywhere, but not differentiable at its vertex</p>
<p>Alright, suppose I have the derivative of the function f' and a question asks me to find f(x). I simply find the antiderivative of f' and that will give me f(x).</p>
<p>Suppose another question asks me to find the area under the graph of f(x) over the interval 'I'. I integrate f(x) and use the first fundamental theorem of calculus to find the area. What does this area represent? Is it just the area under the curve of f(x) or does it have something to do with F(x) with F being the antiderivative of f?</p>
<p>Thanks for the replies so far!</p>
<p>Krabble:
For your post#6:
1. Yes. A definite integral of a one-dimensional function like F(x) from (say) x=1 to x=2 gives you the area under F(x) from x=1 -> 2. </p>
<ol>
<li>Right again. Finding the antiderivative of f'(x) (or, equivalently, finding its indefinite integral) will give you a function f(x). Strictly speaking, it will give you f(x) + a constant C, since</li>
</ol>
<p>d( f(x) + C) /dx = d(f(x))/dx + d(C)/dx = f'(x) + 0 = f'(x)</p>
<p>Krabble:
For your post#8, Q2:</p>
<p>Suppose the interval I represents x=a to x=b. The number you get will be the area under f(x), from x=a to x=b; it's also F(b) - F(a). Note that 'positive' and 'negative' areas will tend to cancel each other out; an example would be f(x) = sin(x) , evaluated from x=0 to x=2 pi.</p>
<p>Thanks! You've been very helpful!</p>
<p>I think I have another question, but if I can word it clearly, I'll post that up as well. Again, thank you for the help!!</p>
<p>So if I'm given the area under a curve of a function f', I can find the equation of the antiderivative of f'? Can I just set the interval to arbitrary end points and somehow figure out the equation if I'm only looking for the equation for function f? No wait, a point on f is given.</p>
<p>I'm going to attempt to answer my own question here, so feel free to correct me!</p>
<p>I want to find the point F(c).</p>
<p>Since a point on F (antiderivative of f) is given as (x, F(x)), I can set my interval for the definite integral of f from x to c. I have this area given to me as a number. I can simply set up my algebraic equation as Area = F(x) - F(c). Since I have F(x) given to me AND the area, I can simply find F(c)?</p>
<p>I think the way you're writing the question is a bit confusing. Don't use 'x' as one of end-points of your interval.</p>
<p>Assuming F(x) is the antiderivative of f(x), and that you are given F(x) as a function (e.g. F(x) = sin(x) + 2), an interval from x=b (known value) to c (unknown value), and the area under f(x) from x=b to x=c, then</p>
<p>Area = F(c) - F(b) or F(c) = Area - F(b)
Since you know Area and F(b) as numerical values, then yes: you can get a numerical value for F(c). To find the value of c, you can then try to solve for it e.g. if F(x) = sin(x) + 2, and F(c) works out to 2.3,
set sin(c) + 2 = 2.3
or sin(c) = 0.3
c = sin_inverse(0.3)</p>
<p>How easy this last step is depends on the structure of F(x). If it's something nasty like F(x) = sin(x) + x^2 - 5 exp(x), you're up the creek.</p>