Need help with the First Fundamental Theorem of Calculus?

Question

I’m having a problem with an equation used in the First Fundamental Theorem of Calculus.  I’ve found it in a textbook (author from MIT), on Wikipedia  and on several web sites.

F is antiderivative

f is function

F(x) = (integral) (from a to x) f(t) dt

My first thought when I saw this was that it would only be true if a=0.  I decided to test it with a simple function f(t) = 2t, a simple range [a,b] = [1,5] and setting the point of interest, x, to be equal to 3 (x=3).

f(t) = 2t

F(x) = t^2

x = 3

F(3) = 9

(integral) (from 1 to 3) 2t dt = t^2 | (from 1 to 3) = 9 - 1 = 8

9 does not equal 8.

Is there a flaw in my logic?  (showing me where I went wrong would constitute an answer)

Has this problem been previously acknowledged? (showing an internet reference to this problem would also constitute an answer)

David · Answer

an anti-derivative is a function, not a number.

suppose we define a function by:

F(x) = ∫[a,x] 2t dt

then F(x) is NOT t^2 (first of all, t is a parameter, not the variable of the function. the variable is x).

a little thought should convince you that

F(a) = ∫[a,a] 2t dt = 0 (an integral over 0 distance is 0).

so in your example, where you chose a = 1, that means:

F(1) = 0. now clearly, F(x) = x^2 does not satisfy F(1) = 0.

what you are missing is that:

F(x) should be x^2 - 1.

x^2 and x^2 -1 are both anti-derivatives of 2x.

anti-derivatives are not unique, if F(x) is an anti-derivative of f(x),

so is F(x) + C, where C can be any constant (number).

it is indeed true that F(x) = x^2 -1 satisifies F(3) = 8.

more generally, if F(x) is an anti-derivative of f(x),

then ∫[a,x] f(t) dt = F(x) - F(a) (note the second term !!!!).

now F(x) = x^2 -1 and G(x) = x^2 are both anti-derivatives, let's compute

G(3) - G(1) and F(3) - F(1):

G(3) - G(1) = (3)^2 - (1)^2 = 9 - 1 = 8.

F(3) - F(1) = (3^2 - 1) - (1^2 - 1) = 8 - 0 = 8.

now, for simple powers:

∫[0,x] t^n dt = x^(n+1)/(n+1) - 0 = x^(n+1)/(n+1)

whereas 

∫[a,x] t^n dt = x^(n+1)/(n+1) - a^(n+1)/(n+1)

note that the starting point of the integral, a, works

as an adjustment factor, to "pin down" the constant term of the anti-derivative.

this is why ∫[a,x] f(t) dt is called a DEFINITE integral.

∫ f(t) dt, on the other hand, is "indefinite", and one has to write the anit-derivative as:

F(x) + C. <---don't forget the C (sometimes we don't know what "C" is!).

Mathmom · Answer

You made a common mistake

Since F(x) is antiderivative of f(x) and f(x) = 2x, then 
F(x) = x^2 + C (and not x^2)

Using f(t) = 2t, and a = 1, we get:

F(x) = ∫[from 1 to x] f(t) dt
F(x) = ∫[from 1 to x] 2t dt
F(x) = t^2 |[from 1 to x] 
F(x) = x^2 - 1

Now we get:
F(3) = 3^2 - 1 = 8 (which does match your result ∫₁³ 2t dt = 8)

and F'(x) = 2x = f(x)

====================

Please note that

∫[from a to x] f(t) dt and ∫[from b to x] f(t) dt will give different results for a ≠ b

These results will differ ONLY in the constant. Therefore, both will give valid antiderivatives of f

? · Answer

If F(a,x) = integr60f42a513a57ecc111d234a1a775ad5te[60f42a513a57ecc111d234a1a775ad560f42a513a57ecc111d234a1a775ad560f42a513a57ecc111d234a1a775ad5] f(t) dt60f42a513a57ecc111d234a1a775ad5 then F'(a,x) = f(a,x) yet yours is t60f42a513a57ecc111d234a1a775ad5n60f42a513a57ecc111d234a1a775ad5 inste60f42a513a57ecc111d234a1a775ad5d of 60f42a513a57ecc111d234a1a775ad560f42a513a57ecc111d234a1a775ad5 so that you should 60f42a513a57ecc111d234a1a775ad5pply the ch60f42a513a57ecc111d234a1a775ad5in rule 60f42a513a57ecc111d234a1a775ad5fter utilizing the concept. the following is going: Y' = root(6t60f42a513a57ecc111d234a1a775ad5n60f42a513a57ecc111d234a1a775ad5 + root(t60f42a513a57ecc111d234a1a775ad5n60f42a513a57ecc111d234a1a775ad5)) * sec^2 a,x

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Need help with the First Fundamental Theorem of Calculus?

3 Answers