Integral problem for calculus nerds...?

Question

I have:

F(x)= S(sqrt(2t^2+1)) on the interval [2, x].

Now, say I want to take the derivative of this. What do I have to tag on to sqrt(2t^2+1) to make it right? Why?

My whole problem is with the interval. I know I just need to substitute x for t, but I would only just leave it if it were on [0, x]. What about [2, x]?

Answer 1

The derivative IS sqrt(2x^2+1) --- you don't have to "tag" anything onto it! All you have to do is to substitute ' x ' for ' t.'

The point is that if:

F(x) = S[f(t) dt] on any interval [c, x],

where "S" means "the integral of" and ' x ' is the UPPER LIMIT of the interval for ' t ', then the Fundamental Theorem of the Calculus says that:

dF(x)/dx = f(x) --- simply the function being integrated, evaluated at
t = x.

(If, on the other hand, it's G(x) = S[g(t) dt] on the interval [x, c], then one has correspondingly dG(x)/dx = - g(x). The " - " sign comes in because the LIMITS have been inverted; ' t ' = ' x ' is now the LOWER limit, not the UPPER limit.)

[Dear old Puggy. He got the wrong sign again --- not for the first time --- in his haste to be first. He INVERTED the limits, which is why he got the NEGATIVE of the correct answer. Now watch him change it to the right answer, after he'll have seen mine !]

Live long and prosper.


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Re. Your question under "Additional details" :

"My whole problem is with the interval. I know I just need to substitute x for t, but I would only just leave it if it were on [0, x]. What about [2, x]?"

The LOWER LIMIT simply doesn't matter. Why not? : Because the DIFFERENCE in the function F(x) that you're differentiating,
on either of TWO intervals (i) [b, x] or (ii) [c, x] is +/- the integral on [b, c] of [f(t) dt]. BUT THAT'S A CONSTANT, so gives you NOTHING on differentiating!

Another way of SEEING this MORE FUNDAMENTALLY is to think about WHY dF(x)/dx IS f(x):

Let theORIGINAL integral be EXTENDED to an interval [c, x + dx].

F(x + dx) = F(x) + S[f(t) dt],

the latter integral on the interval [x, x + dx] ONLY! Then you'll have, at least for a continuous function f(t):

F(x + dx) = FIx) + f(x + theta*dx) dx where 0 < theta < 1

[This is by somebody or other's theorem, but you'll have to forgive me at this point --- it's been ~ 50 years, and I've forgotten the name, sorry.]

So in the limit, [F(x + dx) - F(x)]/dx = f(x + theta*dx) --> f(x).

Thus, we've both PROVED the Fundamental Theorem of Calculus, AND, in so doing, confirmed that in that theorem, ALL the action is at "the end of the interval where ' x ' IS."

There's ABSOLUTELY NOTHING going on of any interest whatsoever, as far as this theorem and this result are concerned, at the other end.

QED

Answer 2

F(x) = Integral ( sqrt(2t^2 + 1) ) on the interval [2, x]

You apply the fundamental theorem of Calculus. Remember that there is _NO_ integration involved; the only thing that's involved is the chain rule.

F'(x) = sqrt(2(x)^2 + 1) {1} - sqrt(2(2)^2 + 1) {0}
F'(x) = sqrt(2x^2 + 1) - 0
F'(x) = sqrt(2x^2 + 1)

Edit: Thanks Dr. Spock for the correction.

To answer your other question, you would get the exact same answer if it were 0 or 2 as the first of the integral bounds.

Answer 3

let f1(t) = sqrt(2*t^2 +1)

Now suppose Sf1(t) = f2(t) + c (indefinite integration)

Also, from above, f1(t) = f2'(t), by definition ------------------- (1)

Now,
F(x) = Sf1(t) in [2,x]

= (f2(t) + c)@x - (f2(t) + c)@2

= f2(x) + c - f2(2) - c

= f2(x) - f2(2)

Therefore,
F'(x) = f2'(x) - 0

= f2'(x) = f1(x) (from (1) above)

= sqrt(2*x^2 +1)

Answer 4

replace log base 2 to ln by potential of the bottom-replace for logarithms: log_2_(x-a million) = ln(x-a million)/ln(2) Use u-substitution : u= ln(x-a million) du = a million/(x-a million) dx The fundamental will change into ? (2/ln2)*udu = (a million/ln2)u^2 + c = (a million/ln2)(ln(x-a million))^2 + c