Question

FTC

Answer 1

If:

\(\int\limits_0^a \ f(x) ~ dx = C\)

then why not

\(\dfrac{d}{da} \left( \int\limits_0^a \ f(x) \, dx = C \right)\)

\(\implies f(a) = 0 \)

I know why because I already messed it up

Answer 2

Differentiating with respect to a constant? :o
Is that a thing?

I know not of this crazy math of which you speak :D

Answer 3

What I always do is I just make up a new function and use that as the "already evaluated integral thingy":

\[\frac{d}{da} [F(a)-F(0)] = \frac{d}{da}(C)\]

So now if you're saying that for sure this integral is equal to a constant, then you are saying that implies:

\[\frac{d}{da} F(a) = 0\]

which might not be what you want, although if this is true then this means \(f(x)=0\) since it's the only way you can change the upper bound on the integral, a, and not change the area under the integral.

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Alternatively you could plug it into the actual definition of derivative (sorta scary but fun!)

\[\frac{d}{da} \int_0^a f(x)dx = \lim_{h \to 0} \frac{\int_0^{a+h} f(x)dx - \int_0^af(x)dx}{h}\]

Now the integral on the right simplifies, and we can rewrite it as:

\[=\lim_{h \to 0} \frac{1}{h} \int_a^{a+h} f(x) dx\]

Now from here, as h becomes infinitesimally small, it will basically divide out dx. This will just leave us with f(x) evaluated at just one point, x=a. I think there might be a more proper way of taking care of this, but I think it's intuitive and works out to get the right answer so I haven't poked it too much to find out since I never really use this but hey it came up so why not share? lol

Answer 4

the specific example that made me ask was an integration like this

\(\int\limits_0^a 3 e^{\frac{x}{2}} + 1 ~ dx = 10\)

plus a question like: what is a?? [or something like that]

and therein lay my mistake because I thought this:

\(\dfrac{d}{da} \left( \int\limits_0^a 3 e^{\frac{x}{2}} + 1 ~ dx = 10 \right)\) would be true

ie
\( 3 e^{\frac{a}{2}} + 1 = 0\) would be true


but there's a howler in there.