Question

Fundamental theorem of Calculus question

Answer 1

\[\int\limits_{2}^{x} 3t^2 dt\]

Answer 2

find g'(x)

Answer 3

So the Fundamental Theorem of Calculus, Part 1 (FTC 1) states,\[\large \frac{d}{dx} \int\limits_a^x f(t) dt \quad = \quad \frac{d}{dx}\left[F(x)-F(a)\right]\]Where F is the anti-derivative of f.
Then taking the derivative gives us,\[\large f(x)-0\]

See amood's picture for the actual steps though :)

Answer 4

I got to the point x^3-8
how do you get from that to 3x^2??

Answer 5

So you integrated and got x^3-8, good.
From there you have to take the derivative of your answer.

Answer 6

oh ok

Answer 7

You're basically, integrating, then undoing the integration by differentiating.

The variable ends up changing due to what was in our limits of integration :D

Answer 8

so that answer is always the same as the beginning but the only difference is the variable?

Answer 9

If it has this basic format yes.

If you get a different problem that is similar but with an X in the upper AND lower limits, then your answer will look a little bit different.

Because instead of subtracting a constant (which becomes 0 when we differentiate), you're subtracting another function of x.
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And also you'll get something slightly different when your boundary contains MORE THAN just x.

Like if it contained x^2 in the upper boundary you would have to apply the CHAIN RULE when you take the derivative.

So in those 2 cases you won't get exactly back what you started with.

Answer 10

thanks for helping me understand why it happens