I don't understand the intuition behind this. Why can we just plug in $x$ for $t$ here and that gives us the result? I thought I was understanding the Fundamental Theorem of Calculus, but I don't see how it applies here. I thought the Theorem mainly stated that the area under a function can be found by taking the the value of the anti derivative over the specified interval. It doesn't make sense to me why we just plug in $x$ and voila that's our answer.

$$\frac {d}{dx} \int_{a}^{x} (t^3 + 1) \ dt = x^3 + 1$$

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    $\begingroup$ If $F(x)$ is the antiderivative function of $f(x)=t^3+1$, then $\int_a^x t^3+1dt=F(x)-F(a)$. So the L.H.S becomes $\frac{dy}{dx}F(x)$ which is $x^3+1$. $\endgroup$ – Frenzy Li Aug 15 '12 at 3:36
  • $\begingroup$ Evaluate that integral, and take the derivative with respect to $x$. $\endgroup$ – Ink Aug 15 '12 at 3:51

Since the function $\,t^3+1\,$ is continuous (and derivable) everywhere, it has a primitive function $\,G(t)\,$in any finite interval. Using the FTC , write

$$\int_a^x(t^3+1)dt=G(x)-G(a)\\\Longrightarrow \frac{d}{dx}{\left(\int_a^x(t^3+1)dt\right)}=\frac{d}{dx}{(G(x)-G(a))}=G'(x)=x^3+1$$

  • $\begingroup$ So we just get rid of the G(a) because we are differentiating with respect to x? $\endgroup$ – mr real lyfe Aug 15 '12 at 3:41
  • $\begingroup$ @Peter Tamaroff, thanks. $\endgroup$ – DonAntonio Aug 15 '12 at 3:41
  • $\begingroup$ Of course, @ordinary. $\endgroup$ – DonAntonio Aug 15 '12 at 3:42
  • $\begingroup$ Sorry I am very new to this $\endgroup$ – mr real lyfe Aug 15 '12 at 3:42
  • $\begingroup$ @ordinary The integral has a fixed lower bound $a$ and a changing upper bound $x$. $G(a)$ is therefore constant, and differentiating it would yield $0$. $\endgroup$ – Frenzy Li Aug 15 '12 at 3:43

The Fundamental Theorem of Calculus doesn't talk about geometrical results, but about the "fundamental" relation between the operation of integration and that of differentiation. Namely, it says the following:

THEOREM. Let $f$ be a continuous over $[a,b]$. Define $F$ on $[a,b]$ by

$$F(x)=\int_a^x f(t) dt$$

Then $F$ is differentiable, and $F'(x)=f(x)$.

The corollary is

COROLLARY Let $f$ be continuous over $[a,b]$ and $f=g'$ for some $g$.


$$\int_a^b f(t)dt=g(b)-g(a)$$

Note we can find this reversed in the books (One is the theorem and the other the corollary, or one is called FTC 1 and the other FTC 2). I recommend you read these two questions some users already asked about FTC:

  • $\begingroup$ $F$ is differentiable at the points where $f$ is continuous. $\endgroup$ – lhf Dec 28 '17 at 13:56
  • $\begingroup$ @Masacroso Yes, of course. There. ;) $\endgroup$ – Pedro Tamaroff Dec 28 '17 at 17:53

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