Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question
3  
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 $x^3+1$. –  FrenzY DT. Aug 15 '12 at 3:36
    
Evaluate that integral, and take the derivative with respect to $x$. –  Vectk Aug 15 '12 at 3:51
add comment

2 Answers

up vote 5 down vote accepted

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

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

share|improve this answer
    
So we just get rid of the G(a) because we are differentiating with respect to x? –  mr real lyfe Aug 15 '12 at 3:41
    
@Peter Tamaroff, thanks. –  DonAntonio Aug 15 '12 at 3:41
    
Of course, @ordinary. –  DonAntonio Aug 15 '12 at 3:42
    
Sorry I am very new to this –  mr real lyfe Aug 15 '12 at 3:42
    
@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 $0$. –  FrenzY DT. Aug 15 '12 at 3:43
show 2 more comments

The Fundamental Theorem of Calculus doesn't not 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 an integrable function 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 $g$.

Then

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

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:

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.