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For extra credit for my class we are supposed to explain or describe to my teacher "The Fundamental Theorem of Calculus".

Now I understand Calculus has a lot to do with integrals, differentiating, finding curves and the area between curves by using integrals.

But my teacher wants us to show us him an example using mathematical examples and such.

I was wondering if someone could maybe help guide me or give me some examples I could use to work off of for this?

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Wikipedia has a good discussion, in some parts rather technical. –  André Nicolas Jul 10 '12 at 1:45
    
Yeah i looked into it but i want a more unique idae –  soniccool Jul 10 '12 at 1:47
    
The standard example is area. Let $F(x)$ be the area under $y=t^n$, above the $t$-axis, from $t=a$ to $t=x$. It is not hard to argue from the definition of the derivative that $F'(x)=x^n$. If we can find an antiderivative of $x^n$, this gives us a formula for the area. That solves a problem that people had worked on, with only partial success, since ancient times. (To be fair, it was solved in a different way in the early 17th century, a few years before the Fundamental Theorem.) –  André Nicolas Jul 10 '12 at 1:57
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3 Answers

The Fundamental Theorem of Calculus states:

For a differentiable function $f$, $\int_a^b f'(x)dx=f(b)-f(a)$.

Dividing both sides by $b-a$ gives us $$\frac{1}{b-a}\int_a^b f'(x)dx=\frac{f(b)-f(a)}{b-a}.$$ You can think of $f'(x)$ as the slope of the tangent line at $x$, and note that $\frac{f(b)-f(a)}{b-a}$ is the slope of the line connecting $(a,f(a))$ and $(b,f(b))$. This way, the FToC can be rewritten as:

The slope of the line connecting $(a,f(a))$ and $(b,f(b))$ is equal to the average of the slope of the tangent line of $f$ at $x$ as $x$ goes from $a$ to $b$.

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I like this. I'd not thought of the FTC that way. –  Chris Taylor Jul 10 '12 at 7:23
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Suppose someone came and told you:

"Hey, I've integrated a function $f(t)$ over the interval $[a,x]$ and I have got the following

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

I know an explicit formula for $F(x)$ [enter your choice of $F$ here], but sadly I have forgotten what $f$ was. What can I do?"

Then you can say "Use the Fundamental Theorem of Calculus. That will tell you that

$$f(x)=F'(x)$$

and you're done!"

But the guy might be curious and ask, "Hey, but how does that work?!"

For the explanation, check this

The importance of the FTC is that it connects the notion of integral and derivative, which wasn't something obvious back then. FTC is saying that the process of integration and the process of integration are inverses of each other. In fact, if we define two sets

$$A=\text{the set of all continuous functions on }[a,b] \;\,,f:[a,b]\to \Bbb R$$

and $C\subset A$

$$C=\text{the set of all continuous functions, with continuous derivative on }[a,b] \text{ and } f(a)=0$$

Then the functions $I$ and $D$ defined as

$$I:A\to C$$ such that

$$I(f)(x)=\int_a^xf(t)dt\,\; ;a\leq x \leq b$$

and

$$D : C \to A$$ such that

$$D(f)(x)=f'(x)$$

are in fact inverse mappings.

Note that

$$(I \circ D(f))(x)=\int_a^xf'(t)dt=f(x)-f(a)=f(x)$$

since $f(a)=0$, and

$$(D \circ I(f))(x)=\frac{d}{dx} \left(\int_a^xf(t)dt\right)=f(x)$$

You can see the "proof" that they are indeed inverse mappings relies wholely on the FTC! Note that $D$ and $I$ take a function $f$ as an input and output a function $I(f)$ or $D(f)$. The fact we write $(x)$ next to them is to remind they are still functions of $x$.


ADD: For the practical uses, FTC is the most efficient tool to evaluate definite integrals. When definite integrals are defined, we can use lower and upper sums, Riemann sums and such to find some of them, such as

$$\eqalign{ & \int\limits_a^b {{x^p}dx} = \frac{{{b^{p + 1}} - {a^{p + 1}}}}{{p + 1}} \cr & \int\limits_a^b {{e^x}dx} = {e^b} - {e^a} \cr} $$

but other integrals become more and more laborious. The FTC allows us to evaluate integrals such as

$$\int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 + {x^2}} }}} $$

Surely, there are far more interesting examples, but I can't come up with one now.

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So you want to use examples. Look at the graph of $y=\sin x$ between $0$ and $\pi$. Sine is a positive function in that interval. Let $x$ be some number between $0$ and $\pi$, and draw the vertical line from that point on the $x$-axis up to the graph. The height of that line is $\sin x$. Imagine that $x$ is moving to the right, so the area under the curve between $0$ and $x$ is getting bigger. Let's ask how fast it's getting bigger. The size of the moving boundary of the region is $\sin x$, the rate at which the boundary is moving is $dx/dt$. The size of the boundary times the rate at which the boundary moves is the rate at which the area grows, so it's $dA/dt$. So $$ \frac{dA}{dt} = (\sin x)\frac{dx}{dt}. $$ Hence $$ \frac{dA}{dx} = \sin x. $$ There's the fundamental theorem, applied to this function.

From this we can say $A = -\cos x + C$.

From the picture, we see that $A=0$ when $x=0$, so $0=-\cos0+C$. Since $\cos0=1$, we have $C=1$. Hence $A=-\cos x+1$.

Applying that when $x=\pi$, we see that the whole area under the curve is $-\cos\pi+1$, i.e. it is $2$.

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