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Fundamental Theorem of Calculus

We've explored areas under curves that form familiar geometric shapes and also how to approximate areas under any curve.

But surely we don't have to draw groups of rectangles every time we want to find the area under a curve! There must be an easier way!

You're right, there is, and this powerful and beautiful approach that we're about to explore allows us to dive deep into what is known as integral calculus and The Fundamental Theorem of Calculus.

The integral and the area function

Let's explore the concept of using infinitely many rectangles to determine the area under a curve. 

We'll use the same curve that we've been exploring, $y=x^2+1$y=x2+1

If you slide the value of $n$n, you can either under-approximate or over-approximate.

While we can create infinitely many rectangles here, we can go up as far as 100 rectangles.

Under-approximating we find that we get $1.33$1.33 $units^2$units2

Over-approximating we find that we get $1.34$1.34 $units^2$units2

The more and more rectangles we create, the closer we get to finding the area.

Formally this can be written as:

If we think about creating infinitely many rectangles, we arrive at what Leibniz formulated in the 17th century. Instead of writing what we're doing using the notation above, we now have:


for our specific scenario using the function $y=x^2+1$y=x2+1

Or in general:

The Fundamental Theorem of Calculus - Part 1

Now that we understand that finding the area bounded by a curve and the $x$x axis over a given domain involves the use of anti-differentiation, we're ready to understand the first part of the Fundamental Theorem of Calculus (the FTC).

Don't be scared by the name! It sounds a lot more complicated than it is!

The first part of the FTC says:

$F(x)$F(x) represents our area function or our antiderivative or our primitive function.

So in everyday language, the first part of the FTC says, to find the area under the curve between the $x=a$x=a and $x=b$x=b, first find the area function and call it $F(x)$F(x). Subtract the value of $F(x)$F(x) at the lower limit from the value at the higher limit, that is, $F(b)-F(a)$F(b)F(a).

Let's look at our function again, and this time let's see what sort of area function, $F(x)$F(x) we get when we change the area under the curve.

As you slowly drag the slider, you should see that the value of the area under our quadratic function between the quadratic and the x-axis) is resulting in a cubic graph.

Hence our area function, $F(x)$F(x), is simply the primitive or antiderivative of our function $f(x)$f(x).

So going back to our original problem, we can once and for all find the area under our curve!

So our approximations were very close!

The Fundamental Theorem of Calculus - Part 2

In the second part of our theorem, we move away from area and now we're talking about calculus as a whole.

What happens if you differentiate something you've just integrated?

What we find is that we end up with the function we started with. Seems intuitive doesn't it? And this is where the name "Fundamental Theorem of Calculus" comes from. It is an absolutely important and powerful fact that differentiation and integration are inverse operations, in fact, it's fundamental!

Let me show you a nice proof of this idea.

We started here with our limit between $x=a$x=a and $x$x, which can vary so that we can generalise our findings.

In order to not get confused, we've called our initial function $f\left(t\right)$f(t) so we can see what's happening more clearly.

So in general we can say:


Now let's consider the case where the upper limit is a function of $x$x. By working through the example, we can derive the second general case.


Find $\frac{d}{dx}\int_1^{2x}3t^2dt$ddx2x13t2dt.

If we let $v=\int_1^{2x}3t^2dt$v=2x13t2dt then we want to find $\frac{dv}{dx}$dvdx. Using the substitution $u=2x$u=2x, we obtain the more familiar form $v=\int_1^u3t^2dt$v=u13t2dt.
We will use the form of the FTC given above to find $\frac{dv}{du}$dvdu.
Then to solve the original expression we will need to use the chain rule $\frac{dv}{dx}=\frac{dv}{du}\times\frac{du}{dx}$dvdx=dvdu×dudx
From the FTC, if $v=\int_1^u3t^2dt$v=u13t2dt, then $\frac{dv}{du}=3u^2$dvdu=3u2.
Secondly, if $u=2x$u=2x, then $\frac{du}{dx}=2$dudx=2.
Hence, using the chain rule, $\frac{dv}{dx}=\frac{dv}{du}\times\frac{du}{dx}$dvdx=dvdu×dudx
                                                    $=$=$3\left(2x\right)^2\times2$3(2x)2×2               (Notice that this is in the form $f(2x)\times\frac{d}{dx}(2x)$f(2x)×ddx(2x))

Therefore, $\frac{d}{dx}\int_1^{2x}3t^2dt=24x^2$ddx2x13t2dt=24x2.
Notice that we can also obtain this result directly by applying the FTC to the upper limit function $2x$2x, and multiplying by the derivative, with respect to $x$x, of $2x$2x. This method is summarised by the second general case:


Worked Example

Example 1

Differentiate, with respect to $x$x

Using the FTC we obtain:







More Worked Questions

Question 1

Calculate $\int_{-1}^2\left(9x^2+1\right)dx$21(9x2+1)dx.

Question 2

Suppose $\int_{-1}^2f\left(x\right)dx=4$21f(x)dx=4 and $\int_2^8f\left(x\right)dx=8$82f(x)dx=8.

  1. Find the value of $\int_{-1}^8f\left(x\right)dx$81f(x)dx.

  2. Find the value of $\int_8^{-1}f\left(x\right)dx$18f(x)dx.

  3. Find the value of $\int_{-1}^22f\left(x\right)dx+\int_2^83f\left(x\right)dx$212f(x)dx+823f(x)dx.

Question 3

Determine $\frac{d}{dx}\int_7^{x^4}\sqrt{2+t^3}dt$ddxx472+t3dt.



Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods


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