The fundamental theorem of calculus is a cornerstone of calculus and provides us with a connection between derivatives and integrals in two ways:
The first part states that differentiation is the opposite of integration and implies the existence of the anti-derivative for any continuous function.
The second part gives us a means to calculate the definite integral of a function over an interval by finding an anti-derivative of function and implies that integration is the opposite of differentiation up to the addition of a constant.
The theorem links the algebraic process of anti-differentiation to the geometrical interpretation of integration and gives us a way to calculate areas under curves for continuous functions.
Part 1:
For a function f\left(x\right) continuous over the interval \left(a, b\right), if we define F\left(x\right) by the integral 
then:
Part 2:
For a function f\left(x\right) continuous over a closed interval \left[a, b\right] with an anti-derivative function F\left(x\right):
In this investigation we will explore a geometric proof of the second part of the fundamental theorem of calculus. Let's begin by exploring an area function for some simple functions.
Exploration
Given a function f\left(t\right) which is continuous on the interval a \le t\le b. Let's define the area function A\left(x\right) to be the area under the curve f\left(t\right) between t=a and t=x, where a\le x \le b. That is A\left(x\right) is the shaded area shown below:
We will use the notation for the definite integral as the notation for the area, that is:
At this stage this is purely notation and we aim to show that the process of finding the area is indeed equivalent to finding an anti-derivative and then evaluating and finding the difference between the end points.
Example 1
Consider the function f\left(t\right)=3, the graph of y=f\left(t\right) is shown below. We want to find the area function A\left(x\right) which gives us the area under the graph from a starting point t=a to t=x.
- What is the width of the shaded area in terms of a and x? ________
- What is the height of the shaded area? ________
Hence,
| A\left(x\right) | = |  |
|
| = | \text{width}\times \text{height} |
|
|
| = | \editable{} \times \editable{} |
Substitute in our width and height found above |
|
| = | \editable{} - \editable{} |
Expand |
For the function F\left(t\right)=3t, write A\left(x\right) in terms of F\left(a\right) and F\left(x\right).
| A\left(x\right) | = | \editable{} - \editable{} |
How does the function F\left(t\right) relate to the function f\left(t\right)?
Example 2
Consider the function f\left(t\right)=2t, the graph of y=f\left(t\right) is shown below. We want to find the area function A\left(x\right) which gives us the area under the graph from a starting point t=a to t=x.
Given the shaded area forms a trapezium, the area of which is \text{Area}=\frac{1}{2}\left(\text{Side 1}+\text{Side 2}\right)h, where the two sides are the parallel sides and h is the perpendicular height:
- What is the perpendicular height of the trapezium in terms of a and x? ________
- What is the length of the side on the left? ________
- What is the length of the side on the right? ________
Hence,
| A(x) | = |  |
|
| = | \frac{1}{2}\left(\text{side 1 + side 2}\right)h |
|
|
| = | \frac{1}{2} \editable{} \times \editable{} |
Substitute in our sides and height found above |
|
| = | \left(\editable{}\right)^{2} - \left(\editable{}\right)^{2} |
Expand |
For the function F\left(t\right)=t^2, write A\left(x\right) in terms of F\left(a\right) and F\left(x\right).
| A\left(x\right) | = | \editable{} - \editable{} |
How does the function F\left(t\right) relate to the function f\left(t\right)?
We have seen for two relatively simple functions that the area function appears to be related the anti-derivative of the function we are finding the area under. Let's now look at a geometrical proof for the general case of a function f\left(t\right) such that f\left(t\right) \ge 0.
Proof
Let f\left(t\right) be a continuous positive function on the interval \left[a,b\right].
And let A\left(x\right) be the area under the curve between t=a and t=x. That is A\left(x\right) is the shaded area shown below:
Again, using the notation for the definite integral as the notation for the area, we have:
Hence, we have:
| A\left(a\right) | = |  |
The area from a to itself |
| = | \editable{} |
Simplify - what is the area under a single point? |
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|
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| A\left(b\right) | = |  |
Complete the integral by adding the endpoints |
From the diagram shown below, the area from t=a to t=x would be the value of A\left(x\right) and the area from t=a to t=x+h would be the value of A\left(x+h\right).
Hence, the area of the narrow purple strip from t=x to t=x+h, in terms of the function A(x) is given by:
\editable{} - \editable{}
We could bound the area of this strip using a rectangle that just fits within the strip (underestimate) to one that just fits over the strip (overestimate).
| Underestimate | Overestimate |
|---|---|
 |
 |
- What is the width of the rectangle in both cases?________
- What is the height of the rectangle which underestimates the area of the strip? ________
- What is the height of the rectangle that overestimates the area of the strip? ________
Hence, we can bound the area of the strip as follows:
(Complete the equation below with the areas of the rectangles using lengths and width found above)
| \text{Area smaller rectangle} | \le | \text{Area of strip} | \le | \text{Area of larger rectangle} |
| \editable{} | \le | A\left(x+h\right)-A\left(x\right) | \le | \editable{} |
Dividing the inequality through by h we get:
| \editable{} | \le | \frac{A\left(x+h\right)-A\left(x\right)}{h} | \le | \editable{} |
Taking the limit of each term as h \rightarrow 0, we obtain:
| \lim_{h\rightarrow 0}\editable{} | \le | \lim_{h\rightarrow 0}\frac{A\left(x+h\right)-A\left(x\right)}{h} | \le | \lim_{h\rightarrow 0}\editable{} |
- Does the term in the middle look familiar?
- Can we simplify the first term? Since f\left(x\right) does not depend on h, \lim_{h \rightarrow 0}f\left(x\right)=f\left(x\right)
- Can we simplify the last term? Since f\left(x\right) is a continuous function, \lim_{h \rightarrow 0}f\left(x+h\right)=f\left(x\right)
Thus we have:
| f\left(x\right) | \le | A'\left(x\right) | \le | f\left(x\right) |
And since A'\left(x\right) is sandwiched between f\left(x\right) on both sides, we must have:
A'(x)=f(x)
We have just shown that A(x) is an anti-derivative of f(x). So the area function only differs from the a given anti-derivative, say F(x), by a constant.
\therefore\ A(x)=F(x)+C
We can solve for C, given that we know A(a)=0:
| A(a) | = | F(a)+C |
| 0 | = | F(a)+C |
| \therefore\ C | = | \editable{} |
Hence, we have:
| A(x) | = |  |
| = | \editable{} - \editable{} |
Letting x=b we obtain the familiar form of the fundamental theorem of calculus:
We have now shown the result to be true for positive continuous functions f(t). This can be extended to all continuous functions to include those that lie all or partially below the horizontal axis. In these cases the area function is replaced with the signed area function which evaluates the net area as the areas above the horizontal axis and subtracting those those regions which fall below.