From the investigation we have seen that the fundamental theorem of calculus plays a key role in 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 $f\left(x\right)$f(x) continuous over the interval $\left(a,b\right)$(a,b), if we define $F\left(x\right)$F(x) by the integral 
, then:
Part 2:
For a function $f\left(x\right)$f(x) continuous over a closed interval $\left[a,b\right]$[a,b] with an antiderivative function $F\left(x\right)$F(x):
The second part of the theorem can be applied to evaluate indefinite integrals and find areas under curves. This theorem also highlights the connection between differentiation and integration as reverse processes. This property can help us simplify calculations that appear quite complex if we recognise we are taking the integral of a derivative function or if we are taking the derivative of an integral.
Worked examples
Example 1
Find $A'(x)$A′(x), given that 
.
Think: We are trying to find the derivative of an integral. If we look at the first part of the fundamental theorem of calculus we can see this will result in the function inside the integral with the variable $t$t, replaced with $x$x.
Do: $A'(x)=\frac{4x}{\sqrt{x^2-5}}$A′(x)=4x√x2−5. Let's also briefly look at how this would result from using the second part of the fundamental theorem of calculus. Let $F(t)$F(t) be an antiderivative of $f(t)$f(t), where $f(t)=\frac{4t}{\sqrt{t^2-5}}$f(t)=4t√t2−5, then we have:
| $A(x)$A(x) | $=$= | $F(x)-F(3)$F(x)−F(3) |
From the fundamental theorem of calculus |
|
|
|||
| $\therefore\ A'(x)$∴ A′(x) | $=$= | $F'(x)-F'(3)$F′(x)−F′(3) |
|
| $=$= | $F'(x)$F′(x) |
Since the derivative of a constant is $0$0 |
|
| $=$= | $f(x)$f(x) |
As $F(x)$F(x) was an antiderivative of $f(x)$f(x) |
|
| $=$= | $\frac{4x}{\sqrt{x^2-5}}$4x√x2−5 |
|
Reflect: As the derivative of a constant is zero, notice that the lower bound of the integral is irrelevant. That is, the exact value of the constant $a$a in the equation has no impact on the outcome.
Example 2
Determine 
.
Think: We are trying to find the derivative of an integral. The fundamental theorem of calculus tells us the result will be the function inside the integral with the variable $t$t, replaced with $x$x.
Do:
EXAMPLE 3
Calculate 
.
Think: We are trying to find the definite integral of a derivative. The process of integration will undo the process of differentiation up to the addition of a constant. So we can jump to the second step of evaluating the inner function $f(x)=\frac{x^4}{x^2+1}$f(x)=x4x2+1 at the end points of the integral and subtracting.
Do:
Practice questions
Question 1
The function $f$f has an antiderivative $F$F, and $F\left(3\right)=4$F(3)=4.
Express $\int_3^xf\left(t\right)dt$∫x3f(t)dt in terms of $F$F and $x$x.
Now, calculate $\frac{d}{dx}\int_3^xf\left(t\right)dt$ddx∫x3f(t)dt.
Question 2
Consider the expression $\frac{d}{dx}\int_k^x\frac{1}{\sqrt{t}}dt$ddx∫xk1√tdt.
What restrictions must be on the value of $k$k for the integration to be possible?
$k<0$k<0
A$k\le0$k≤0
B$k>0$k>0
C$k\ge0$k≥0
DEvaluate $\int_k^x\frac{1}{\sqrt{t}}dt$∫xk1√tdt.
Hence, evaluate $\frac{d}{dx}\int_k^x\frac{1}{\sqrt{t}}dt$ddx∫xk1√tdt and express your answer without using powers.
Question 3
Calculate the value of $\int_0^{\frac{\pi}{2}}\frac{d}{dx}\sin\left(3x-\frac{\pi}{4}\right)dx$∫π20ddxsin(3x−π4)dx.
The signed area function
In proving the fundamental theorem of calculus we considered the geometrical interpretation of the definite integral as the signed area between a curve and the $x$x-axis, and defined the signed area function as:
Specifically, given a function $f\left(t\right)$f(t) which is continuous on the interval $a\le t\le b$a≤t≤b. We define the signed area function $A\left(x\right)$A(x) to be the value of the net area under the curve $f\left(t\right)$f(t) between $t=a$t=a and $t=x$t=x, where $a\le x\le b$a≤x≤b.
For a positive function this will represent the shaded area shown below:
For a function which lies completely or partially below the horizontal axis, the value of the signed area function represents the total area above the horizontal axis minus the total area below the horizontal axis. For example, for the function below $A(x)=A_1-A_2$A(x)=A1−A2.
From the second part of the fundamental theorem of calculus we have that the signed area function for a function $f\left(t\right)$f(t) continuous over a closed interval $\left[a,b\right]$[a,b] with an antiderivative function $F\left(t\right)$F(t), can be evaluated as:
Worked example
Example 4
Consider the function $f(t)=8-2t$f(t)=8−2t for $t\ge0$t≥0 and the signed area function 
.
(a) State the function $A(x)$A(x) in terms of $x$x.
Think: Use the fundamental theorem of algebra to evaluate the the integral.
Do:
(b) Hence, evaluate $A(3)$A(3) and give a geometrical interpretation.
Think: Substitute $x=3$x=3 into the formula for $A(x)$A(x).
Do:
| $A(3)$A(3) | $=$= | $8(3)-(3)^2$8(3)−(3)2 |
| $=$= | $15$15 |
Interpretation: This will be the value of the signed area from $t=0$t=0 to $t=3$t=3, as the graph is above the horizontal axis for this interval this can be interpreted as the area shown below and could be confirmed by finding the area geometrically. Thus, the shaded area is $15$15 units2.
(c) For what values of $x$x is the function $A(x)$A(x) increasing?
Think: Visualise the area formed under the graph as $x$x increases from zero. This area will continue to grow until the graph falls below the horizontal axis and the area under the curve is subtracted from the area above.
Do: The function will increase until $x=4$x=4. Thus, the function is increasing on the interval $0
(d) Find when $A(x)=0$A(x)=0 and give a geometrical interpretation of this.
Think: Solve for when $A(x)=0$A(x)=0 and visualise this value on the graph.
Do: If $A(x)=0$A(x)=0, we have:
| $8x-x^2$8x−x2 | $=$= | $0$0 |
| $x(8-x)$x(8−x) | $=$= | $0$0 |
| $x$x | $=$= | $0$0 or $8$8 |
Interpretation: For $x=0$x=0 we would have 
which indicates the area under the curve from $t=0$t=0 to itself. Thus, the result indicates the area under a point is zero.
For $x=8$x=8, we can see from the graph below that the function gives the result of the signed area from $t=0$t=0 to $t=8$t=8, from the resulting net area of zero we can conclude that $A_1=A_2$A1=A2.
Practice questions
Question 4
Consider the function $f\left(t\right)=2t$f(t)=2t, where $t\ge0$t≥0.
Let $A\left(x\right)$A(x) be the area function which represents the area bound by $f\left(t\right)$f(t) and the horizontal axis from $0$0 to $x$x. That is, $A\left(x\right)=\int_0^xf\left(t\right)dt$A(x)=∫x0f(t)dt.
Find an expression for $A\left(x\right)$A(x).
Hence determine the area under the function $f\left(t\right)$f(t) from $t=0$t=0 to $t=4$t=4.
Question 5
Given that $F\left(x\right)=\int_1^xf\left(t\right)dt$F(x)=∫x1f(t)dt, with $F\left(4\right)=78$F(4)=78 and $F''\left(x\right)=6x$F′′(x)=6x, find $f\left(t\right)$f(t).
Question 6
Consider the function $f\left(t\right)=2t-6$f(t)=2t−6, and the signed area function $A\left(x\right)=\int_0^xf\left(t\right)dt$A(x)=∫x0f(t)dt.
Complete the following table of values:
$x$x $0$0 $1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $A\left(x\right)$A(x) $0$0 $-5$−5 $\editable{}$ $\editable{}$ $\editable{}$ $-5$−5 $\editable{}$ $7$7 For what values of $x$x is the function $A\left(x\right)$A(x) increasing?
Give your answer as an inequality.
Sketch a graph of $y=A\left(x\right)$y=A(x) for $0\le x\le7$0≤x≤7 below.
Loading Graph...At what values of $x$x is $A\left(x\right)=0$A(x)=0?
Enter both values on the same line, separated by a comma.
What is the significance of these values of $x$x?
These values are the points where the function $f\left(t\right)$f(t) crosses the $x$x-axis.
AThe gradient of the curve $f\left(t\right)$f(t) is horizontal at both of these points.
BThere is no area under the curve $f\left(t\right)$f(t) at either of these points.
CAt these $x$x-values, there are equal amounts of area on either side of the $x$x-axis under the curve $f\left(t\right)$f(t).
DState the function $A\left(x\right)$A(x) in terms of $x$x.
When a curve forms a familiar geometric shape, it's relatively easy to determine the area enclosed between the curve and the $x$x-axis.
But what if the curve you've graphed doesn't make a nice geometric shape? One approach is then to approximate the area between the curve and the $x$x-axis.
Under-approximating the area under a curve
Let's consider finding the area between the curve $y=x^2+1$y=x2+1 and the $x$x-axis from $x=0$x=0 to $x=1$x=1.
We're really stepping into the shoes of the Leibniz and Newton here as we put on our mathematician thinking caps!
How do you approximate a curved area? Using circles? But how?
Let's go back to primary school. How did you first experiment with the idea of the area of a shape? You counted the number of square centimetres inside the shape, often giving an approximation.
We'll take a similar approach, but instead of counting squares, we'll find the areas of rectangles.
In this first approach we'll under-approximate the area and to do so we'll draw a series of $5$5 rectangles that sit underneath the curve. Why $5$5 rectangles? No particular reason, it's just a nice place to start.
To calculate the total area we find the area of each rectangle and add them all up.
We can see the width of each is $0.2$0.2, but what is the height? To find the height, we need to substitute the $x$x value into the function to calculate the $y$y value and hence the height of each rectangle.
$Area=0.2\left(0^2+1\right)+0.2\left(0.2^2+1\right)+0.2\left(0.4^2+1\right)+0.2\left(0.6^2+1\right)+0.2\left(0.8^2+1\right)$Area=0.2(02+1)+0.2(0.22+1)+0.2(0.42+1)+0.2(0.62+1)+0.2(0.82+1)
$Area=0.2\left(1+1.04+1.16+1.36+1.64\right)$Area=0.2(1+1.04+1.16+1.36+1.64)
$Area=1.24$Area=1.24 $units^2$units2
If we wanted a better approximation and less white space between the rectangles and the curve, what could we do? We could increase the number or rectangles, making the width of each smaller.
Let's try this with$10$10rectangles.
$Area=0.1\left(1+1.01+1.04+1.09+1.16+1.25+1.36+1.49+1.64+1.81\right)$Area=0.1(1+1.01+1.04+1.09+1.16+1.25+1.36+1.49+1.64+1.81)
$Area=1.285$Area=1.285 $units^2$units2
We could do more and more rectangles to get an even better approximation. Experiment with the applet below to see what happens to the under-approximation of the area.
Over-approximating the area under a curve
We can also choose to over-approximate instead of under-approximate our area.
Again I've chosen $5$5 rectangles, and notice the placement of the right hand side of the rectangle so that the rectangles are drawn over the curve.
$Area=0.2\left(1.04+1.16+1.36+1.64+2\right)$Area=0.2(1.04+1.16+1.36+1.64+2)
$Area=1.4325$Area=1.4325 $units^2$units2
The middle way for approximating area
We can also use a middle-way here and draw rectangles so that the middle of the rectangle touches the curve, as shown above.
$Area=0.2\left(1.01+1.09+1.25+1.47+1.81\right)$Area=0.2(1.01+1.09+1.25+1.47+1.81)
$Area=1.33$Area=1.33 $units^2$units2
Practice questions
Question 7
The function $f\left(x\right)=5x$f(x)=5x is defined on the interval $\left[0,6\right]$[0,6].
Graph $f\left(x\right)$f(x).
Loading Graph...Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into three sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into three subintervals of equal length and using rectangles whose widths are equal to the range of the intervals and whose heights are equal to the function values of the right corner of the rectangles
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into six sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into six sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the right side of the sub-intervals.
What is the actual area $A$A?
Question 8
The interval $\left[0,8\right]$[0,8] is partitioned into four sub-intervals $\left[0,2\right]$[0,2], $\left[2,4\right]$[2,4], $\left[4,6\right]$[4,6], and $\left[6,8\right]$[6,8].
| $x$x | $0$0 | $2$2 | $4$4 | $6$6 | $8$8 |
|---|---|---|---|---|---|
| $y$y | $11$11 | $5$5 | $10$10 | $5$5 | $5$5 |
Approximate the area $A$A using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Now approximate the area $A$A using rectangles for each sub-interval whose heights are equal to the function values of the right side of the sub-intervals.
Question 9
Approximate $\int_0^88xdx$∫808xdx by using four rectangles of equal width whose heights are the values of the function at the midpoint of each rectangle.