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Steven Prascius
Mrs. Tallman
AP Calculus
3 March 2014
Relationship between the Derivative and the Integral
The fundamental Theorem of Calculus relates two very important aspects of Calculus
together: The derivative and the integral. Both the derivative and integral are central parts of
Calculus with nearly all parts of Calculus using them in some way. Because nearly all parts of
Calculus relate to the derivative and the integral it is important to understand their properties.
One part of the Fundamental Theorem of Calculus is the derivative. The derivative is the
instantaneous rate of change of any function f(x), with respect to the x-value that corresponds to
that point f(x). For example, if given a function showing the relationship between distance and
time the derivative would be the instantaneous rate of change of distance over time at any given
x-value of f(x), or in other words would be the exact velocity at that given time. The derivative
of a function can numerically be found by taking the limit of the average rate of change over the
interval from c as x approaches c. The derivative can be found graphically by finding the slope
of function.
The second part of the Fundamental Theorem of Calculus is the integral. The integral
can either be a definite integral or an indefinite integral. The indefinite integral for a function,
f(x), is simply the opposite of the derivative or the antiderivative. In other words the indefinite
integral is the value of a function before a derivative is taken. For example, if given the value of
the derivative of function f’(x) the indefinite integral would be equal to the original function f(x).
The definite integral, however, relates to finding the area under the curve of a given function for
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𝑏
a set interval. The notation for a definite integral is ∫𝑎 𝑓(𝑥) ∗ 𝑑𝑥 where a and b are the start and
end points of an interval and f(x) is the function. The actual order of a and b is important
𝑏
𝑎
𝑏
because ∫𝑎 𝑓(𝑥) ∗ 𝑑𝑥 does not equal∫𝑏 𝑓(𝑥) ∗ 𝑑𝑥. The integral of ∫𝑎 𝑓(𝑥) ∗ 𝑑𝑥 is the same as
𝑎
∫𝑏 𝑓(𝑥) ∗ 𝑑𝑥 but negative. This can be seen when solving the integrals. The integral of
𝑎
𝑏
∫𝑏 𝑓(𝑥) ∗ 𝑑𝑥 is g(a) – g(b) whereas the integral of ∫𝑎 𝑓(𝑥) ∗ 𝑑𝑥 is g(b) – g(a) (with g(x) being
the integral of f(x)) which shows that the integrals produce the same numerical values but just
have a different sign.
Since the integral is the opposite of the derivative integrating cancels out differentiation.
If given the derivative of a function, taking the integral would result in getting the original
function. For example, if a function shows the relationship of velocity and time taking the
derivative would give a graph of acceleration over time whereas the integral would give a graph
of distance over time. Another example can be seen when considering a function of acceleration
over time. The derivative of this function would be the instantaneous rate of change of
acceleration over time whereas the integral would be velocity over time. The opposite nature of
the integral and derivative can also be seen when determining the units of a given graph.
The units of the derivative of any given graph can be found by dividing the dependent
variable, f(x), by the independent variable x of a graph. For example, if given a graph of distance
(feet) over time (seconds) the derivative would be in feet per second because the dependent
variable feet is divided by the independent variable seconds. The units of the integral, however,
are found by multiplying the dependent variable by the independent variable. If given a graph of
velocity (feet/second) over time (seconds) the integral would be in feet because the dependent
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variable feet per second is multiplied by the independent variable seconds which cancels out the
seconds leaving only feet as the unit.
Relating to the graphs of integrals and derivatives are critical points. A critical point is a
point on a graph where the derivative is zero or undefined. These critical points can be local
maximums, local minimums, global maximums, global minimums, or points of inflection on the
original function. Local max and mins are the max and mins of a certain portion of a function
whereas the global max and mins are the max and mins for the entire function. Points of
inflection are spots on a function where the function changes concavity, or the direction the
function faces. These critical points give important information regarding the appearance of the
original function and can be found using derivatives. The max and mins are found by taking the
derivative of the original function and finding at what x-values the derivative is equal to zero.
Once the x-values are determined the slope of the derivative is then looked at to determine if it is
positive or negative. If the slope of the derivative is negative where the critical point is the point
is a maximum, and if the slope is positive where the critical point is the point is a minimum.
Points of inflection are found by taking both the first and second derivatives of the original
function and then finding at what x-values the second derivative is equal to zero. These values
correspond to where the points of inflections are on a graph.
The signs of the first and second derivative also give important information about the
original function. The sign of the first derivative determines whether the original function is
increasing or decreasing. If the first derivative is positive (above the x-axis) the original function
is increasing. This makes sense because the first derivative is the rate of change of the function
and if the rate of change is positive then the function must be increasing. Because of this is
makes sense that when the first derivative is negative (below the x-axis) the original function is
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decreasing. When the first derivative is zero it shows that the original function has either a max
or min at that point. The sign of the second derivative on the other hand determines the
concavity of the original function. If the second derivative is positive then the original function
is concave upwards and if the second derivative is negative then the original function is concave
downwards. If the second derivative is zero it shows that there is a point of inflection on the
original function at that point. Even though the second derivative is used to determine the
concavity and locations of the points of inflection on the original function, the first derivative can
determines these things to. The concavity of the original function can be determined by
examining the slope of the first derivative. If the slope of the first derivative is positive then the
original function is concave up and if the slope of the first derivative is negative then the original
function is concave down. This makes sense because the second derivative is the rate of change
of the first derivative so if the slope of the first derivative is positive then the second derivative is
positive giving the original function an upward concavity with the same going for a negative
slope. The points of inflection of the original function can be determined by finding the max and
mins of the first derivative. These max and mins of the first derivative give the location of the
points of inflection because they are where the slope of the derivative changes which causes a
change in concavity of the original function which is what a point of inflection is.
An example of using critical points and the signs of the first and second derivative to find
information about the original function can be seen when considering figure 1.
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Figure 1. Graph of f’(x)
Figure 1 shows the derivative of the function f(x) in the interval from negative 5 to 5. By
considering the derivative of the original function the local maximums and minimums of the
original function can be found. The local maximums of f(x) occur when the derivative is zero
and has a negative slope (which shows the function is concave down) and the minimums occur
when the derivative is zero and has a positive slope. By looking at the derivative it is clear that
the local maximums of f(x) must occur when x is 4 and -3 because that is where the derivative is
zero and where the slope of the derivative is negative. Also, it is clear that the absolute
minimum for f(x) must occur when x is 1 because that is where the derivative is zero and has a
positive slope. This minimum is the absolute minimum on the graph because it is the only place
a min occurs. The derivative of the original function can also be used to find the points of
inflection of f(x). The points of inflections occur when the first derivative has a slope of zero.
By looking at the graph it is clear that the points of inflection for f(x) must occur when x=-4,-1,
and 2 because that is where the slope of the derivative is zero and also where the f(x) changes
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concavity because of the change in direction of slope. Finally, the concavity of the original
function f(x) can be found by looking at the sign of the slope of the first derivative. The original
function is concave up from negative 5 to negative 4 and from negative 1 to 2 because that is
where the slope of the derivative is positive. Also, f(x) has a positive slope from -5 to -3 and
from -1 to -4 because that is where the derivative is positive which gives the original function a
positive slope because the derivative is the slope of the original function
The Fundamental Theorem of Calculus gives the relationship between the derivative and
integral. The fundamental theorem of calculus comes in two forms which both relate integrals
and derivatives to one another. The first part of the Fundamental Theorem of Calculus is an
𝑏
algebraic method for calculating definite integral and states ∫𝑎 𝑓(𝑥)𝑑𝑥 = g(b) – g(a) where g’(x)
= f(x). This part of the theorem explains how a definite integral is found and states the derivative
of the integral of f(x) is equal to the original function f(x). In order to calculate a definite
integral the antiderivative g(x) of the original function must be found. One the antiderivative is
found the limits a and b are then substituted into g(x) as g(b) and g(a) and then subtracted giving
the value of the definite integral. This part of the Fundamental Theorem of Calculus can be
applied to the function found in figure 1. If we are given function g which is equal to g(x) =
𝑥
∫1 𝑓 ′ (𝑡)𝑑𝑡 and we have to find g(3) using the function in figure 1 we would find the definite
integral of f’(x) from 1 to 3. When you take an integral of a derivative you get the original
function f(x) so we are left with g(x) = f(x) integrated from 1 to 3. So to find g(3) we substitute
the limits into f(x) and subtract them and get f(3)-f(1) = 3.9-3=0.9.
The second part of the Fundamental Theorem of Calculus is the derivative of an integral
form. This part of the theorem states that if the function is continuous on a closed interval [a,b]
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𝑥
then g(x) = ∫𝑎 𝑓(𝑡)𝑑𝑡 ,when x is between a and b, is continuous on the closed interval [a,b] and
differentiable on the open interval (a,b) , and g’(x) = f(x). This theorem states that the derivative
of an integral is equal to the original function and that the derivative can be found by substituting
x for t in the function f(t) and then multiplying the now f(x) by the derivative of x. This theorem
is used when there is a variable for one of the limits of a definite integral. This theorem works
because the constant limit becomes zero because its derivative is zero leaving only the variable
part of the limit. This part of the Fundamental Theorem of Calculus can also be shown in an
𝑥
example from figure 1. If we were again given g(x) = ∫1 𝑓 ′ (𝑥)𝑑𝑥 and the function f’(x) and had
to find g’(3) we would use this second part of the Fundamental Theorem. We would begin by
integrating g(x) which would give us g(x) = f(x) because an integral of a derivative is just the
original function. We would then take the derivative of both sides giving us g’(x) = f’(x) and
plug in 3 as our x-value and look at the graph of f’(x) to determine what that is equal to. When
looking at the graph it is clear that f’(3) and g’(3) are equal to 1. Going off this example is
finding g’’(3) given the same information. Since g’(x) = f’(x) we would take the second
derivative giving us g’’(x) = f’’(x) where would then plug in 3 as x to find g’’(3). Since the
function in figure 1 is f’(x) we must find the slope or rate of change at x=3 to find out what
g’’(3) equals. By finding the slope between the points (2,2) and (4,0) we come up with a slope
of -1 which is what g’’(3) is equal to. Yet another example of how the second part of the
Fundamental Theorem of Calculus is used can be see when considering the function of w(x) =
𝑔(𝑥)
∫1
𝑓(𝑡)𝑑𝑡 and the values found in table 1. If we had to find the value of w’(3) we would use
Derivative of the Integral Form of the Fundamental Theorem because the upper limit of the
function is a variable and the lower limit is a constant. When we solve for w’(3) we start by
taking the derivative of w(x) which comes out to w’(x) = f(g(x))*g’(x) and then plug in 3 as x
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and solve for w’(3) which comes out to be w’(3)=f(4)*2=-1*2=-2. So for this example w’(3) = 2.
Another theorem relating to integrals and derivatives is the Intermediate Value Theorem.
The theorem states that if a function is continuous for all x-values in the closed interval [a,b] and
y is a number between f(a) and f(b) there is a number x = c in the interval (a,b) for which f(c) =
y. This theorem basically means that given a y-value in the interval a to b there is a x-value that
corresponds with that y-value which can be seen in figure 2 below
Figure 2. Representation of the Intermediate Value Theorem
An example of the Intermediate Value Theorem can be seen when considering table 1 below.
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Table 1
Function Values
Table 1 shows the functions F and G which are differentiable for all real numbers and
where G is strictly increasing. Also, the function H is given by H(x) = F(G(x))-6.
The Intermediate Value Theorem can be used to explain why there must be a value r for 1<r<3
such that H(r) =-5. In order to explain this, though, we must first determine if H(x) is continuous
to see if the IVT applies. Since functions G and F are differentiable for all real numbers that
means that H(x) is continuous and that the IVT applies. Now, the values of H(1) and H(3) must
be found. When plugged into the table these values come out to be H(1) =F(G(1))-6=3 and
H(3)=F(G(3))-6=-7. Since H(x) is continuous and because -5 is between 3 and -7 there must be a
value of r for 1<r<3 such that H(r) = -5.
Relating to the Intermediate Value Theorem is the Mean Value Theorem. The Mean
Value Theorem states that if f(x) is continuous on the closed interval [a,b] and differentiable on
the open interval (a,b) then there exists at least one point, c, in [a,b] where f’(c)=
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
. This
basically means that given a continuous function there is a point c that has an instantaneous rate
of change, or f’(c), that is equal to the average rate of change from a to b. In order for this
theorem to apply, though, the function must be both continuous and differentiable. If the
function is both continuous and differentiable then there must be one point c such that its
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instantaneous rate of change is equal to the average rate of change from a to b. This theorem is
demonstrated in figure 3 below.
Figure 3. Mean Value Theorem
Figure 3 shows a graph representing the Mean Value Theorem. As seen in the figure
point c on the function has an instantaneous rate of change (shown as the tangent line) that is
equal to the average rate of change between a and b (shown as the secant line).
An example of the Mean Value Theorem can be seen when again considering table 1 and the
function H(x). The Mean Value Theorem explains why there must be a value c for 1<c<3 such
that H’(c)= -5. As previously mentioned, H(x) is both continuous and differentiable so that the
Mean Value applies. Since the Mean Value Theorem applies then f’(c)=
out to be -5 =
−7−3
3−1
𝐻(𝑏)−𝐻(𝑎)
𝑏−𝑎
which comes
or -5 = -5. Since the instantaneous rate of change f’(c) is the same as the
average rate of change
𝐻(𝑏)−𝐻(𝑎)
𝑏−𝑎
and the function in continuous and differentiable then there
must be a value c for 1<c<3 that corresponds to f’(c).
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Another example that can be solved using the values in table 1 is to find an equation for
the line tangent to the graph of y =G-1(x) at x = 2 where G-1 is the inverse function of G. In order
to find the line tangent to the graph of y =G-1(x) at x = 2 we must first understand the
relationship between G-1 and G. Since G-1 is the inverse to G then when G(1) = 2 the inverse
G-1 is G-1(2)=1 because the x and the y get switched. When trying to find the line tangent to y
=G-1(x) we must find the rate of change at that point which is just the derivative of G-1(2)
because that is the slope of the line in the equation. The derivative of an inverse is found with
the equation G-1’(x) = 1/(G’(G-1(x)) which comes out to be G-1’(2) = 1/(G’(G-1(2)) or G-1’(x) =
1/(G’(1)) which is equal to 1/5. Now that the slope of 1/5 is found you simply plug it in along
with the x and y values into the line equation to get y = 1/5(x-2) +1 which is the equation for the
line tangent to the graph of y =G-1(x) at x = 2. The final example involving table 1 is to find
H’(3) if H(x) = x B(x) where B(x) = F-1(x). You begin by understanding the relationship
between F-1(x) and F(x). Since F-1(x) is the inverse of F(x) when F(1) = 3 the inverse F-1(x) is
F-1(3) = 1. You then take the derivative of H(x) which comes out to be H’(x) = 1*F-1(x) + x*F '1
(x) or H’(x) = 1*F-1(x) + x*1/(F’(F-1(x)) if you substitute F-1(x) in for B(x). You then solve for
H’(3) by plugging in 3 for x and 1 for F-1(x). This comes out to be H’(3) = 1+ 3*1/(F’(1)) or
H’(3) = 1 + ¾ or 1.75.
The Fundamental Theorem of Calculus relates the integral and the derivative to each
other. These two concepts are crucial to the understanding of Calculus and allow you to solve
problems that would be impossible to solve otherwise.
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Works Cited
Dawkins, Paul. "Pauls Online Notes : Calculus I - The Mean Value Theorem." Pauls
Online Notes : Calculus I - The Mean Value Theorem.
Http://tutorial.math.lamar.edu/, n.d. Web. 02 Mar. 2014
Bohun, Sean. "Intermediate Value Theorem." Associate Professor Sean Bohun. Associate
Professor Sean Bohun, n.d. Web. 02 Mar. 2014.