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BASIC CALCULUS REFRESHER
1. Introduction.
This is a very condensed and simplified version of basic calculus, which is a prerequisite for many courses in Mathematics, Statistics, Engineering, Pharmacy, etc. It is not comprehensive, and absolutely not intended to be a substitute for a one-year freshman course in differential and integral calculus. You are strongly encouraged to do the included Exercises to reinforce the ideas. Important mathematical terms are in boldface ; key formulas and concepts are boxed and highlighted
2. Exponents – Basic Definitions and Properties
For any real number base x , we define powers of x : x^0 = 1, x^1 = x , x^2 = x x , x^3 = x x x , etc. (The exception is 0^0 , which is considered indeterminate .) Powers are also called exponents.
Examples: 5^0 = 1, ( 11.2)^1 = 11.2, (8.6)^2 = 8.6 8.6 = 73.96, 10^3 = 10 10 10 = 1000, ( 3)^4 = ( 3) ( 3) ( 3) ( 3) = 81.
Also, we can define fractional exponents in terms of roots , such as x 1/2^ = x , the square root of x.
Similarly, x 1/3^ =
3 x , the cube root of x , x 2/3^ = (
3 x )
2 , etc. In general, we have xm/n^ = (
n x )
m , i.e., the n th^ root of x , raised to the m th^ power.
Examples: 641/2^ = 64 = 8, 643/2^ = ( 64 )
3 = 8^3 = 512, 641/3^ =
3 64 = 4, 642/3^ = (
3 64 )
2 = 4^2 = 16.
Finally, we can define negative exponents: x r^ =
xr^
. Thus, x^1 =
x^1 , x^2 =
x^2 , x 1/2^ =
x 1/^
x
, etc.
Examples: 10 1 =
101 = 0.1, 7^
2 =^1
49 , 36^
6 , 9^
( 9 )
5
35 =^
Properties of Exponents
xa^ xb^ = xa+b^ Examples: x^3 x^2 = x^5 , x 1/2^ x 1/3^ = x 5/6, x^3 x 1/2^ = x 5/
xa xb^ = xa^ b^ Examples:
x^5 x^3 = x^2 ,
x^3 x^5 = x^2 ,
x^3 x 1/^ = x 5/
- ( xa ) b^ = xab^ Examples: ( x^3 )^2 = x^6 , ( x 1/2)^7 = x 7/2, ( x 2/3)5/7^ = x 10/
( 1.5, 0)
(–1.8, 7) (0, 7) (2.5, 7)
(0, 3)
Descartes ~ 1640
3. Functions and Their Graphs Input x Output y
If a quantity y always depends on another quantity x in such a way that every value of x corresponds to one and only one value of y , then we say that “ y is a function of x ,” written y = f ( x ); x is said to be the independent variable , y is the dependent variable. (Example: “Distance traveled per hour ( y ) is a function of velocity ( x ).”) For a given function y = f ( x ), the set of all ordered pairs of ( x , y )- values that algebraically satisfy its equation is called the graph of the function, and can be represented geometrically by a collection of points in the XY -plane. (Recall that the XY -plane consists of two perpendicular copies of the real number line – a horizontal X -axis , and a vertical Y -axis – that intersect at a reference point (0, 0) called the origin , and which partition the plane into four disjoint regions called quadrants. Every point P in the plane can be represented by the ordered pair ( x , y ), where the first value is the x -coordinate – indicating its horizontal position relative to the origin – and the second value is the y -coordinate – indicating its vertical position relative to the origin. Thus, the point P (4, 7) is 4 units to right of, and 7 units up from, the origin.)
Examples: y = f ( x ) = 7; y = f ( x ) = 2 x + 3; y = f ( x ) = x^2 ; y = f ( x ) = x 1/2; y = f ( x ) = x^1 ; y = f ( x ) = 2 x.
The first three are examples of polynomial functions. (In particular, the first is constant , the second is linear , the third is quadratic .) The last is an exponential function; note that x is an exponent!
Let’s consider these examples, one at a time.
y = f ( x ) = 7: If x = any value, then y = 7. That is, no matter what value of x is chosen, the value of the height y remains at a constant level of 7. Therefore, all points that satisfy this equation must have the form ( x , 7), and thus determine the graph of a horizontal line, 7 units up. A few typical points are plotted in the figure.
Exercise: What would the graph of the equation y = 4 look like? x = 4? y = 0? x = 0?
y = f ( x ) = 2 x + 3: If x = 0, then y = f (0) = 2(0) + 3 = 3, so the point (0, 3) is on the graph of this function. Likewise, if x = 1.5, then y = f ( 1.5) = 2( 1.5) + 3 = 0, so the point ( 1.5, 0) is also on the graph of this function. (However, many points, such as (1, 1), do not satisfy the equation, and so do not lie on the graph.) The set of all points ( x , y ) that do satisfy this linear equation forms the graph of a line in the XY -plane, hence the name.
Exercise: What would the graph of the line y = x look like? y = x? The absolute value function y = | x |?
f
y = f ( x ) = x^1 =
x :^ This is a bit more delicate.^ Let’s first restrict our attention^ to^ positive
domain x -values, i.e., x > 0. If x = 1, then y = f (1) =
1 = 1, so the point (1, 1) lies on the graph of this function. Now from here, as x grows larger (e.g., x = 10, 100, 1000, etc.), the values of the
height y =
x e.g.‚^
10 = 0.1‚^
100 = 0.01‚^
1000 = 0.001‚^ etc.^ become^ smaller ,^ although they never actually reach 0. Therefore, as we continue to move to the right, the graph approaches the X -axis as a horizontal asymptote , without ever actually touching it. Moreover, as x gets smaller from the point (1, 1) on the graph (e.g., x = 0.1, 0.01, 0.001, etc.), the values of the height
y =
x e.g.‚^
0.1 = 10‚^
0.01 = 100‚^
0.001 = 1000‚^ etc.^ become^ larger.^ Therefore, as we continue to move to the left, the graph shoots upwards, approaching the Y -axis as a vertical asymptote , without ever actually touching it. (If x = 0, then y becomes infinite (+ ), which is undefined as a real number, so x = 0 is not in the domain of this function.) A similar situation exists for negative domain x -values, i.e., x < 0. This is the graph of a hyperbola , which has two symmetric branches , one in the first quadrant and the other in the third.
Exercise: How does this graph differ from that of y = f ( x ) = x^2 =
x^2 ? Why?
p < 0
0 < p < 1
p = 1
y = x p
p = 0
p > 1
NOTE: The preceding examples are special cases of power functions , which have the general form y = x p , for any real value of p , for x > 0. If p > 0, then the graph starts at the origin and continues to rise to infinity. (In particular, if p > 1, then the graph is concave up , such as the parabola y = x^2. If p = 1, the graph is the straight line y = x. And if 0 < p < 1, then the graph is concave down , such as
the parabola y = x 1/2^ = x .) However, if p < 0, such as y = x^1 =
x , or^ y^ =^ x^
2 =^1
x^2 , then the^ Y -axis acts as a vertical asymptote for the graph, and the X- axis is a horizontal asymptote.
Exercise: Why is y = xx^ not a power function? Sketch its graph for x > 0.
Exercise: Sketch the graph of the piecewise-defined functions
x^2 , if x 1 f ( x ) = x^3 , if x > 1
This graph is the parabola y = x^2 up to and including the point (1, 1), then picks up with the curve y = x^3 after that. Note that this function is therefore continuous at x = 1, and hence for all real values of x.
x^2 , if x 1 g ( x ) = x^3 + 5, if x > 1.
This graph is the parabola y = x^2 up to and including the point (1, 1), but then abruptly changes over to the curve y = x^3 + 5 after that, starting at (1, 6). Therefore, this graph has a break, or “jump discontinuity,” at x = 1. (Think of switching a light from off = 0 to on = 1.) However, since it is continuous before and after that value, g is described as being piecewise continuous.
Newton, Leibniz ~ 1680
4. Limits and Derivatives
We saw above that as the values of x grow ever larger , the values of
x become ever^ smaller.
We can’t actually reach 0 exactly, but we can “sneak up” on it, forcing
x to become^ as close to 0 as we like , simply by making x large enough. (For instance, we can force 1/ x < 10 500 by making
x > 10^500 .) In this context, we say that 0 is a limiting value of the
x values, as^ x^ gets arbitrarily large. A mathematically concise way to express this is a “limit statement”:
Many other limits are possible, but we now wish to consider a special kind. To motivate this, consider again the parabola example y = f ( x ) = x^2. The average rate of change between the two points P (3, 9) and Q (4, 16) on the graph can be calculated as the slope of the secant line connecting
them, via the previous formula: m sec =
y x
= 7. Now suppose that we slide to a new
point Q (3.5, 12.25) on the graph , closer to P (3, 9). The average rate of change is now m sec =
y x = 12.25 9 3.5 3 = 6.5, the slope of the new secant line between P and Q. If we now slide to a new point
Q (3.1, 9.61) still closer to P (3, 9), then the new slope is m sec =
y x =^
3.1 3 =^ 6.1, and so on. As Q approaches P , the slopes m sec of the secant lines appear to get ever closer to 6 – the slope m tan of the tangent line to the curve y = x^2 at the point P (3, 9) – thus measuring the instantaneous rate of change of this function at this point P (3, 9). (The same thing also happens if we approach P from the left side.) We can actually verify this by an explicit computation: From fixed point P (3, 9) to
any nearby point Q (3 + x , (3 + x )^2 ) on the graph of y = x^2 , we have m sec =
y x
(3 + x )^2 x
9 + 6 x + ( x )^2 x
6 x + ( x )^2 x
x (6 + x ) x = 6 + x. (We can check this formula against
the m sec values that we already computed: if x = 1, then m sec = 7 ; if x = 0.5, then m sec = 6.5 ; if x = 0.1, then m sec = 6.1 .) As Q approaches P – i.e., as x approaches 0 – this quantity m sec = 6 + x approaches the quantity m tan = 6 as its limiting value, confirming what we initially suspected.
lim
x^ x^ = 0.
X
Y
P
Q
tangent line
secant lines
X
Y
P (0, 0), m tan = 0
P (3, 9), m tan = 6
P (4, 16), m tan = 8
P ( 2, 4), m tan = 4
P ( 5, 25), m tan = 10
Suppose now we wish to find the instantaneous rate of change of y = f ( x ) = x^2 at some other point P on the graph, say at P (4, 16) or P ( 5, 25) or even P (0, 0). We can use the same calculation as we did above: the average rate of change of y = x^2 between any two generic points P ( x , x^2 ) and
Q ( x + x , ( x + x )^2 ) on its graph, is given by m sec =
y x =
( x + x )^2 x^2 x =^
x^2 + 2 x x + ( x )^2 x^2 x
=
2 x x + ( x )^2 x
x (2 x + x ) x = 2 x + x. As Q approaches P – i.e., as x approaches 0 – this
quantity approaches m tan = 2 x “in the limit,” thereby defining the instantaneous rate of change of the function at the point P. (Note that if x = 3, these calculations agree with those previously done for m sec and m tan .) Thus, for example, the instantaneous rate of change of the function y = f ( x ) = x^2 at the point P (4, 16) is equal to m tan = 8, at P ( 5, 25) is m tan = 10, and at the origin P (0, 0) is m tan = 0.
In principle, there is nothing that prevents us from applying these same ideas to other functions y = f ( x ). To find the instantaneous rate of change at an arbitrary point P on its graph, we first calculate the average rate of change between P ( x , f ( x )) and a nearby point Q ( x + x , f ( x + x )) on
its graph, as measured by m sec =
y x
f ( x + x ) f ( x ) x
. As Q approaches P – i.e., as x approaches
0 from both sides – this quantity becomes the instantaneous rate of change at P , defined by:
This object – denoted compactly by
dy dx –^ is called the^ derivative^ of the function^ y^ =^ f ( x )^ , and is
also symbolized by several other interchangeable notations:
d f(x) dx ,^
d dx [ f^ ( x )],^ f^ ( x ), etc. (The last is sometimes referred to as “prime notation.”) The process of calculating the derivative of a function is called differentiation. Thus, the derivative of the function y = f ( x ) = x^2 is equal to the function dy dx = f ( x ) = 2 x. This can also be written more succinctly as
d ( x^2 ) dx = 2 x ,
d dx ( x^2 ) = 2 x , or dy = 2 x dx.
The last form expresses the so-called differential dy in terms of the differential dx , which can be used to estimate a small output difference y in terms of a small input difference x , i.e., y ≈ 2 x x.
m tan = lim
y x = lim
f ( x + x ) f ( x ) x x 0 x 0
Properties of Derivatives
1. For any constant c , and any differentiable function f ( x ),
d dx [ c^ f ( x )] =^ c^
df dx
[ c f ( x )] = c f ( x )
For any two differentiable functions f ( x ) and g ( x ),
2. Sum and Difference Rules d dx [^ f ( x )^ g ( x ) ] =^
df dx
dg dx
[ f ( x ) g ( x ) ] = f ( x ) g ( x )
3. Product Rule [ f ( x ) g ( x ) ] = f ( x ) g ( x ) + f ( x ) g ( x )
Example: If y = x^11 e^6 x , then
dy dx = (11^ x
(^10) )( e 6 x ) + ( x (^11) )(6 e 6 x ) = (11 + 6 x ) x (^10) e 6x.
4. Quotient Rule
[
f ( x ) g ( x )
] =
f ( x ) g ( x ) f ( x ) g ( x ) [ g ( x )]^2 provided g ( x ) 0
Example: If y =
e^4 x x^7 + 8 , then^
dy dx =^
( x^7 + 8 ) 4 e^4 x^ ( 7 x^6 ) e^4 x ( x^7 + 8 )^2 =^
( 4 x^7 7 x^6 + 32 ) e^4 x ( x^7 + 8 )^2.
5. Chain Rule NOTE: See below for a more detailed explanation.
[ f ( g ( x ) )] = f ( g ( x ) ) g ( x )
Example: If y = ( x 2/3^ + 2 e^9 x )^6 , then
dy dx = 6 ( x
2/3 + 2 e 9 x ) 5 ( 2
3 x^
1/3 18 e 9 x ).
Example: If y = e x ²/2, then
dy dx
= e x ²/2^ ( 22 x ) = x e x ²/2.
The graph of this function is related to the “bell curve” of probability and statistics. Note that you cannot calculate its derivative by the “exponential rule” given above, because the exponent is a function of x , not just x itself (or a constant multiple ax )!
Example: If y = ln(7 x^10 + 8 x^6 – 4 x + 11), then
dy dx =^
7 x^10 + 8 x^6 – 4 x + 11 (70 x
(^9) + 48 x (^5) – 4).
Example: If y = x 3/2^7 x^4 + 10 e^3 x^ 5, then dy dx =^
2 x
1/2 (^28) x (^3 30) e 3 x.
Example: If y = 5 x^3 , then
dy dx = 5 (3^ x
(^2) ) = 15 x (^2).
Example: If y =
9 x , then^
dy dx =^
9 (1) =^
Example: If y = 3 e^2 x , then
dy dx =^ 3 (2^ e^
2 x ) = 6 e 2 x.
The Chain Rule , which can be written several different ways, bears some further explanation. It is a rule for differentiating a composition of two functions f and g , that is, a function of a function y = f ( g ( x ) ). The function in the first example above can be viewed as composing the “outer” function f ( u ) = u^6 , with the “inner” function u = g ( x ) = x 2/3^ + 2 e^9 x. To find its derivative, first take the derivative of the outer function (6 u^5 , by the Power Rule given above), then multiply that by the derivative of the inner, and we get our answer. Similarly, the function in the second example can be viewed as composing the outer exponential function f ( u ) = eu^ (whose derivative, recall, is itself), with the inner power function u = g ( x ) = x^2 /2. And the last case can be seen as composing the outer logarithm function f ( u ) = ln( u ) with the inner polynomial function u = g ( x ) = 7 x^10 + 8 x^6 – 4 x + 11.
Hence, if y = f ( u )‚ and u = g ( x )‚ then
dy dx =^ f^ ( u )^ g^ ( x ) is another way to express this procedure.
The logic behind the Chain Rule is actually quite simple and intuitive (though a formal proof involves certain technicalities that we do not pursue here). Imagine three cars traveling at different rates of speed over a given time interval: A travels at a rate of 60 mph, B travels at a rate of 40 mph, and C travels at a rate of 20 mph. In addition, suppose that A knows how fast B is traveling, and B knows how fast C is traveling, but A does not know how fast C is traveling. Over this time span, the average rate of change of A , relative to B , is equal to the ratio of their respective distances traveled,
i.e.,
A
B
2 , three-halves as much. Similarly, the average rate of change of^ B ,^ relative to C , is
equal to the ratio
B
C
1 , twice as much. Therefore, the average rate of change of^ A ,^ relative to C , can be obtained by multiplying together these two quantities, via the elementary algebraic identity A C
A
B
B
C
, or
, three times as much. (Of course, if A has direct knowledge of C , this
would not be necessary, for
A
C
1.^ As it is,^ B^ acts an intermediate link in the chain, or an auxiliary function , which makes calculations easier in some contexts.) The idea behind the Chain Rule is that what is true for average rates of change also holds for instantaneous rates of change, as the time interval shrinks to 0 “in the limit,” i.e., derivatives.
With this insight, an alternate – perhaps more illuminating – equivalent way to write the Chain Rule
is
dy dx =^
dy du
du dx which, as the last three examples illustrate, specialize to a^ General Power Rule ,
General Exponential Rule , and General Logarithm Rule for differentiation, respectively:
Exercise: Recall from page 5 that y = xx^ is not a power function. Obtain its derivative via the technique of implicit (in particular, logarithmic ) differentiation. (Not covered here.)
If y = u p , then
dy dx =^ p u^
p 1 du dx.^ That is,^ d ( u^
p ) = p u p (^1) du.
If y = e u , then
dy dx =^ e^
u du dx.^ That is,^ d ( e^
u ) = e u (^) du.
If y = ln( u ), then
dy dx =
u
du
dx.^ That is,^ d [ln( u )]^ =
u du.
Notice the extremely rapid convergence to the root. In fact, it can be shown that the small error (between each value and the true solution) in each iteration is approximately squared in the next iteration, resulting in a much smaller error. This feature of quadratic convergence is a main reason why this is a favorite method. Why does it work at all? Suppose P 0 is a point on the graph of f ( x ), whose x -coordinate x 0 is reasonably close to a root. Generally speaking, the tangent line at P 0 will then intersect the X -axis at a value much closer to the root. This value x 1 can then be used as the x -coordinate of a new point P 1 on the graph, and the cycle repeated until some predetermined error tolerance is reached. Algebraically formalizing this process results in the general formula given above.
If a function f ( x ) has either a relative maximum (i.e., local maximum ) or a relative minimum
(i.e., local minimum ) at some value of x , and if f ( x ) is differentiable there, then its tangent line must be horizontal, i.e., slope m tan = f ( x ) = 0. This suggests that, in order to find such relative extrema (i.e., local extrema ), we set the derivative f ( x ) equal to zero, and solve the resulting algebraic equation for the critical values of f , perhaps using a numerical approximation technique like Newton’s Method described above. ( But beware : Not all critical values necessarily correspond to relative extrema! More on this later…)
Example (cont’d): Find and classify the critical points of y = f ( x ) = x^3 – 21 x^2 + 135 x – 220.
We have f ( x ) = 3 x^2 – 42 x + 135 = 0. As this is a quadratic (degree 2) polynomial equation, a numerical approximation technique is not necessary. We can use the quadratic formula to solve this explicitly, or simply observe that, via factoring, 3 x^2 – 42 x + 135 = 3 ( x – 5)( x – 9) = 0. Hence there are two critical values , x = 5 and x = 9. Furthermore, since f (5) = 55 and f (9) = 23, it follows that the corresponding critical points on the graph of f ( x ) are (5, 55) and (9, 23).
Once obtained, it is necessary to determine the exact nature of these critical points. Consider the first critical value, x = 5, where f = 0. Let us now evaluate the derivative f ( x ) at two nearby values that bracket x = 5 on the left and right, say x = 4 and x = 6. We calculate that:
m tan(4) = f (4) = +15 > 0, which indicates that the original function f is increasing at x = 4,
m tan(5) = f (5) = 0, which indicates that f is neither increasing nor decreasing at x = 5,
m tan(6) = f (6) = – 9 < 0, which indicates that the original function f is decreasing at x = 6.
Hence, as we move from left to right in a local neighborhood of x = 5, the function f ( x ) rises, levels off, then falls. This indicates that the point (5, 55) is a relative maximum for f , and demonstrates an application of the “ First Derivative Test ” for determining the nature of critical points. In an alternate method, the “ Second Derivative Test ,” we evaluate f ( x ) = 6 x – 42 at the critical value x = 5, i.e.,
f (5) = – 12 < 0, which indicates that the original function f is concave down (“spills”) at this value.
Hence, this also shows that the point (5, 55) is a relative maximum for f , consistent with the above.
Exercise: Show that: (1) f (8) < 0, f (10) > 0 First Derivative Test (2) f (9) > 0. Second Derivative Test
In either case , conclude that the point (9, 23) is a relative minimum for f.
f decreases
f concave down (^) f concave up
f increases f increases
Finally, notice that f ( x ) = 6 x – 42 = 0 when x = 7, and in a local neighborhood of that value,
f (6) = – 6 < 0, which indicates that the original function f is concave down (“spills”) at x = 6,
f (7) = 0, which indicates that f is neither concave down nor concave up at x = 7,
f (8) = +6 > 0, which indicates that the original function f is concave up (“holds”) at x = 8.
Hence, across x = 7, there is change in concavity of the function f ( x ). This indicates that (7, 39) is a point of inflection for f.
The full graph of f ( x ) = x^3 – 21 x^2 + 135 x – 220 is shown below, using all of this information.
As we have shown, this function has a relative maximum (i.e., local maximum ) value = 55 at x = 5, and a relative minimum (i.e., local minimum ) value = 23 at x = 9. But clearly, there are both higher and lower points on the graph! For example, if x 11, then f ( x ) 55; likewise, if x 3, then f ( x ) 23. (Why?) Therefore, this function has no absolute maximum (i.e., global maximum ) value, and no absolute minimum (i.e., global minimum ) value. However, if we restrict the domain to an interval that is “closed and bounded” (i.e., compact ), then both absolute extrema are attained. For instance, in the interval [4, 10], the relative extreme points are also the absolute extreme points, i.e. the function attains its global maximum and minimum values of 55 (at x = 5) and 23 (at x = 9), respectively. However, in the interval [4, 12], the global maximum of the function is equal to 104, attained at the right endpoint x = 12. Similarly, in the interval [0, 12], the global minimum of the function is equal to – 220, attained at the left endpoint x = 0.
We formally define a new function
F ( x ) = Area under the graph of f in the interval [ a , x ].
Clearly, because every value of x results in one and only one area (shown highlighted above in blue), this is a function of x, by definition! Moreover, F must also have a strong connection with f itself. To see what that connection must be, consider a nearby value x + x. Then,
F ( x + x ) = Area under the graph of f in the interval [ a , x + x ],
and take the difference of these two areas (highlighted above in green):
F ( x + x ) F ( x ) = Area under the graph of f in the interval [ x , x + x ]
= Area of the rectangle with height f ( z ) and width x (where z is some value in the interval [ x , x + x ])
= f ( z ) x.
Therefore, we have F ( x + x ) F ( x ) x = f ( z ).
Now take the limit of both sides as x 0. We see that the left hand side becomes the derivative of F ( x ) (recall the definition of the derivative of a function, previously given) and, noting that z x as x 0, we see that the right hand side f ( z ) becomes (via continuity) f ( x ). Hence,
F ( x ) = f ( x ) ,
i.e., F is an antiderivative of f.
Therefore, we formally express… F ( x ) = a
x f ( t) dt ,
where the right-hand side a
x f ( t) dt represents the definite integral of the function f from a to x.
(In this context, f is called the integrand .) More generally, if F is any antiderivative of f , then the
two functions are related via the indefinite integral : f ( x ) dx = F ( x ) + C , where C is an arbitrary
constant.
Example 1: F ( x ) = 1 10 x
(^10) + C (where C is any constant) is the general antiderivative of f ( x ) = x (^9) ,
because F ( x ) = 1 10 (10 x
(^9) ) + 0 = x (^9) = f ( x ).
We can write this relation succinctly as x^9 dx = 1 10 x
10 + C.
Example 2: F ( x ) = 8 e x /8^ + C (where C is any constant) is the general antiderivative of f ( x ) = e x /8,
because F ( x ) = 8 ( 1 8 e^
x /8) + 0 = e x /8 (^) = f ( x ).
We can write this relation succinctly as e x /8^ dx = 8 e x /8^ + C.
NOTE: Integrals possess the analogues of Properties 1 and 2 for derivatives, found on page 10. In particular, the integral of a constant multiple of a function, c f(x), is equal to that constant multiple c, times the integral of the function f(x). Also, the integral of a sum (respectively, difference) of two functions is equal to the sum (respectively, difference) of the integrals. (The integral analogue for products corresponds to a technique known as integration by parts ; not reviewed here.) These are extremely important properties for the applications that follow.
From the previous two examples, it is evident that the differentiation rules for power and exponential functions can be inverted (essentially by taking the integral of both sides) to the General Power Rule , General Logarithm Rule , and General Exponential Rule for integration:
NOTE: In order to use these formulas correctly, du must be present in the integrand (up to a constant multiple ). To illustrate…
u p + p + 1
u p^ du = ln | u | + C , if p = 1
eu^ du = eu^ + C
Properties of Integrals
1. For any constant c , and any integrable function f ( x ),
[ c f ( x )] dx = c f ( x ) dx
For any two integrable functions f ( x ) and g ( x ),
2. Sum and Difference Rules
[ f ( x ) g ( x )] dx = f ( x ) dx g ( x ) dx
a
b f ( x ) dx
Finally, all these results can be summarized into one elegant statement, the Fundamental Theorem
of Calculus for definite integrals : a
b f ( x ) dx = F ( b ) F ( a ). (Advanced techniques of integration
- such as integration by parts , trigonometric substitution , partial fractions , etc. – will not be reviewed here.)
Example 7: 0
1 x^3 (1 x^4 )^2 dx
This definite integral represents the amount of area under the curve f ( x ) = x^3 (1 – x^4 )^2 , from x = 0 to x = 1.
Method 1. Expand and integrate term-wise: 0
1 x^3 (1 2 x^4 + x^8 ) dx = 0
1 ( x^3 2 x^7 + x^11 ) dx
= [
x^4 4
2 x^8 8 +^
x^12
12 ]^
1 0
= [
2 (1)^8
12 ]^ [^
2 (0)^8
12 ]^ =^
12 –^ 0 =^
Method 2. Use the power function formula (if possible): If u = 1 – x^4 , then du = 4 x^3 dx , and x^3 is indeed present in the integrand. Recall that the x -limits of integration should also be converted to u -limits: when x = 0, we get u = 1 – 04 = 1; when x = 1, we get u = 1 14 = 0.
x= 0
x= 1 (1 x^4 )^2 ( 4 x^3 ) dx =
u= 1
u= 0 u^2 du =
0
1 u^2 du =
4 [
u^3
3 ]^
1 0
4 [
3 ]^ =^
NOTE: Numerical integration techniques, such as the Trapezoidal Rule , are sometimes used also.
u^2 du
7. Differential Equations
As a first example, suppose we wish to find a function y = f ( x ) whose derivative
dy dx is given , e;g.,
dy dx = x^2. Formally, by separation of variables – y on the left, and x on the right – we can rewrite this ordinary differential equation (or o.d.e. ) as dy = x^2 dx. In this differential form, we can now integrate both sides explicitly to obtain y =
3 x
(^3) + C , where C is an arbitrary
additive constant. Note that the “solution” therefore actually represents an entire family of functions; each function corresponds to a different value of C. Further specifying an initial value (or initial condition ), such as y (2) = 5, singles out exactly one of them passing
through the chosen point, in this case, y =
3 x
3 +^7
Now consider the case of finding a function y = f ( x )
whose rate of change
dy dx is proportional to^ y^ itself, i.e., dy dx =^ a^ y ,^ where^ a^ is^ a^ known^ constant^ of proportionality (either positive or negative).
Separating variables produces
y dy^ =^ a^ dx , integrating yields ln( y ) = a x + C , and solving gives y = e ax^ +^ C^ = e ax^ eC , or y = A e ax , where A is an arbitrary (positive) multiplicative constant. Hence this is a family of exponential curves, either increasing for a > 0 (as in unrestricted population growth, illustrated) or decreasing for a < 0 (as in radioactive isotope decay). Specifying an initial amount y (0) = y 0 “when the clock starts at time zero” yields the unique solution y = y 0 e ax.
Finally, suppose that population size y = f ( x ) is restricted
between 0 and 1, such that the rate of change
dy dx is
proportional to the product y (1 – y ), i.e.,
dy dx = a y (1 – y ), where a > 0. With the initial condition y (0) = y 0 , the
solution is given by y =
y 0 e ax y 0 e ax^ + (1 – y 0 ).^ This is known as the logistic curve , which initially resembles the exponential curve, but remains bounded as x gets large.
NOTE: Many types of differential equation exist, including those that cannot be explicitly solved using “elementary” techniques. Like integration above, such equations can be solved via Euler’s Method and other, more sophisticated, numerical techniques.
(2, 5)
( 0 , y 0 )
( 0 , y 0 )
