# 5.1: The Indefinite Integral - Mathematics

Derivatives appear in many physical phenomena, such as the motion of objects. Recall, for example, that given the position function (s(t)) of an object moving along a straight line at time (t), you could find the velocity (v(t)=s'(t)) and the acceleration (a(t)=v'(t)) of the object at time (t) by taking derivatives. Suppose the situation were reversed: given the velocity function how would you find the position function, or given the acceleration function how would you find the velocity function?

In this case calculating a derivative would not help, since the reverse process is needed: instead of differentiation you need a way of performing antidifferentiation, i.e. you would calculate an antiderivative.

Differentiation is relatively straightforward. You have learned the derivatives of many classes of functions (e.g. polynomials, trigonometric functions, exponential and logarithmic functions), and with the various rules for differentiation you can calculate derivatives of complicated expressions involving those functions (e.g. sums, powers, products, quotients). Antidifferentiation, however, is a different story.

To see some of the issues involved, consider a simple function like (f(x)=2x). Of course you know that (ddx(x^2) = 2x), so it seems that (F(x)=x^2) is the antiderivative of (f(x)=2x). But is it the only antiderivative of (f(x))? No. For example, if (F(x)=x^2+1) then (F'(x)=2x=f(x)), and so (F(x)=x^2+1) is another antiderivative of (f(x)=2x). Likewise, so is (F(x)=x^2+2). In fact, any function of the form (F(x)=x^2 + C), where (C) is some constant, is an antiderivative of (f(x)=2x).

Another potential issue is that functions of the form (F(x)=x^2 + C) are just the most obvious antiderivatives of (f(x)=2x). Could there be some other completely different function—one that cannot be simplified into the form (x^2 + C)—whose derivative also turns out to be (f(x) =2x)? The answer, luckily, is no:

To prove this, consider the function (H(x) = F(x) - G(x)), defined for all (x) in the common domain (I) of (F) and (G). Since (F'(x) = G'(x) = f(x)), then

[H'(x) ~=~ F'(x) ~-~ G'(x) ~=~ f(x) ~-~ f(x) ~=~ 0] for all (x) in (I), so (H(x)) is a constant function on (I), as was shown in Section 4.4 on the Mean Value Theorem. Thus, there is a constant (C) such that

The practical consequence of the above result can be stated as follows:

So for the function (f(x) = 2x), since (F(x) = x^2) is one antiderivative then all antiderivatives of (f(x)) are of the form (F(x) = x^2 + C), where (C) is a generic constant. Thus, functions do not have just one antiderivative but a whole family of antiderivatives, all differing only by a constant. The following notation makes all this easier to express:

The large S-shaped symbol before (f(x)) is called an integral sign. Though the indefinite integral (int f(x)~dx) represents all antiderivatives of (f(x)), the integral can be thought of as a single object or function in its own right, whose derivative is (f'(x)):

You might be wondering what the integral sign in the indefinite integral represents, and why an infinitesimal (dx) is included. It has to do with what an infinitesimal represents: an infinitesimal “piece” of a quantity. For an antiderivative (F(x)) of a function (f(x)), the infinitesimal (or differential) (d!F) is given by (d!F = F'(x),dx = f(x),dx), and so

[F(x) ~=~ int,f(x)~dx ~=~ int,d!F ~.] The integral sign thus acts as a summation symbol: it sums up the infinitesimal “pieces” (d!F) of the function (F(x)) at each (x) so that they add up to the entire function (F(x)). Think of it as similar to the usual summation symbol (Sigma) used for discrete sums; the integral sign (int) takes the sum of a continuum of infinitesimal quantities instead.

Finding (or evaluating) the indefinite integral of a function is called integrating the function, and integration is antidifferentiation.

Example (PageIndex{1}): antideriv1

Solution

Evaluate (displaystyleint,0~dx).

Solution: Since the derivative of any constant function is 0, then (int,0~dx = C), where (C) is a generic constant.

Note: From now on (C) will simply be assumed to represent a generic constant, without having to explicitly say so every time.

Example (PageIndex{1}): antideriv2

Solution

Evaluate (displaystyleint,1~dx).

Solution: Since the derivative of (F(x) = x) is (F'(x) = 1), then (int,1~dx = x + C).

Example (PageIndex{1}): antideriv3

Solution

Evaluate (displaystyleint,x~dx).

Solution: Since the derivative of (F(x) = frac{x^2}{2}) is (F'(x) = x), then (int,x~dx = frac{x^2}{2} + C).

Since (ddx,left(frac{x^{n+1}}{n+1} ight) = x^n) for any number (n e -1), and (ddx,(ln,abs{x}) = frac{1}{x} = x^{-1}), then any power of (x) can be integrated:

The following rules for indefinite integrals are immediate consequences of the rules for derivatives:

The above rules are easily proved. For example, the first rule is a simple consequence of the Constant Multiple Rule for derivatives: if (F(x) = int,f(x)~dx), then

[ddx(k,F(x)) ~=~ k,ddx(F(x)) ~=~ k,f(x) quadRightarrowquad int,k;f(x)~dx ~=~ k,F(x) ~=~ k,int,f(x)~dx ~.quadcheckmark] The other rules are proved similarly and are left as exercises. Repeated use of the above rules along with the Power Formula shows that any polynomial can be integrated term by term—in fact any finite sum of functions can be integrated in that manner:

[antideriv4] Evaluate (displaystyleint,(x^7 - 3x^4)~dx).

Solution: Integrate term by term, pulling constant multiple outside the integral:

[int,(x^7 - 3x^4)~dx ~=~ int,x^7~dx ~-~ 3int,x^4~dx ~=~ frac{x^8}{8} ~-~ frac{3x^5}{5} ~+~ C]

[antideriv5] Evaluate (displaystyleint,sqrt{x}~dx).

Solution: Use the Power Formula:

[int,sqrt{x}~dx ~=~ int,x^{1/2}~dx ~=~ frac{x^{3/2}}{3/2} ~+~ C ~=~ frac{2x^{3/2}}{3} ~+~ C]

[antideriv6] Evaluate (displaystyleint,left(dfrac{1}{x^2} + dfrac{1}{x} ight)~dx).

Solution: Use the Power Formula and integrate term by term:

[int,left(frac{1}{x^2} + frac{1}{x} ight)~dx ~=~ int,left(x^{-2} + frac{1}{x} ight)~dx ~=~ frac{x^{-1}}{-1} ~+~ ln,abs{x} ~+~ C ~=~ -frac{1}{x} ~+~ ln,abs{x} ~+~ C]

The following indefinite integrals are just re-statements of the corresponding derivative formulas for the six basic trigonometric functions:

Since (ddx(e^x) = e^x), then:

[antideriv7] Evaluate (displaystyleint,(3sin,x ~+~ 4cos,x ~-~ 5e^x)~dx).

Solution: Integrate term by term:

[egin{aligned} int,(3sin,x ~+~ 4cos,x ~-~ 5e^x)~dx ~&=~ 3int,sin,x~dx ~+~ 4int,cos,x~dx ~-~ 5int,e^x~dx

[10pt] &=~ -3cos,x ~+~ 4sin,x ~-~ 5e^x ~+~ Cend{aligned}]

Example (PageIndex{1}): gravity

Solution

Recall from Section 1.1 the example of an object dropped from a height of 100 ft. Show that the height (s(t)) of the object (t) seconds after being dropped is (s(t) = -16t^2 + 100), measured in feet.

Solution: When the object is dropped at time (t=0) the only force acting on it is gravity, causing the object to accelerate downward at the known constant rate of 32 ft/s2. The object’s acceleration (a(t)) at time (t) is thus (a(t) = -32). If (v(t)) is the object’s velocity at time (t), then (v'(t) = a(t)), which means that

[v(t) ~=~ int a(t)~dt ~=~ int -32~dt ~=~ -32t ~+~ C] for some constant (C). The constant (C) here is not generic—it has a specific
value determined by the initial condition on the velocity: the object was at rest at time (t=0). That is, (v(0) = 0), which means

[0 ~=~ v(0) ~=~ -32(0) ~+~ C ~=~ C quadRightarrowquad v(t) ~=~ -32t] for all (t ge 0). Likewise, since (s'(t) = v(t)) then

[s(t) ~=~ int v(t)~dt ~=~ int -32t~dt ~=~ -16t^2 ~+~ C] for some constant (C), determined by the initial condition that the object was 100 ft above the ground at time (t=0). That is, (s(0) = 100), which means

[100 ~=~ s(0) ~=~ -16(0)^2 ~+~ C ~=~ C quadRightarrowquad s(t) ~=~ -16t^2 ~+~ 100] for all (t ge 0).

The formula for (s(t)) in Example

Example (PageIndex{1}): gravity

Solution

can be generalized as follows: denote the object’s initial position at time (t=0) by (s_0), let (v_0) be the object’s initial velocity (positive if thrown upward, negative if thrown downward), and let (g) represent the (positive) constant acceleration due to gravity. By Newton’s First Law of motion the only acceleration imparted to the object after throwing it is due to gravity:

[a(t) ~=~ -g quadRightarrowquad v(t) ~=~ int a(t)~dt ~=~ int -g~dt ~=~ -gt ~+~ C] for some constant (C): (v_0 = v(0) = -g(0) + C = C). Thus, (v(t) = -gt + v_0) for all (t ge 0), and so

[s(t) ~=~ int v(t)~dt ~=~ int left(-gt ~+~ v_0 ight)~dt ~=~ - frac{1}{2}gt^2 ~+~ v_0t ~+~ C] for some constant (C): (s_0 = s(0) = - frac{1}{2}g(0)^2 + v_0(0) + C = C). To summarize:

Note that the units are not specified—they just need to be consistent. In metric units, (g = 9.8) m/s2, while (g = 32) ft/s2 in English units.

Thinking of an indefinite integral as the sum of all the infinitesimal “pieces” of a function—for the purpose of retrieving that function—provides a handy way of integrating a differential equation to obtain the solution. The key idea is to transform the differential equation into an equation of differentials, which has the effect of treating functions as variables. Some examples will illustrate the technique.

Example (PageIndex{1}): intdecay

Solution

For any constant (k), show that every solution of the differential equation (dydt = ky) is of the form (y = Ae^{kt}) for some constant (A). You can assume that (y(t) > 0) for all (t).

Solution: Put the (y) terms on the left and the (t) terms on the right, i.e. separate the variables:

[frac{dy}{y} ~=~ k,dt] Now integrate both sides (notice how the function (y) is treated as a variable):

[egin{aligned} int,frac{dy}{y} ~&=~ int k,dt

[6pt] ln,y + C_1 ~&=~ kt + C_2 quad ext{($C_1$ and $C_2$ are constants)} ln,y ~&=~ kt + C quad ext{(combine $C_1$ and $C_2$ into the constant $C$)} y ~&=~ e^{kt+C} ~=~ e^{kt} cdot e^C ~=~ A e^{kt}end{aligned}] where (A = e^C) is a constant. Note that this is the formula for radioactive decay from Section 2.3.

Example (PageIndex{1}): intidealgas

Solution

Recall from Section 3.6 the equation of differentials

[dfrac{dP}{P} ~+~ dfrac{dV}{V} ~=~ dfrac{dT}{T}] relating the pressure (P), volume (V) and temperature (T) of an ideal gas. Integrate that equation to obtain the original ideal gas law (PV = RT), where (R) is a constant. .

Solution: Integrating both sides of the equation yields

[egin{aligned} int,dfrac{dP}{P} ~+~ int,dfrac{dV}{V} ~&=~ int,dfrac{dT}{T}

[6pt] ln,P ~+~ ln,V ~&=~ ln,T ~+~ C quad ext{($C$ is a constant)} ln,(PV) ~&=~ ln,T ~+~ C PV ~&=~ e^{ln,T + C} ~=~ e^{ln,T} cdot e^{C} ~=~ T,e^C ~=~ RTend{aligned}] where (R = e^C) is a constant.

The integration formulas in this section depended on already knowing the derivatives of certain functions and then “working backward” from their derivatives to obtain the original functions. Without that prior knowledge you would be reduced to guessing, or perhaps recognizing a pattern from some derivative you have encountered. A number of integration techniques will be presented shortly, but there are many indefinite integrals for which no simple closed form exists (e.g. (int e^{x^2},dx) and (int sin(x^2),dx)).

[sec5dot1]

For Exercises 1-15, evaluate the given indefinite integral.

3

(displaystyleint,left(x^2 ~+~ 5x ~-~ 3 ight)~dx)

(displaystyleint,3 cos,x~dx)

(displaystyleint,4 e^x~dx)

3

(displaystyleint,left(x^5 ~-~ 8x^4 ~-~ 3x^3 ~+~ 1 ight)~dxvphantom{dfrac{3e^x}{5}})

(displaystyleint,5 sin,x~dxvphantom{dfrac{3e^x}{5}})

(displaystyleint,dfrac{3e^x}{5}~dx)

3

(displaystyleint,dfrac{6}{x}~dx)

(displaystyleint,dfrac{4}{3x}~dx)

(displaystyleint,left(-2 sqrt{x}, ight)~dxvphantom{dfrac{6}{x}})

3

(displaystyleint,dfrac{1}{3 sqrt{x}}~dx)

(displaystyleint,left(x ~+~ x^{4/3} ight)~dx)

(displaystyleint,dfrac{1}{3 sqrt[3]{x}}~dx)

3

(displaystyleint,3sec,x; an,x~dx)

(displaystyleint,5 sec^2x~dx)

(displaystyleint,7csc^2x~dx)

Prove the sum and difference rules for indefinite integrals: (int (f(x) pm g(x)),dx ;=; int f(x),dx ;pm; int g(x),dx)

Integrate both sides of the equation

[frac{dP}{P} ~+~ frac{d!M}{M} ~=~ frac{dT}{2T}] to obtain the ideal gas continuity relation: (dfrac{PM}{sqrt{T}} =) constant.

[exer:projmax0] Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity (v_0) is (frac{v_0^2}{2g}).

1. The function (f) is assumed to be differentiable at (x), in this case. If not then the points where (f) is not differentiable can be excluded without affecting the integral.↩
2. For a proof and fuller discussion of all this, see Ch.1-2 in Knopp, M.I., Theory of Area, Chicago: Markham Publishing Co., 1969. The book attempts to define precisely what an “area” actually means, including that of a rectangle (showing agreement with the intuitive notion of width times height).↩
3. The theorem can be proved for the weaker condition that (f) is merely continuous on (ival{a}{b}). See p.173-175 in Parzynski, W.R. and P.W. Zipse, Introduction to Mathematical Analysis, New York: McGraw-Hill, Inc., 1982.↩
4. Created by the physicist P.A.M. Dirac (1902-1984), who won a Nobel Prize in physics in 1933. The function is neither real-valued nor continuous at (x=0). The “graph” in Figure [fig:dirac] is perhaps misleading, as (infty) is not an actual point on the (y)-axis. One interpretation is that (delta) is an abstraction of an instantaneous pulse or burst of something, preceded and followed by nothing. To learn more about this fascinating and useful function, see §15 in Dirac, P.A.M., The Principles of Quantum Mechanics, 4th ed., Oxford, UK: Oxford University Press, 1958.↩
5. See pp.140-141 in Buck, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965.↩

## Integral of $int ^_int ^>_>dydx$

You are making the substitution wrong. If $u=(x-1)^2$ , then $du=2(x-1)dx$ , and the factor $(x-1)/2$ becomes simply $1/4$ , not $frac14sqrt u$ .

First integrate the function with respect to $y$ :

Now, suppose $u = (x-1)^2$ . Defferntiating gives you $frac = 2(x-1)$ or $du = 2(x-1)dx$ . The limits are: When $x = 1$ , $u = 0$ an when $x = 5$ , $u = 4^2 = 16$ . Nowe, apply $u = (x-1)^2$ , $(x-1)dx = frac12 du$ , and limits in the equation:

## Indefinite Integrals of Polynomials

We are now going to look at a technique for finding the indefinite integrals of the simplest type of functions - polynomials!

• Proof: From the fundamental theorem of calculus part 1, we get that $fracint f(x) : dx = f(x)$ . If we can differentiate both sides of $int bx^n : dx = frac<>>+ C$ and prove their equality, then we have proven the equality of the integral.

We will now apply this important property to some examples.

## Georgia Institute of Technology School of Mathematics | Georgia Institute of Technology | Atlanta, GA

Definite and indefinite integrals, techniques of integration, improper integrals, infinite series, applications.

MATH 1550 or MATH 1551 or MATH 1501 or MATH 15X1 or MATH 1X51.

Thomas, Calculus: Early Transcendentals, (14th ed.)

Flow chart describing textbook choices for Fall 2019.

Topic Text Sections Lectures
Review of differentiation and indefinite integration Ch. 3, 4.8 2
Riemann Sums and the fundamental theorem of calculus 5.1-5.4 4
Integration by substitution, area 5.5-5.6 3
Transcendental functions: logarithms, exponentials 7.1-7.2 2
Techniques of integration 8.2-8.5, 8.7 7
L&rsquoHôpital&rsquos rule, improper integrals 4.5, 8.8 4
First-order linear differential equations 9.2 1
Infinite sequences and series 10.1-10.2 3
Convergence tests, power series 10.3-10.7 6
Taylor polynomials and Taylor approximation 10.8-10.9 4
Applications: volumes, length, work, center of mass 6.1-6.6 6

#### GT Math Resources

Georgia Institute of Technology
North Avenue, Atlanta, GA 30332
Phone: 404-894-2000

## 5.1: The Indefinite Integral - Mathematics

We have seen how we can approximate the area under a non-negative valued function over an interval $[a,b]$ with a sum of the form $sum_^n f(x^*_i) Delta x_i$, and how this approximation gets better and better as our $Delta x_i$ values become very small.

Of course, when these widths $Delta x_i$ of the sub-intervals of the related partition become small, we necessarily have many more terms to our sum (i.e., thinner rectangular strips require more rectangles to cover the original area).

As such, good approximations generally happen when they are a sum of many, many terms -- each of which contributes a very small value.

This idea of approximating something of interest with a large (in the limit, infinite) sum of many, many small values (in the limit, each being infinitesimal) has broad application and forms the next "big idea" in the study of calculus.

We will drive much of our exploration into this idea with the question: "What is the area under this curve?" -- much like we drove the development of the derivative with the question: "What is the slope of the tangent line to a curve at a given point?".

However, there is more than one way to interpret this next big idea. This should not be surprising. Think how the derivative can also be construed as an instantaneous rate of change between two variables, and how thinking about it in that way greatly widened the contexts in which the derivative could appear. Here too, the ideas revealed by asking about the area under a curve can also be seen in other contexts.

Before we move forward, let us consider one of these..

Suppose the upward velocity of a piston t seconds after it starts to move is given by $v(t) = sin(t)$ m/s. If the piston is 3 meters high when it starts to move, how high will it be 4 seconds later?

Suppose we attempted to approximate the answer to the question above. We might recall that

distance = rate $imes$ time

but that is only true when the rate is a constant velocity - and clearly the velocity in the problem above is changing over time. (Note: sometimes, the piston even moves backwards!)

However, if the time interval was just shorter. the velocity wouldn't be changing by as much, right? Consider the difference in velocity at the following two times:

$v(1) doteq 0.84147 quad v(1.0001) doteq 0.84152$

So maybe, for just that brief interval of time, distance = rate $imes$ time could give a decent approximation of the small change in position during that tiny time interval.

$extrm doteq 0.8415 imes (1.0001 - 1)$

Note, in the interest of getting an even better approximation, we tried to split the difference here with regard to the rate used - something between 0.84147 and 0.84152. There is more to say about this - but for now, recall that we are at least assured (by the Intermediate Value Theorem, in this case) that at some $x^*$ we have $v(x^*) = 0.8415$.

Consequently, if we denote the small change in position that happens over this tiny time interval by $Delta s$, and if we denote the change in time (i.e., time elapsed) from the start to the end of that tiny time interval by $Delta t$, then we can say

$Delta s doteq v(x^*) imes Delta t$

Now, to approximate the total net change in position from $t=0$ to $t=4$, we could find all of these "little-changes-in-position" and just add them up! (Keep in mind, some of these could be negative, indicating a distance traveled backwards.)

Consider the following possible way to cut up the time from $t=0$ to $t=2$ into nine tiny intervals no longer than 0.6 seconds -- and this is just one of many, many ways to do this.

$egin i & i^ extrm < time interval >& extrm < range of$v(t)$values >& Delta t 1 & (0,0.4) & 0 extrm < to >0.39 & 0.4 2 & (0.4,1) & 0.39 extrm < to >0.84 & 0.6 3 & (1,1.3) & 0.84 extrm < to >0.96 & 0.3 4 & (1.3,1.9) & 0.96 extrm < to >1 & 0.6 5 & (1.9, 2.4) & 0.67 extrm < to >0.95 & 0.5 6 & (2.4,2.6) & 0.51 extrm < to >0.67 & 0.2 7 & (2.6,3.15) & -0.01 extrm < to >0.51 & 0.55 8 & (3.15, 3.55) & -0.4 extrm < to >-0.01 & 0.4 9 & (3.55,4) & -0.75 extrm < to >-0.4 & 0.45 end$

Now, which $v(x^*)$ value we should use for each little tiny time interval could be debated. We might want to use the minimum value of the function in each range, thereby ensuring we do not over-estimate how high the piston will be. We might instead want to use the maximum value of the function in each range, so that we don't under-estimate the piston's height. Alternatively, we might decide to split the difference and use some $v(x^*)$ value between the two of these. While the choice at this point will affect our approximation, this becomes less and less of a concern as our time intervals become more and more narrow (i.e., there will be a smaller and smaller range of $v(x^*)$ values from which to choose). For the purposes of simply making a decision about such things for this example, let us suppose we use the value $f,(x^*)$ where $x^*$ is chosen to be the right endpoint of each time interval.

So, based on the above intervals and the choice of $v(x^*)$ values just described, an approximation for the total change in position the piston experiences from $t=0$ to $t=2$ is given by

A couple of comments about the construction of the above sum are in order:

The $4^$ term above, (0.94)(0.6), does indeed correspond to the product of the height of the function at the right endpoint of the time interval in question $v(1.9) doteq 0.94$ and the width of that time interval, 0.6. This may, however, not be apparent in the table, as $v(t)$ attains a maximum value of 1 within this time interval

Importantly, notice that this sum takes the form

where $n=9$ in this case, and even the biggest $Delta t_i$ is fairly small (i.e., no more than 0.6) -- and this sum generally represents a better and better approximation to the true change in position for our piston as we shrink the size of these individual time intervals (i.e., as the maximum $Delta t_i$ gets small).

Recall the summation we used to approximate the area under some non-negative valued function on the interval $[a,b]$:

Notice how the summation to approximate the net change in height of the piston is structurally absolutely identical to the summation immediately above!

So finding the net change over some interval of time in the height of a piston whose velocity is governed by some function can be answered with the same process as finding the area under a curve over some interval. As earlier suggested, there is a more general process at work here.

Bernhard Riemann With this in mind, let us define for a given partition, $a=x_0 lt x_1 lt x_2, ldots x_n=b$ of $[a,b]$ a sum of the following form to be a Riemann sum (named after the nineteenth century German mathematician Bernhard Riemann),

where for each $i$, $x_i^*$ is some chosen value in the $i^$ subinterval, $[x_,x_i]$, and $Delta x_i = x_i - x_$.

In both of the cases we have considered, recall how the approximations these sums represented generally got better as the norm of the partition used became small.

Hoping to find the "best" approximation (e.g., the actual net change in height of the piston, or the actual area under the curve), let us then try to find the limit of the Riemann sum as the norm of the partition goes to zero (i.e., as $||Delta|| ightarrow 0$).

When this limit exists, we say that $f(x)$ is Riemann integrable on the interval in question and define the definite integral of $f,(x)$ from $a$ to $b$, to be the value of this limit, denoted by

You are probably wondering why the same symbol $int$ is used here in connection to Riemann sums, when it is also used to describe a set of antiderivatives for a function. What do these two ideas have in common?

"Cycle Design" store in Germany First, one should know that the symbol $int$ initially appeared in the work of German mathematician Gottfried Wilhelm Leibniz in 1675 in his private writings. It is based on an archaic variant of the letter "s" that can still be seen in signs and logos in Nordic and German-speaking countries.

Leibniz used this elongated latin "S" to highlight the nature of the sum in question. While sigma $(Sigma)$, the greek version of the letter "S", was used to denote a finite sum of finite values -- Leibniz wanted to stress that the integral should be thought of as an infinite sum of infinitesimals.

In a similar manner, we trade the upper case delta $(Delta)$, the greek version of an uppercase "D", for a lowercase "d" -- the former representing a finite difference, which is large in comparison to the "small" infinitesimal difference the latter represents.

With regard to the similarity to the notation for antidifferentiation (i.e., finding the indefinite integral) there is an intimate an intimate connection between these two ideas -- which on the surface appear very different. This connection is described by the Fundamental Theorem of Calculus and will be discussed soon.

As one final comment about the notation -- as the definite integral tells us something about the behavior of the function over an interval $[a,b]$ and that information gets somewhat lost in the limit of the Riemann sum through the consideration of the partition used and the norm of that partition - we make the connection more prominant by putting the $a$ and $b$ as a subscript and superscript, respectively, next to the $int$ symbol.

As a matter of vocabulary, the $x$-value written as a subscript (here, $a$) is called the lower limit of integration, while the one written as a superscript (here, $b$) is called the upper limit of integration.

Often, to evaluate a definite integral directly from its limit of a Riemann sum definition, we choose a convenient partition, one in which all of the $Delta x_i ## Integration of Quadratic Equation in the Square Root - Examples Evaluate the following with respect to "x". (6 - x)(x - 4) = 6x - 24 - x 2 + 4x (√a 2 -x 2 ) dx = (x/2)(√a 2 -x 2 )+(a 2 /2) sin -1 (x/a) + c Evaluate the following with respect to "x". √[9 - (2x + 5) 2 ] = √[3 2 - (2x + 5) 2 ] dx (√a 2 -x 2 )dx=(x/2)(√a 2 -x 2 )+(a 2 /2) sin -1 (x/a)+c = (2x + 5)/2√[9 - (2x + 5) 2 ] + (9/2) sin -1 [(2x + 5)/3] + c Evaluate the following with respect to "x". (√x 2 +a 2 ) dx = (x/2)(√x 2 +a 2 )+(a 2 /2) log (x+√(x 2 +a 2 )+c = ((2x+1)/3)√[81 + (2x + 1) 2 ]+(9/2) log (2x+1)+√[81 + (2x + 1) 2 ] + c Evaluate the following with respect to "x". (√x 2 -a 2 )dx=(x/2)(√a 2 -x 2 )-(a 2 /2)log(x+ √a 2 -x 2 ) + c = ((x-1)/2)√[(x + 1) 2 - 4] - (4/2) log (x - 1 - √[(x + 1) 2 - 4] + c ## Integrals of polynomials Finding the integral of a polynomial involves applying the power rule, along with some other properties of integrals. Take a look at the example to see how. ### Example Find: (displaystyleint 2x^3 + 4x^2 ext< dx>) ### Solution We will write out every step here so that you can see the process. After some practice, you will probably just write the answer down immediately. First, remember that integrals can be broken up over addition/subtraction and multiplication by constants. Therefore: (egin displaystyleint 2x^3 + 4x^2 ext < dx>&= displaystyleint 2x^3 ext < dx>+ displaystyleint 4x^2 ext< dx> &= 2displaystyleint x^3 ext < dx>+ 4displaystyleint x^2 ext< dx>end) Now apply the power rule by adding 1 to each exponent, and then dividing by the same number. When you do this, the integral symbols are dropped since you have “taken the integral”. (2displaystyleint x^3 ext < dx>+ 4displaystyleint x^2 ext < dx>= 2left(dfrac><3+1> ight) + 4left(dfrac><2+1> ight) + C) Now, simplify the expression to find your final answer. Now, let’s look at how this kind of integral would be with skipping some of the more straightforward steps. ### Example Find: (displaystyleint -3x^2 + x – 5 ext< dx>) ### Solution This one is a little different. We have an (x) by itself and a constant. For the (x) by itself, remember that the exponent is 1. For the constant, remember that the integral of a constant is just the constant multiplied by the variable. For example, the integral of 2 with respect to (x) is (2x). (displaystyleint -3x^2 + x – 5 ext < dx>= -3left(dfrac<3> ight) + dfrac <2>– 5x + C) Simplifying, the final answer is: ( = box[border: 1px solid black padding: 2px]<-x^3 + dfrac <2>– 5x + C>) Evaluate the following integral:$int an ^ <-1>x : dx$Let$u = an ^ <-1>x$and$dv = dx$. Therefore$du = frac<1> <1 + x^2>dx$and$v = x$. Thus it follows by the integration by parts identity that: We can now use substitution for the remaining integral. Let$p = 1 + x^2$. Then$dp = 2x : dx$, so we can make the following substitution: egin int an ^ <-1>x : dx = x tan^ <-1>x - int frac<1> <2>cdot frac<1> : dx int an ^ <-1>x : dx = x tan^ <-1>x - frac<1> <2>int frac<1> : dx int an ^ <-1>x : dx = x tan^ <-1>x - frac<1> <2>ln p int an ^ <-1>x : dx = x tan^ <-1>x - frac<1> <2>ln (1 + x^2) int an ^ <-1>x : dx = x tan^ <-1>x - ln sqrt <(1 + x^2)> end ## Sum rule of Integration The integral of sum of functions equals to sum of their integrals is called the sum rule of integration. ### Introduction Let$f(x)$and$g(x)$represent two functions in$x$, then the indefinite integral of each function with respect to$x$can be written in mathematical form as follows. The sum of functions is written as$f(x)+g(x)$in mathematics. The integral of sum of the functions with respect$x$is written in the following mathematical form in integral calculus. According to integral calculus, the integral of sum of two or more functions is equal to the sum of their integrals. The following equation expresses this integral property and it is called as the sum rule of integration.$impliesdisplaystyle intdx,=,displaystyle intdx $displaystyle intdx$

The sum rule of indefinite integration can also be extended to sum of infinite number of functions.

#### Example

Evaluate $displaystyle int<(3+4x),>dx$

Now, use the integral sum rule to evaluate the integral of sum of the functions.

$implies$ $displaystyle int<(3+4x),>dx$ $,=,$ $displaystyle int<3,>dx$ displaystyle int<4x,>dximpliesdisplaystyle int<(3+4x),>dx,=,3x+c_1 $4displaystyle intdx$

$implies$ $displaystyle int<(3+4x),>dx$ $,=,$ $3x+c_1$ 4 imes Big(dfrac<2>+c_2Big)impliesdisplaystyle int<(3+4x),>dx,=,3x+c_1 $4 imes dfrac<2>+4 imes c_2$

$implies$ $displaystyle int<(3+4x),>dx$ $,=,$ $3x+c_1$ 2 imes x^2+4c_2impliesdisplaystyle int<(3+4x),>dx,=,3x+c_1 $2x^2+4c_2$