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4.0: Antidervatives and Indefinite Integration (Revisited)


We have spent considerable time considering the derivatives of a function and their applications. In the following chapters, we are going to starting thinking in "the other direction." That is, given a function (f(x)), we are going to consider functions (F(x)) such that (F'(x) = f(x)). There are numerous reasons this will prove to be useful: these functions will help us compute areas, volumes, mass, force, pressure, work, and much more.

Given a function (y=f(x)), a differential equation is one that incorporates (y), (x), and the derivatives of (y). For instance, a simple differential equation is:

[y' = 2x.]

Solving a differential equation amounts to finding a function (y) that satisfies the given equation. Take a moment and consider that equation; can you find a function (y) such that (y' = 2x)?

Can you find another?

And yet another?

Hopefully one was able to come up with at least one solution: (y = x^2). "Finding another" may have seemed impossible until one realizes that a function like (y=x^2+1) also has a derivative of (2x). Once that discovery is made, finding "yet another" is not difficult; the function (y = x^2 + 123,456,789) also has a derivative of (2x). The differential equation (y' = 2x) has many solutions. This leads us to some definitions.

Definition (PageIndex{1}): Antiderivatives and Indefinite Integrals

Let a function (f(x)) be given. An antiderivative of (f(x)) is a function (F(x)) such that (F'(x) = f(x)).

The set of all antiderivatives of (f(x)) is the indefinite integral of (f), denoted by

[int f(x) dx.]

Make a note about our definition: we refer to an antiderivative of (f), as opposed to the antiderivative of (f), since there is always an infinite number of them. We often use upper-case letters to denote antiderivatives.

Knowing one antiderivative of (f) allows us to find infinitely more, simply by adding a constant. Not only does this give us more antiderivatives, it gives us all of them.

Theorem (PageIndex{1}): Antiderivative Forms

Let (F(x)) and (G(x)) be antiderivatives of (f(x)). Then there exists a constant (C) such that

[G(x) = F(x) + C.]

Given a function (f) and one of its antiderivatives (F), we know all antiderivatives of (f) have the form (F(x) + C) for some constant (C). Using Definition (PageIndex{1}), we can say that

[int f(x) dx = F(x) + C.]

Let's analyze this indefinite integral notation.

Figure (PageIndex{1}) shows the typical notation of the indefinite integral. The integration symbol, (int), is, in reality, an "elongated S," representing "take the sum." We will later see how sums and antiderivatives are related.

The function we want to find an antiderivative of is called the integrand. It contains the differential of the variable we are integrating with respect to. The (int) symbol and the differential (dx) are not "bookends" with a function sandwiched in between; rather, the symbol (int) means "find all antiderivatives of what follows," and the function (f(x)) and (dx) are multiplied together; the (dx) does not "just sit there."

Let's practice using this notation.

Example (PageIndex{1}): Evaluating indefinite integrals

Evaluate (displaystyle int sin x dx.)

Solution

We are asked to find all functions (F(x)) such that (F'(x) = sin x). Some thought will lead us to one solution: (F(x) = -cos x), because (frac{d}{dx}(-cos x) = sin x).

The indefinite integral of (sin x) is thus (-cos x), plus a constant of integration. So:

[int sin x dx = -cos x + C.]

A commonly asked question is "What happened to the (dx)?" The unenlightened response is "Don't worry about it. It just goes away." A full understanding includes the following.

This process of antidifferentiation is really solving a differential question. The integral

[int sin x dx]

presents us with a differential, (dy = sin x dx). It is asking: "What is (y)?" We found lots of solutions, all of the form (y = -cos x+C).

Letting (dy = sin x dx), rewrite

[int sin x dx quad ext{as}quad int dy.]

This is asking: "What functions have a differential of the form (dy)?" The answer is "Functions of the form (y+C), where (C) is a constant." What is (y)? We have lots of choices, all differing by a constant; the simplest choice is (y = -cos x).

Understanding all of this is more important later as we try to find antiderivatives of more complicated functions. In this section, we will simply explore the rules of indefinite integration, and one can succeed for now with answering "What happened to the (dx)?" with "It went away."

Let's practice once more before stating integration rules.

Example (PageIndex{2}): Evaluating indefinite integrals

Evaluate (int (3x^2 + 4x+5) dx).

Solution

We seek a function (F(x)) whose derivative is (3x^2+4x+5). When taking derivatives, we can consider functions term--by--term, so we can likely do that here.

What functions have a derivative of (3x^2)? Some thought will lead us to a cubic, specifically (x^3+C_1), where (C_1) is a constant.

What functions have a derivative of (4x)? Here the (x) term is raised to the first power, so we likely seek a quadratic. Some thought should lead us to (2x^2+C_2), where (C_2) is a constant.

Finally, what functions have a derivative of (5)? Functions of the form (5x+C_3), where (C_3) is a constant.

Our answer appears to be

[int (3x^2+4x+5) dx = x^3+C_1+2x^2+C_2+5x+C_3.]

We do not need three separate constants of integration; combine them as one constant, giving the final answer of

[int (3x^2+4x+5) dx = x^3+2x^2+5x+C.]

It is easy to verify our answer; take the derivative of (x^3+2x^3+5x+C) and see we indeed get (3x^2+4x+5).

This final step of "verifying our answer" is important both practically and theoretically. In general, taking derivatives is easier than finding antiderivatives so checking our work is easy and vital as we learn.

We also see that taking the derivative of our answer returns the function in the integrand. Thus we can say that:

[frac{d}{dx}left(int f(x) dx ight) = f(x).]

Differentiation "undoes" the work done by antidifferentiation.

Theorem 27 gave a list of the derivatives of common functions we had learned at that point. We restate part of that list here to stress the relationship between derivatives and antiderivatives. This list will also be useful as a glossary of common antiderivatives as we learn.

Theorem (PageIndex{2}): Derivatives and Antiderivatives

Common Differentiation RulesCommon Indefinite Integration Rules
  1. (frac{d}{dx}ig(cf(x) ig) = ccdot f'(x))
  2. (frac{d}{dx}ig(f(x)pm g(x) ig) =f'(x)pm g'(x))
  3. (frac{d}{dx}ig(C ig) = 0)
  4. (frac{d}{dx}ig(x ig) = 1)
  5. (frac{d}{dx}ig(x^n ig) = ncdot x^{n-1})
  6. (frac{d}{dx}ig(sin x ig) = cos x)
  7. (frac{d}{dx}ig(cos x ig) = -sin x)
  8. (frac{d}{dx}ig( an x ig) = sec^2 x)
  9. (frac{d}{dx}ig(csc x ig) = -csc xcot x)
  10. (frac{d}{dx}ig(sec x ig) = sec x an x)
  11. (frac{d}{dx}ig(cot x ig) = -csc^2 x)
  12. (frac{d}{dx}ig(e^ x ig) = e^x)
  13. (frac{d}{dx}ig(a^x ig) = ln acdot a^x)
  14. (frac{d}{dx}ig(ln x ig) = frac1 x)
  1. (int ccdot f(x) dx = ccdot int f(x) dx)
  2. (int ig(f(x)pm g(x)ig) dx =int f(x) dxpm int g(x) dx)
  3. (int 0 dx = C)
  4. (int 1 dx = int dx = x+C)
  5. (int x^n dx =frac{1}{n+1}x^{n+1}+ C)
  6. (int cos x dx = sin x+C)
  7. (int sin x dx = -cos x+C)
  8. (int sec^2 x dx = an x+C)
  9. (int csc xcot x dx = -csc x+C)
  10. (int sec x an x dx = sec x+C)
  11. (int csc^2 x dx = -cot x+C)
  12. (int e^x dx = e^x+C)
  13. (int a^x dx = frac{1}{ln a}cdot a^x+C)
  14. (int frac{1}x dx = ln |x|+C)

We highlight a few important points from Theorem (PageIndex{2}):

  • Rule #1 states (int ccdot f(x) dx = ccdot int f(x) dx). This is the Constant Multiple Rule: we can temporarily ignore constants when finding antiderivatives, just as we did when computing derivatives (i.e., (frac{d}{dx}ig(3x^2ig)) is just as easy to compute as (frac{d}{dx}ig(x^2ig))). An example:

[int 5cos x dx = 5cdotint cos x dx = 5cdot (sin x+C) = 5sin x + C.[
In the last step we can consider the constant as also being multiplied by 5, but "5 times a constant" is still a constant, so we just write "(C),".

  • Rule #2 is the Sum/Difference Rule: we can split integrals apart when the integrand contains terms that are added/subtracted, as we did in Example (PageIndex{2}). So:

[egin{align} int(3x^2+4x+5) dx &= int 3x^2 dx + int 4x dx + int 5 dx &= 3int x^2 dx + 4int x dx + int 5 dx &= 3cdot frac13x^3 + 4cdot frac12x^2+5x+C &= x^3+2x^2+5x+C end{align}]
In practice we generally do not write out all these steps, but we demonstrate them here for completeness.

  • Rule #5 is the Power Rule of indefinite integration. There are two important things to keep in mind:
    1. Notice the restriction that (n eq -1). This is important: (int frac{1}{x} dx eq ) "(frac{1}{0}x^0+C)"; rather, see Rule #14.
    2. We are presenting antidifferentiation as the "inverse operation" of differentiation. Here is a useful quote to remember: "Inverse operations do the opposite things in the opposite order."
      When taking a derivative using the Power Rule, we first multiply by the power, then second subtract 1 from the power. To find the antiderivative, do the opposite things in the opposite order: first add one to the power, then second divide by the power.
  • Note that Rule #14 incorporates the absolute value of (x). The exercises will work the reader through why this is the case; for now, know the absolute value is important and cannot be ignored.

Initial Value Problems

In Section 2.3 we saw that the derivative of a position function gave a velocity function, and the derivative of a velocity function describes acceleration. We can now go "the other way:" the antiderivative of an acceleration function gives a velocity function, etc. While there is just one derivative of a given function, there are infinite antiderivatives. Therefore we cannot ask "What is the velocity of an object whose acceleration is (-32)ft/s(^2)?", since there is more than one answer.

We can find the answer if we provide more information with the question, as done in the following example. Often the additional information comes in the form of an initial value, a value of the function that one knows beforehand.

Example (PageIndex{3}): Solving initial value problems

The acceleration due to gravity of a falling object is (-32) ft/s(^2). At time (t=3), a falling object had a velocity of (-10) ft/s. Find the equation of the object's velocity.

Solution

We want to know a velocity function, (v(t)). We know two things:

  1. The acceleration, i.e., (v'(t)= -32), and
  2. the velocity at a specific time, i.e., (v(3) = -10).

Using the first piece of information, we know that (v(t)) is an antiderivative of (v'(t)=-32). So we begin by finding the indefinite integral of (-32):

[int (-32) dt = -32t+C=v(t).]

Now we use the fact that (v(3)=-10) to find (C):

[egin{align} v(t) &= -32t+C v(3) &= -10 -32(3)+C &= -10 C &= 86 end{align}]

Thus (v(t)= -32t+86). We can use this equation to understand the motion of the object: when (t=0), the object had a velocity of $v(0) = 86$ ft/s. Since the velocity is positive, the object was moving upward.

When did the object begin moving down? Immediately after (v(t) = 0):

[-32t+86 = 0 quad Rightarrowquad t = frac{43}{16} approx 2.69 ext{s}.]

Recognize that we are able to determine quite a bit about the path of the object knowing just its acceleration and its velocity at a single point in time.

Example (PageIndex{4}): Solving initial value problems

Find (f(t)), given that (f''(t) = cos t), (f'(0) = 3) and (f(0) = 5).

Solution

We start by finding (f'(t)), which is an antiderivative of (f''(t)):

[int f''(t) dt = int cos t dt = sin t + C = f'(t).]

So (f'(t) = sin t+C) for the correct value of (C). We are given that (f'(0) = 3), so:

[f'(0) = 3 quad Rightarrow quad sin 0+C = 3 quad Rightarrow quad C=3.]

Using the initial value, we have found (f'(t) = sin t+ 3.)

We now find (f(t)) by integrating again.

[f(t)=int f'(t) dt = int (sin t+3) dt = -cos t + 3t + C.]

We are given that (f(0) = 5), so

[egin{align} -cos 0 + 3(0) + C &= 5 -1 + C &= 5 C &= 6 end{align}]

Thus (f(t) = -cos t + 3t + 6).

This section introduced antiderivatives and the indefinite integral. We found they are needed when finding a function given information about its derivative(s). For instance, we found a position function given a velocity function.

In the next section, we will see how position and velocity are unexpectedly related by the areas of certain regions on a graph of the velocity function. Then, in Section 5.4, we will see how areas and antiderivatives are closely tied together.


Antiderivatives and Its Application.

Actually, antiderivatives is the reserve of derivative.

If f(x) and F(x) are the functions of x such that $frac<< m>><<< m>>>$ (F(x)) then the antiderivative of f(x) with respect to x is the function F(x) and is written as:

Or, $mathop smallint olimits < m>left( < m> ight)$.dx = F(x)

If c is constant quantity and derivative of a constant quantity is zero, then,

Hence, $mathop smallint olimits < m>left( < m> ight)$.dx = F(x) + c.

Since, c is an unknown constant and it may be any real number, so there are a number of antiderivatives of a function whoch are different from one another by a constant. Let us see the following example:

If y = x 2 be the given function then,

Or, $frac<< m>><<< m>>>$(x 2 ) = 2x, then $mathop smallint olimits 2< m>.$dx = x 2

If a is constant, $frac<< m>><<< m>>>$(x 2 + a) = 2x, then $mathop smallint olimits 2< m>.< m>$ = x 2 + a and so on.

Above examples show that there are a number of antiderivatives of 2x which are different from one another by a constant.

If f(x) and F(x) are two functions such that $frac<< m>><<< m>>>$(F(x)) = f(x) then $frac<< m>><<< m>>>$(F(x) + c) = f(x) as derivative of a constant term is zero, so

Or, $mathop smallint olimits < m>left( < m> ight)$.dx = F(x) + c

Where, c is an arbitrary constant. Since this constant is the unknown quantity, so it is known as the indefinite integral.

The integral of a function independent of arbitrary constant is known as definite integral.

= $mathop smallint olimits left( <2< m> + 3> ight)> + 8> ight)^5>$.dx

= $frac<1><3>mathop smallint olimits left( <6< m> + 9> ight)> + 8> ight)^5>$.dx = $frac<1><3>mathop smallint olimits left< > + 8> ight) + 1> ight>> + 8> ight)^5>$.dx


1. Constant function

The applet shows a graph on the left of the integrand f ' (x) = 2, a constant function. Below is the graph of the antiderivative: Think of the graph on the left (the integrand) as representing the slope of the graph on the right (the antiderivative). Note that since the left-hand graph is constant, so is the slope of the right hand graph, and we get a line with slope 2. Move the C slider what happens to the graph? If you work backwards, thinking that the graph on the left is the derivative of the graph on the right, you see why changing C has no effect on the left-hand graph.


This is actually a good question and one that is awfully confusing without a proper background in analysis.

An antiderivative of $f(x)$ on a set $X$ is a function $F(x)$ such that $F'(x)=f(x)$ for all $xin X$ . Note that antiderivatives are certainly not unique, which is why a C$ is added (derivative of constant is zero).

The first fundamental theorem of calculus says that if $f$ is integrable on $[a,x]$ and an antiderivative $F(x)$ exists for all $xin[a,b]$ , then $int_a^x f=F(x)-F(a)$ . We call $F(x)=F(a)+int_a^x f(y),mathrmy$ an indefinite integral. Obviously, $F'(x)=f(x)$ by the second fundamental theorem if $f$ is continuous at $x$ .

Where things get subtle is considering $F(x)$ in general when $f$ is integrable, but not necessarily continuous. It is a perfectly legitimate, continuous function. However, it's not necessarily an antiderivative of $f$ if $f$ is not continuous at $x$ .

For example, consider $f(x)=egin1,&quad xin[0,1] 2,& quad xin(1,2]end$

There is no antiderivative function $F(x)$ on $[0,2]$ . One might suspect that $ F(x)=eginx,&quad xin[0,1] 2x,& quad xin(1,2]end $ works. However, while each piece of $F(x)$ serves as an antiderivative on $[0,1]$ and $(1,2]$ respectively, this piecewise function is not an antiderivative on $[0,2]$ , even though $F(x)=F(0)+int_0^x f(y),mathrmy$ is continuous. Another way to see it is since $int_0^2 f=1+2=3$ , which is not equal to $F(2)-F(0)=4-0=4$ , we would be contradicting the fundamental theorem to claim $F$ is an antiderivative.

Now consider when $f$ is continuous. Then it's always true that $F(x)=C+int_a^x f(y),mathrmy$ serves the role of an antiderivative. The reason this notation is common is because this holds even if the antiderivative does not have an elementary form. For example, consider $f(x)=e^$ , which is continuous on any $[a,b]$ . Then it's true that $F(x)=int_a^x e^,mathrmy$ satisfies $F'(x)=f(x)$ , but the notation is more instructive since an elementary form does not exist.

I hope some of this helps. The above also underscores the fact that all differentiable functions are continuous, and that a function can be differentiable without the derivative being continuous.


3 Answers 3

$F$ is not the antiderivative of $f$ on the whole interval $[0,2]$, because its derivative doesn't exist at $1$. So the hypothesis of the $2$nd fundamental theorem of integral calculus is not satisfied.

Using the antiderivative to compute an integral is the (second) Fundamental Theorem of Calculus. Said theorem requires the function $f$ which you want to integrate to be continuous on the whole interval (where you want to integrate). Your $f$ is not continuous in $[0,2]$ and hence the theorem doesn't work. In fact, $f$ cannot have an antiderivative at point $x=1$ because $f$ has a simple discontinuity at point $x=1$.

We can find a function $G$ such that $G'=f$ in [0,1] (considering left hand derivatives at $x=1$) and $G'=f$ in [1,2] considering right hand derivatives at $x=1$, and such that $G$ computes the area below $f$ in the interval $[0,x]$

When considering the second half $[1,2]$, you must not forget the area below $f$ in $[0,1]$. With this in mind put:

$G(x)=eginx quad quad ext xin [0,1] 1+2(x-1) ext xin [1,2]end$

Where the $x-1$ comes from considering the rectangle whose base is the segment $[1,x]$ for $x>1$

Then $G(2)-G(0)=1+2(2-1)-0=3=int_<0>^<2>f(x)dx$, as desired.

Your $F$ is not really an antiderivative, because we don't have $F'(x)=f(x)$ everywhere -- in fact $F'(1)$ doesn't exist at all!

Even worse, $F$ is not even an indefinite integral, because it has a jump discontinuity at $1$.

If you add an appropriate constant to one of the two cases in the definition of $F$, you can get rid of the jump discontinuity, and then it will actually be an indefinite integral (but still not an antiderivative) -- if we define "indefinite integral" to mean a function that allows us to compute definite integrals by the $int_a^b f(x),dx = F(b)-F(a)$ rule. (On the other hand, it seems to be more common to define "indefinite integral" simply as a synonym for "antiderivative", and then getting rid of the jump doesn't produce one, of course).

In fact $f$ can't have any antiderivative because derivatives always satisfy the intermediate value property (by Darboux's theorem), but $f$ doesn't do that.


Example

The semicolons are to separate one instruction from the next, and they become necessary now that we’re doing real programming. Line 1 of this program defines the variable n, which will take on all the values from 1 to 100. Line 2 says that we haven’t added anything up yet, so our running sum is zero so far. Line 3 says to keep on repeating the instructions inside the square brackets until n goes past 100. Line 4 updates the running sum, and line 5 updates the value of n. If you’ve never done any programming before, a statement like might seem like nonsense — how can a number equal itself plus one? But that’s why we use the := symbol it says that we’re redefining , not stating an equation. If was previously 37, then after this statement is executed, n will be redefined as 38. To run the program on a Linux computer, do this (assuming you saved the pro- gram in a file named ):

Here the % symbol is the computer’s prompt. The result is 5,050, as expected. One way of stating this result is
The capital Greek letter , sigma, is used because it makes the “s” sound, and that’s the first sound in the word “sum.” The below the sigma says the sum starts at 1, and the 100 on top says it ends at 100. The is what’s known as a dummy variable: it has no meaning outside the context of the sum. Figure 4.1 shows the graphical interpretation of the sum: we’re adding up the areas of a series of rectangular strips. (For clarity, the figure only shows the sum going up to 7, rather than 100.)

Now how about an integral? Figure 4.2 shows the graphical interpretation of what we’re trying to do: find the area of the shaded triangle. This is an example we know how to do symbolically, so we can do it numerically as well, and check the answers against each other. Symbolically, the area is given by the integral. To integrate the function , we know we need some function with a in it, since we want something whose derivative is , and differentiation reduces the power by one. The derivative of would be rather than , so what we want is . Let’s compute the area of the triangle that stretches along the axis from 0 to 100: .

Figure 4.3 shows how to accomplish the same thing numerically. We break up the area into a whole bunch of very skinny rectangles. Ideally, we’d like to make the width of each rectangle be an infinitesimal number , so that we’d be adding up an infinite number of infinitesimal areas. In reality, a computer can’t do that, so we divide up the interval from to into rectangles, each with finite width . Instead of making H infinite, we make it the largest number we can without making the computer take too long to add up the areas of the rectangles.


Find the following integrals:

Simplify square roots into fractional exponents and apply addition rules for integrals.

Move constants outside integrals

Integrate using power rule for integration and don't forget the CONSTANT

For more complex powers you can use a substitution before applying the power rule.

[int (x^2 -x)left(x^3 - frac<3><2>x^2)^8 ight) dx ]

Substitute the values inside the bracket and find the derivative of the substitution


4.0: Antidervatives and Indefinite Integration (Revisited)

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    Calculus Revisited #11: Integration by Substitution

    Welcome to Part 11 of our 21-part Calculus Revisited journey! Today we are continuing with integrals - and starting the first of four blog entries concerning techniques of integration. I will try my best to explain each one but I feel the best way to demonstrate these techniques is by example.

    Integration by Substitution

    We have the integral ∫ f(x) dx.

    Find a function u(x) such that

    ∫ f(u) * du is easier to integrate than ∫ f(x) dx.

    For the definite integral

    and the integral is re-written as

    Let's get to the solved problems to demonstrate this technique.

    Problems
    Indefinite Integrals

    Let u(x) = 4 + 5x
    Then du = 5 dx

    Note that √u = √(4 + 5x)
    and dx = du/5

    ∫ √(4 + 5x) dx
    = ∫ √u * 1/5 du
    = 1/5 * ∫ √u du
    = 1/5 * (u^(3/2))/(3/2) + C
    = 1/5 * 2/3 * u^(3/2) + C
    = 2/15 * u^(3/2) + C
    When substitution is used, please don't not forget to state the answer in the original variable, in this case, x. u = 4 + 5x
    = 2/15 * (4 + 5x)^(3/2) + C

    Let u = 1 + 2 cos x
    Then du = - 2 sin x dx

    Note - du/2 = sin x dx (which matches the numerator)

    Rewriting the integral:
    ∫ sin x / √(1 + 2 cos x) dx
    = ∫ 1/√u * -1/2 du
    = -1/2 * ∫ u^(-1/2) du
    = -1/2 * u^(1/2) * 2 + C
    = -u^(1/2) + C
    = -√u + C
    Remember. the answer must contain the original variable!
    = -√(1 + 2 cos x) + C

    The right substitutions make integration much easier. The next two problems illustrate this.

    Let u = sin x
    Then du = cos dx (note the cos x in the integral)

    ∫ sin^4 x * cos x dx
    = ∫ u^4 du
    = u^5/5 + C
    = sin^5 x/5 + C

    Let u = e^x
    Then du = e^x dx (matches the numerator)

    ∫ e^x / (e^(2x) + 1) dx
    = ∫ du/(u^2 + 1)
    = atan u + C
    = atan (e^(2x)) + C

    If you use the substitution technique with indefinite integrals, do NOT forget to account for the limits ! For definite integrals, you will recalculate the limits to account for the substitution. The next two problems illustrate this point.

    Note du/2 = dx (try to get the term with the dx to match some part of the integral)

    Accounting for the limits:
    New Upper Limit = 2 * 1 = 2
    New Lower Limit = 2 * -1 = -2

    = 1/2 * (sin(2) + sin(2)) = sin(2) ≈ 0.90930

    Finding the right substitution may be tricky at times. Algebraic manipulation may be necessary. Here is a more difficult problem.

    Let u = √t - 1
    Then du = 1/(2√t) dt

    To help match the integral, (2√t) du = dt

    Note that 2√t = 2(u + 1) = 2u + 2 (Look at the substitution and solve for √t).

    We can't forget about the limits, since we are working with a definite integral.
    u = √t - 1

    New Upper Limit: 𕔍 - 1 = 4 - 1 = 3
    New Lower Limit: 𕔈 - 1 = 2 - 1 = 2

    The antiderivative is u^2 + 4u + 2 ln u

    = ( (2)^2 + 4(2) + 2 ln (2) ) - ( (1)^2 + 4(1) + 2 ln (1) )

    = ( 4 + 8 + 2 ln 2 ) - ( 1 + 4 + 2 * 0 )

    Best thing to do if you are in a calculus course is to practice, practice, practice!

    I hope these six example problems give you an understanding of the substitution technique. Mastery of this technique will come with time and practice - and will speed up integral evaluation.

    Next time, we look at the famous (or maybe infamous) Integration by Parts technique.


    Calculus

    1)Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative. Remember to use ln |u| where appropriate.) f(x) = (1/5)−(3/x) -----> (x/5)-3lnx+C

    Trigonometry

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