# 2. Antiderivatives and The Indefinite Integral

by M. Bourne

### Mini-Lecture

See the

mini-lecture on differentials

We wish to perform the **opposite process** to
differentiation. This is called "antidifferentiation" and later, we will call it
"integration".

### Example 1

If we know that

`(dy)/(dx)=3x^2`

and we need to know the function this derivative came from, then we "undo" the differentiation process. (Think: "What would I have to differentiate to get this result?")

`y = x^3` is ONE antiderivative of `(dy)/(dx)=3x^2`

There are infinitely many other antiderivatives which would also work, for example:

`y = x^3+4`

`y = x^3+pi`

`y = x^3+27.3`

In general, we say `y =
x^3+K` is the
**indefinite integral** of `3x^2`. The number *K* is called the **constant of
integration**.

**Note:** Most math text books use `C` for the constant of integration, but for questions involving electrical engineering, we prefer to write "+*K*", since *C* is normally used for **capacitance** and it can get confusing.

## Notation for the Indefinite Integral

We write: `int3x^2dx=x^3+K` and say in words:

"The integral of 3*x*^{2} with respect to
*x* equals *x*^{3} + *K*."

### The Integral Sign

The `int` sign is an elongated "S",
standing for "sum". (In old German, and English, "s" was often written using this elongated shape.) Later we will see that the integral is the
**sum** of the areas of infinitesimally thin rectangles.

`sum` is the symbol for "sum". It can be used for finite or infinite sums.

`int` is the symbol for the sum of an infinite number of infinitely small areas (or other variables).

This "long s" notation was introduced by Leibniz when he developed the concepts of integration.

### Other Notation for Integrals

**Note:** Sometimes we write a capital letter to signify integration. For example, we write *F*(*x*) to mean
the integral of *f*(*x*). So we have:

`F(x)=intf(x)dx`

### Example 2

Find `int(x^2-5)dx`

Answer

The antiderivative of `x^2` is `x^3/3`, and the antiderivative of `5` is `5x`, so we can write:

`int (x^2-5) dx = frac{x^3}{3}-5x+K`

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We now learn some important general rules for integration.

## A. Integral of a Constant

`intk dx=kx+K`

(`k` and `K` are constants.)

The integral of a constant is that constant times
*x*, plus a constant.

### Example 3

Find `int4 dx`

Answer

Using our new rule, we can simply write:

`int4 dx=4x+K`

Always check by **differentiating** your answer, and you should get back to what the question was asking you to integrate.

Easy to understand math videos:

MathTutorDVD.com

## B. Integral of a Power of *x*

`intx^ndx=(x^(n+1))/(n+1)+K` (This is true as long as `n ≠ -1`)

For the integral of a power of *x*: add 1 to the power
and divide by the new number.

### Example 4

Integrate `intx^5 dx`

Answer

We use our new rule, and obtain:

`intx^5 dx=(x^(5+1))/(5+1)+K`

`=x^6/6+K`

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## The Constant of Integration

Don't forget the "+ *K*" (or, alternatively, "+ *C*"). This constant of integration is **vital** in later applications of the indefinite integral.

### Example 5

Integrate `int 8x^6 dx`

Answer

`int 8x^6 dx`

`8` is a constant, so it can go out the front:

`8 int x^6 dx`

Next, do the integration step by adding 1 to the index and dividing by the new number:

`8 x^(6+1)/(6+1) = 8(x^7)/7`

And of course, we must not forget the constant.

So the final answer is:

`int 8x^6 dx = 8(x^7)/7+K`

Easy to understand math videos:

MathTutorDVD.com

### Example 6

Integrate `dy = (5x^2 - 4x + 3)dx`

Answer

`dy = (5x^2 - 4x + 3)dx`

This is already in differential form, so we can just add the integral signs:

`int dy = int(5x^2 - 4x + 3)dx`

On the left hand side, we simply have:

`int dy = int(1)dy = y` (We are integrating the constant `1` with respect to `y`.)

On the right hand side, we integrate each of the terms, one at a time:

`int(5x^2 - 4x + 3)dx = 5(x^3)/3-4(x^2/2)+3x+K`

`=5(x^3)/3-2x^2+3x+K`

So putting it all together, we have the solution:

`y=5(x^3)/3-2x^2+3x+K`

Easy to understand math videos:

MathTutorDVD.com

### Example 7

`int7x^6dx`

Answer

`int 7x^6 dx = 7 frac{x^7}{7}+K`

` = x^7+K`

### Example 8

`int(3x^2+sqrtx-5/x^3)dx`

Answer

Our first step in this question is to re-write the exponents so it is easier to integrate:

`int (3x^2 + sqrt[x] - frac{5}{x^3})dx = int (3x^2+x^[frac{1}{2}]-5x^-3)dx`

`=x^3+frac{x^[frac{3}{2}]}{3/2}-5frac{x^-2}{-2}+K`

`=x^3+frac{2x^[frac{3}{2}]}{3}+frac{5x^-2}{2}+K`

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### Example 9

`introot(3)(x^2)dx`

Answer

Once again, we re-write the exponents in a more convenient form.

`int root(3)(x^2)dx = intx^(2//3)dx`

`=(x^(2/3+1)/(2/3+1))+K`

`=(x^(5//3))/(5//3)+K`

`=(3root(3)(x^5))/5+K`

Notice the constant of integration, `+K`, in each of these examples.

### Example 10

A particular curve has its derivative given by `(dy)/(dx)=3x^2-2x`**. **

We are told that the curve passes through the point (2, 5). Find the equation of the curve.

Answer

The first step for this problem is to **integrate** the
expression (i.e. find the **antiderivative**). This will give us the expression for `y`.

`int(3x^2-2x)dx=x^3-x^2+K`

So we have `y = x^3− x^2+ K`

This represents a **family of curves**, and depends on the
value of `K` for the *y*-intercept.

We must now find the value of `K` from the information given in the question.

Since the curve passes through `(2, 5)`, we substitute these values into

`y = x^3− x^2+ K`

to give:

`5 = (2)^3 − (2)^2 + K`

`5 = 8 − 4 + K`

So `K = 1`.

So the required curve is `y = x^3− x^2+ 1`

Here is a graph of the curve we found in Example 10:

Notice the curve passes through the point `(2,5)`.

### Example 11

Consider this integration:

`int(2x^4-5)^6x^3dx`

This is different to the other exercises above!

The expression we have to integrate containts `(2x^4-5)^6`, which is a
function of a function, **and** we have that `x^3` at the end. We cannot do this integration using the rules we have learned so far.

In this case, we have to do the reverse of
the Chain Rule, which we met in the section on **differentiation.**

We introduce a new rule for integrating cases like these.

### Mini-Lecture

See the

mini-lecture on substitution.

## C. Power Formula for Integration

`int u^ndu=u^(n+1)/(n+1)+K`

(This is true if `n ≠ -1`)

This requires a substitution step, where *u*(*x*) is some function of *x*.

Now back to the problem to see how to apply this formula.

Integrate `int(2x^4-5)^6x^3dx`.

Answer

We use, as a starting point, the substitution

`u = 2x^4- 5`.

Why? Because `2x^4- 5` is the expression in brackets in the question.

Now differentiating `u` gives: `du = 8x^3 dx`

Our question has only **one** `x^3 dx` (not 8 of them) so we need
to divide both sides by `8`:

`1/8du=x^3 dx`

(Now the right hand side is the same as what we have in the question, `x^3 dx`.)

We can now rewrite our question as:

`int(2x^4-5)^6x^3 dx`

`=1/8intu^6 du`

(Notice I'm not mixing up `x`'s and `u`'s in any one expression here.)

Now we integrate with respect to `u`:

`=1/8xxu^7/7+K`

`=u^7/56+K`

Finally, we express everything in terms of `x`, since that's the variable we started with:

`=((2x^4-5)^7)/56+K`

## More substitution examples

### Example 12

`int(x^3-2)^6(3x^2)dx`

Answer

We use, as a starting point, `u = x^3- 2`.

Now `du = 3x^2 dx`

This time, this is exactly what we have in the question, so there is no need to divide both sides.

We can now write:

`int(x^3-2)^6(3x^2) dx=intu^6 du`

`=u^7/7+K`

`=((x^3-2)^7)/7+K`

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### Example 13

Find `intx/(sqrt(x^2+9))dx` using a substitution.

Answer

Put `u=x^2+9`

Then `du = 2x dx`

The question has just one `x dx` so we divide both sides by 2:

`(du)/2=x dx`

Now

`intx/sqrt(x^2+9)dx = 1/2int(du)/sqrtu`

`=1/2intu^(-1//2)du`

`=1/2(u^(1//2))/(1//2)+K`

`=u^(1//2)+K`

`=sqrt(x^2+9)+K`

### Example 14

Given `y'=sqrt(2x+1`, find the function `y = f(x)` which passes through the point `(0,2)`.

Answer

**NOTE:** `y'` means **the derivative of **`y`, that is `(dy)/dx`.

`y'=sqrt(2x+1`

So

`y=intsqrt(2x+1) dx`

This question requires us to integrate, and in the process, to find the **constant of integration.**

Put `u = 2x + 1`.

Now `du = 2 dx`

So

`1/2du=dx`

So

`y=intsqrt(2x+1) dx`

`=1/2intu^(1//2)du`

`=1/2xx2/3xxu^(3//2)+K`

`=1/3(2x+1)^(3//2)+K`

Now if `x = 0` then `y = 2` (from the question).

`2=(1^(3//2))/3+K`

and this gives `K = 5/3`.

So

`y=((2x+1)^(3//2))/3+5/3`

is the required function.

Easy to understand math videos:

MathTutorDVD.com

Note: You will see "+K" **and** "+C" in this work. Most
textbooks use + C.

It's a good idea to **always** use **+K** if you are answering electrical
problems.

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