Integration

Lesson

An antiderivative is the function obtained by going in the opposite (anti) direction to differentiation.

We define the antiderivative of a function $f(x),F(x)$`f`(`x`),`F`(`x`) if $F'(x)=f(x)$`F`′(`x`)=`f`(`x`).

There is a special symbol $\int$∫

that we use for the antiderivative, that is also called the indefinite integral. (The term integration is the topic in mathematics that includes definite, and indefinite integrals - we will define the difference later).

If $F'(x)=f(x)$`F`′(`x`)=`f`(`x`) then $\int f(x)=F(x)$∫`f`(`x`)=`F`(`x`)

Of course there is a rule (in fact a whole bunch of rules) for this process of antidifferentiation, but I want you to really understand the idea here.

So bare with me as I step you through this little activity.

Find the derivative of the following $5$5 functions.

a) $F(x)=3x^2+4x+5$`F`(`x`)=3`x`2+4`x`+5

b) $F(x)=3x^2+4x+1$`F`(`x`)=3`x`2+4`x`+1

c) $F(x)=3x^2+4x-17$`F`(`x`)=3`x`2+4`x`−17

d)$F(x)=3x^2+4x-3$`F`(`x`)=3`x`2+4`x`−3

e) $F(x)=3x^2+4x+a$`F`(`x`)=3`x`2+4`x`+`a`

What did you notice?

Well, because all the constant terms (the $5,1,-17,-3$5,1,−17,−3 and $a$`a`) all had a derivative of zero, we ended up getting the same derivative for all of them -> $f(x)=6x+4$`f`(`x`)=6`x`+4.

Why does this matter? Well, what if I told you that a function $f(x)$`f`(`x`) is $f(x)=6x+4$`f`(`x`)=6`x`+4 and I asked you to find the antiderivative? How would you know which one it was? Does it have to be one of the $5$5 listed above?

In fact there are many many antiderivatives of the function $f(x)=6x+4$`f`(`x`)=6`x`+4 and these are just some of them $F(x)=3x^2+4x+1$`F`(`x`)=3`x`2+4`x`+1, or $F(x)=3x^2+4x-17$`F`(`x`)=3`x`2+4`x`−17 or $F(x)=3x^2+4x-3$`F`(`x`)=3`x`2+4`x`−3.

The antiderivative of the function $f(x)=6x+4$`f`(`x`)=6`x`+4 is $F(x)=3x^2+4x+C$`F`(`x`)=3`x`2+4`x`+`C` Where $C$`C` is a constant term that we don't yet know! But the $C$`C` is VERY important. Don't forget it!

As we know we can define the derivative using the different notations

$f(x)$`f`(`x`) then the derivative is $f'(x)$`f`′(`x`)

$y$`y` then the derivative is $y'$`y`′

and

$y$`y` then the derivative is $\frac{dy}{dx}$`d``y``d``x`

$y=x^3$`y`=`x`3 is just ONE of the antiderivatives of $\frac{dy}{dx}=3x^2$`d``y``d``x`=3`x`2

There are many many others, in fact infinitely many others.

$y=x^3-1$`y`=`x`3−1

$y=x^3+4$`y`=`x`3+4

$y=x^3+\pi$`y`=`x`3+π

$y=x^3-\sqrt{2}$`y`=`x`3−√2

and so on. (Any vertical translation of the function $y=x^3$`y`=`x`3 will result in the same derivative)

Generally speaking the function $y=x^3+C$`y`=`x`3+`C` is the indefinite integral of $3x^2$3`x`2.

We write this using mathematical symbols like this

$\int3x^2dx=x^3+C$∫3`x`2`d``x`=`x`3+`C`

Which when said verbally is read out as,

"The integral of $3x^2$3

x2 with respect to x is equal to $x^3+C$x3+C. "

The $\int$∫sign, stands for sum. Which is why it looks like a long skinny letter $s$`s`. The integral, as we will learn later, is actually the sum of the areas of many many thin (very very thin) rectangles. There is another symbol in mathematics you may have come across already that also means sum, it is this one $\Sigma$Σ.

$\Sigma$Σ can be used to denote infinite or finite sums

$\int$∫ can be used to denote the sum of an infinite number of small areas.

The inverse property of indefinite integrals:

$\frac{d}{dx}\int f(x)dx=f(x)$`d``d``x`∫`f`(`x`)`d``x`=`f`(`x`)

$\int g'(x)dx=g(x)+C$∫`g`′(`x`)`d``x`=`g`(`x`)+`C`

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply integration methods in solving problems