Integration

Lesson

As we have just seen, there are infinitely many curves that result in the same derivative.

We call this collection of curves the family of antiderivative functions for a given function.

We need to be able to start identifying from a graph and from equations, those that will belong to the same family.

Let's look at some examples.

For a function $f(x)=5x^4$`f`(`x`)=5`x`4, which of the following belong to the family of antiderivative functions.

a) $F(x)=x^5+1$`F`(`x`)=`x`5+1

b) $F(x)=-x^5+1$`F`(`x`)=−`x`5+1

c) $F(x)=x^3-4$`F`(`x`)=`x`3−4

d) $F(x)=5x^5$`F`(`x`)=5`x`5

e) $F(x)=x^5-\frac{1}{2}$`F`(`x`)=`x`5−12

f) $F(x)=6\pi+x^5$`F`(`x`)=6π+`x`5

Firstly we notice that the degree of $f(x)$`f`(`x`) is $4$4. This means that the degree of the antiderivative must be one more - ie $5$5. This means we can discount option $(c)$(`c`), as it is of degree $3$3.

The antiderivative of $f(x)=5x^4$`f`(`x`)=5`x`4 is $F(x)=x^5+C$`F`(`x`)=`x`5+`C`. We can check this by inspecting $F(x)$`F`(`x`) and establishing what $F'(x)$`F`′(`x`) is. This means we need to have $F(x)$`F`(`x`) functions that contain only $x^5$`x`5 terms with a coefficient of $1$1.

Option $(d)$(`d`) for example, $F(x)=5x^5$`F`(`x`)=5`x`5 would yield a derivative of $f(x)=25x^4$`f`(`x`)=25`x`4. Which is clearly quite different to $f(x)=5x^4$`f`(`x`)=5`x`4. Also, we know from our work with power functions that the function $F(x)=5x^5$`F`(`x`)=5`x`5 is a LOT steeper than $F(x)=x^5$`F`(`x`)=`x`5, hence the gradient function of these are not the same. I've plotted both graphs on this plot to remind you.

So we can eliminate $(d)$(`d`) and $(b)$(`b`) as these do not have coefficients of $x^5$`x`5 of $1$1. There coefficients are $5$5 and $-1$−1 respectively.

This leaves us with

a) $F(x)=x^5+1$`F`(`x`)=`x`5+1

e) $F(x)=x^5-\frac{1}{2}$`F`(`x`)=`x`5−12

f) $F(x)=6\pi+x^5$`F`(`x`)=6π+`x`5

The only difference with these is the value of the constant term at the end. Let's have a look at the graph of these three.

Do you remember the effect the constant term has on the graph? That's right, it's a vertical translation. It doesn't effect the gradient at $x=a$`x`=`a` at all.

This interactive demonstrates what I mean by this. Here, you can vertically translate the curve y=x^5. You can also see the gradient of the tangent at point A, remember that this is the value of the derivative at this point. As you move the curve, the gradient remains unchanged.

All of the curves generated by that vertical translation belong to the family of antiderivatives of the function $f(x)=5x^4$`f`(`x`)=5`x`4.

Thus $(a),(e)$(`a`),(`e`) and $(f)$(`f`) are antiderivatives.

a) $F(x)=x^5+1$`F`(`x`)=`x`5+1 YES

b) $F(x)=-x^5+1$`F`(`x`)=−`x`5+1 NO This function has a reflection across the x-axis (indicated by the negative), it's coefficient of $x^5$`x`5 is $-1$−1, not $1$1.

c) $F(x)=x^3-4$`F`(`x`)=`x`3−4 NO This is only degree $3$3, the antiderivative must be of degree $5$5.

d) $F(x)=5x^5$`F`(`x`)=5`x`5 NO This function has a coefficient of $x^5$`x`5 of $5$5, not $1$1.

e) $F(x)=x^5-\frac{1}{2}$`F`(`x`)=`x`5−12 YES

f) $F(x)=6\pi+x^5$`F`(`x`)=6π+`x`5 YES

This applet, shows a family of antiderivatives. It shows $y=x^2$`y`=`x`2, $y=x^2+1$`y`=`x`2+1, $y=x^2-1$`y`=`x`2−1 and $y=x^2-3$`y`=`x`2−3.

You can move the point $A$`A`. As you do you can see how all the tangent lines stay in parallel. This demonstrates that the value of the derivative is the same for the whole family of antiderivatives of $\frac{dy}{dx}=2x$`d``y``d``x`=2`x`

Consider the gradient function $f'\left(x\right)$`f`′(`x`)$=$=$2$2.

The family of the antiderivative, $f\left(x\right)$

`f`(`x`), will be:Exponential

ACubic

BLinear

CQuadratic

DExponential

ACubic

BLinear

CQuadratic

DThe form of the antiderivative will be $f\left(x\right)$

`f`(`x`)$=$=$mx+c$`m``x`+`c`. State the value of $m$`m`.Which of the following functions represent possible values for an antiderivative $f\left(x\right)$

`f`(`x`)?Select all that apply.

$f\left(x\right)=-2x$

`f`(`x`)=−2`x`A$f\left(x\right)=2x+6$

`f`(`x`)=2`x`+6B$f\left(x\right)=2x-3$

`f`(`x`)=2`x`−3C$f\left(x\right)=-2x+6$

`f`(`x`)=−2`x`+6D$f\left(x\right)=-2x$

`f`(`x`)=−2`x`A$f\left(x\right)=2x+6$

`f`(`x`)=2`x`+6B$f\left(x\right)=2x-3$

`f`(`x`)=2`x`−3C$f\left(x\right)=-2x+6$

`f`(`x`)=−2`x`+6D

Consider the gradient function $f'\left(x\right)$`f`′(`x`)$=$=$4x+3$4`x`+3.

The family of the antiderivative, $f\left(x\right)$

`f`(`x`), will be:Linear

AQuadratic

BCubic

CExponential

DLinear

AQuadratic

BCubic

CExponential

DThe form of the antiderivative will be $f\left(x\right)$

`f`(`x`)$=$=$ax^2+bx+C$`a``x`2+`b``x`+`C`. State the value of $a$`a`and $b$`b`in simplified form if necessary.$a$

`a`$=$= $\editable{}$$b$

`b`$=$= $\editable{}$Which of the following functions could possibly represent $f\left(x\right)$

`f`(`x`)? Select all that apply.$f\left(x\right)=4\left(x+3\right)^2+7$

`f`(`x`)=4(`x`+3)2+7A$f\left(x\right)=2x^2+3x$

`f`(`x`)=2`x`2+3`x`B$f\left(x\right)=2x^2+3x+4$

`f`(`x`)=2`x`2+3`x`+4C$f\left(x\right)=2x^2+3$

`f`(`x`)=2`x`2+3D$f\left(x\right)=4\left(x+3\right)^2+7$

`f`(`x`)=4(`x`+3)2+7A$f\left(x\right)=2x^2+3x$

`f`(`x`)=2`x`2+3`x`B$f\left(x\right)=2x^2+3x+4$

`f`(`x`)=2`x`2+3`x`+4C$f\left(x\right)=2x^2+3$

`f`(`x`)=2`x`2+3D

Consider the gradient function $f'\left(x\right)$`f`′(`x`)$=$=$-7x^2$−7`x`2.

The family of the antiderivative, $f\left(x\right)$

`f`(`x`), will be:Exponential

ACubic

BQuadratic

CLinear

DExponential

ACubic

BQuadratic

CLinear

DThe form of the antiderivative will be $f\left(x\right)$

`f`(`x`)$=$=$ax^3+C$`a``x`3+`C`. State the value of $a$`a`.Which of the following functions could possibly represent $f\left(x\right)$

`f`(`x`)?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

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