You should be able to:
• understand the basic informal principles of limits and
convergence
• find the derivative of a polynomial function from first
principles
• use a rule to differentiate algebraic functions containing
sums and differences of kx", nEQ
• differentiate the functions sinx, cosx, tanx, Inx and eX
• the gradient function can be used to find the gradient of
the tangent to the curve at a point and hence the gradient
of the curve at that poi nt.
Example
Let r'(x) = r -;2X2 - l.
(a) Find/'(x).
Apt/j'iAtJ tlu- fDwu
= 3x
we,~I'(x)
• find the second derivative of a function
• interpret the derivative as the gradient function
• interpret the derivative as a rate of change.
You should know:
• the notation f'(x) or ~~ is used to denote the first
derivative of a function. The second derivative is denoted
d2
by f"(x) or d;
• derivatives of
d
• dx kx"
=
-
4x.
(b) Find the gradient of the curve ofJ(x) at the point
(2, -1).
We-a;re-MWto fou£tlu-Jr~t
x = 2, tluvt is,1'(2).
f'(2)
= 3(2j
- 4(2)
oftlu-CM;Y~~
=4
Tkuefor~ tlu-Jrttdie,/1;tof tlu- CM;Y~(a;fUlkeece: tluJr~t
oftlu-ta;fVje-M) a;t x = 2 if 4.
Be prepared
nkx" - 1
d .
• dxsmx
=
d
• dxcoSX
.
= -smx
d
nde:for deriAra;tivu of fDLyfWmAAM,
2
• Work carefully when finding the derivative from first
principles as it is easy to get lost in the algebra.
cosx
1
• -d tanx = -2X
cos x
d
1
d
• -Inx=-and-e=e
dx
x
dx
• Write polynomial functions in an appropriate form before
attempting to find the derivative. For example, the
function f(x) = 6-07" should be expressed asf(x) = 6X2/3
fi rst.
You should be able to:
•
use the derivative to find the equation of a tangent or a
(b) The equation of the tangent to the graph off at
x = p is y = 7x - 9.
normal to a graph
•
(i) Find the value of p.
use the equation of a tangent or a normal to a point on a
7Ive-jY~toftke,tM'l/j'*tfUu,u
graph to find the coordinates of that point.
h4n
7. S~f'(x)
=.0 we:
6x - S = .0 ,0 x = 2
You should know:
•
HMt-atke,ll"~offu
(ii) Find the value of k.
touches the curve at that point but does not cross it
•
the tangent to the graph of f at the point where
Tofoul the. Il"fN{uof k, toe. m;u,d font fod tke,
vorrufoH.A:U1tJY-Il"P,;{uOft-tke.jYfNt!v off whe« x u 2.
Ar tke,fO*t fNUOlie« Oft-tke,tfNltJe,ft-~~t rMufou tke.
UjUAA/#Oft-.
Hence,
x = 0 has
equation
y - f(o) = f'(o) (x - 0)
•
the normal to a curve is the line perpendicular
to the
y=
tangent at a point on the curve
•
2.
the tangent to a curve at a point is the straight line that
the normal to the curve at x = 0 has gradient -
=s
f'~r
COMUjlUft-tty,tke,fO*t (2, s) lce« Oft-tke,jyap4off
rubrtH~t~ *to tke,OYiJ~fIMfdWft-
Example
Consider the functionf(x)
= 3.x2 -
5x
+ k.
= 6x
So,
5 = 3(2j - 5(2) + k
(a) Write downf'(x).
j'(x)
7(2) - 9
k=3
- S
Be prepared
•
Be prepared to work backwards-that
is, find the
coordinates of a point on a graph if you know the gradient
of the tangent or normal to that point.
(b) y
You should be able to:
• use the product rule to differentiate functions of the form
y = uv where u and v are both functions of x
• use the quotient rule to differentiate functions of the form
u
y=-y
• use the chain rule to differentiate composite functions of
the form y = g(f(x)).
= xtanx
TkefVl/MtWIt'f if the.p-rodf,U,t0/ tw-o?MtWM X (uU£
ttv~X. It if ~tpfVl/lto fuu£t~ de.rtv-tvtw-e, ~k
tkese.
jVl/l1NtwMftrst- ikee: slNbstttM~ vorre-vtly tMo tke.jornuvltv
jor t~p-rodud rule.
0/
d
d
-x=1~-t~x=-do:
dx
0/
1
volx-
SlNbstttMiAtj,
You should know:
• if y
= uv, and if both
u and v are differentiable functions
~xttv~X
dx
=x~t~x
d»:
=_x_+t~x
volx
of x, then
dy
dx
=
+ttv~X~x
dx
dv
du
u dx + v dx
(c ) y=- lnx
x
• if y
7hM 1ustw~if MkUtj
1tUf{-uM o/tw-oFtWM
= ~v then
dy _
dx -
du
v--udx
dv
tofuu£t~ww-a,tw-e,
o/t~
tvfUix. we-lawurtktvt
~X
~ l~x =.1 ~tktvt~
X = 1.
dx
X
d»:
SuJJstttVl/tiAtjtMo t~1tUf{-tmt rule, w-e-ktvn
dx
v2
• it is useful to think of the chain rule as the "outside-inside"
rule. First, differentiate the "outside" function, evaluate
it at the "inside" function, then multiply the result by the
derivative of the "inside" function. That is, if y = g(f(x)), then
~~ = g'(f(x)) x
tU
:!:t _
d
x-~x-~x-x
dx
dx -
d
dx
X2
x.1-~x
f' (x).
X
Example
Differentiate each of the following with respect to x.
(a) y
= sin3x
Be prepared
7hM if tvvol11.f'OsttejVl/l1Ntw~~
f~x~3x.
AptJlfUtj t~~r~
~
dx
f~3x
=
cos (sx)
x~
dx
s»
Uf' 0/ tke.jVl/l1NtwM
w-e-~
• The key to successful differentiation
which rule to apply.
is in recognizing
• The phrase "rate of change" suggests that you need to find
a derivative.
= 3vos3x
• You may need to use a combination of rules when finding
the derivative of a function.
l1h
You should be able to:
• use derivatives to analyse the shape and key features of a
graph, including points of inflexion, maxima and minima
and concavity
• the graph of f is concave downwards in an interval if the
tangent lines drawn at every point in the interval are
decreasing in gradient (Figure 2). Consequently, if f'(x) is
decreasing, then f"(x) < 0 for every x in that interval
• interpret and analyse the relationships between the
graphs of i.t' and t:
You should know:
The following graphs can be used to help you interpret the
statements below.
{(x) = X4_
4x3
+5
x
• the graph of fis increasing on an interval if f'(x) > 0 and
decreasing if f'(x) < 0 for every x in the interval
• f"(x) is the rate of change of f'(x) in the same way that f'(x)
is the-rate of change of f(x)
• if at x == c, f'(c) = 0
• the graph of f has a minimum value if f' changes from
negative to positive at c
• the graph of f has a maximum value if f' changes from
positive to negative at c
• the graph of f has no extreme values if t' has the same
sign on both sides of c
• the graph of f is concave upwards in an interval if the
tangent lines drawn at every point in the interval are
increasing in gradient (Figure 1). Consequently, if f'(x) is
increasing, then f"(x) > 0 for every x in that interval
Figure 2 The graph of y = -x' is concave downwards
• a point of inflexion occurs when a graph changes from
concave downwards to concave upwards or from concave
upwards to concave downwards. At these points, f"(x) = o.
Example
The graph of a function g is given in the diagram
below.
y
c
o
E
L-~--~------~--~----~--------------~x
a
b
c
d
e
The gradient function g'(x) has its maximum value at
point B and its minimum value at point D. The tangent
is horizontal at points C and E.
y
10
-4
4
(a) Complete the table below, by stating whether the
first derivative g' is positive or negative, and
whether the second derivative g" is positive or
negative.
Interval
a<x<b
Figure 1 The graph of y
= X2
is concave upwards
e<x<f
g'
g"
..
Itl/t~if/;tMv-aJIAI< X < b, t~jrap4ofj
is iAtNreMUtj, so
j' is POfUWe.-.TMjrap40fj
is aMOVb~urw-IAlYM,
so
/ is aMOPOfUWe.-,
jrap4is u)~MW1U/!MM,
so/ (x) < 0, a.+t.AiMfOif/;t E,
/(x) = 0 tU?t is aMOIAlfOif/;tof i#jtexWtl/. H~
tke.
poif/;t C is t~Only poif/;t wfu,y~j'(x) = 0 M/At/(x) < o.
Itl/t~?MMlTd~<
X < t t~jrap4ofj
is dwreM?tI/j, so
j' is ~MWe.-. TMjrap40f j is Vb~MunU/!MM
if/;
tlU? ?tl/tMIT1AI0
so/ U1!Mth~MWe.-.
W~IAIY~
told tlrAt ike.jr~t
juAtvtWtl/ IrAf IAI~U#(;
ITa£u,e,Mt~poif/;t DJ solex) = o. Also, M DJ t~jrap4
ofj(X) to dwreMUtj so tlrAt j'(x) < o. H~
t~fo?M
fMisjiM bd'f~ of t~j~
ceiteei«;
D
(b) Complete the table below by noting the points on
the graph described by the following conditions.
Be prepared
Conditions
°
g'(x)
= 0, g"(x)
<
g'(x)
< 0, g"(x)
=0
Point
• If there is a point of inflexion, then 1"(x) = O. But the
converse is not necessarily true. Knowing1"(x) = 0 does
not guarantee a point of inflexion.
=
j'(x) = O~IAIMr?Z()tl/tdtMtj~tf..iAu,otl/t~jrap4
ofj. 'l'ItM occars Mt~fOif/;tf C a.+t.AiE.HoWWUj M C the:
• For a point of inflexion to exist at x c, you must show
not only that 1"(c) = 0 but also that the sign of 1" changes
on either side of c.
J
119
You should be able to:
• find the maximum or minimum value of a function
• use the first or second derivative to justify whether a
function has a local maximum or a local minimum
AltAiytf..c.td
We-jWst rued tofoHA,t/.;f(, MM;!lNftNt?Of1,
to I114NX?I*?Ze-.
AD = BC = x, tke« DC = 60 - zx. Af ABeD U a"
rutP.n!J~ t~MM;juMtWf1,VMf.,be-exfYegedtU
If fAte, ut
A(x)
• apply the concept of maximum and minimum to
problems for which there is an optimal solution.
=
• a function has a maximum value at x = cif f'(c) = 0 and
f'(x) changes from positive to negative on either side of c
• a function has a minimum value at x = cif f'(c) = 0 and
f'(x) changes from negative to positive on either side of c
• the second derivative can also be used to justify a
maximum or a minimum value. If f'(c) = 0 and J"(c) < 0,
then f has a local maximum at x = c. If f'(c) = 0 and
J"(c) > 0, then f has a local minimum at x = c
• the derivative can be used in applications where
maximum and minimum values are required. These
are called optimization problems. Examples include
maximizing profit or volume and minimizing costs or
surface area.
Example
The following diagram shows a rectangular area
ABeD enclosed on three sides by 60 m of fencing, and
on the fourth by a wall AB.
= x(60
- zx)
= 60x
- 2X2.
1
TofoItdit~ oalae: x tMt I114NX~ZM tIU. juMtWf1" sM
A'(x) = 0 ~ sow-efor x.
You should know:
• to find the maximum or minimum value of a function, set
f'(x)
0
solacion.
A'(x) = 60 - 4x = 0, sox = 15
TojMtify tIU. rwdt fAte,I*Mf Mw-slvow-tMt x = 15
fYo~
a" I114NX~1NI*
oalue; elvoos~ tnltNe-sdose- P';HA, to
t~Ufr115,
fAte,foItditMt A'(x) > o.
elvoos~ osdae« dos~~to
A'(x) < o.
tlr.u?JM
11~ sutMt
fAte,
'/"1u,refore;t~MM;ju&tWf1,A(X) U1Ue-MUto t~ufr1
x = 15 P.;HA,de-vre-MU
to ike: r?J~ so a" I114NX~ I*Mt
oWUA'~X
= 15.
S~ x rqU'Me-f1,tst~ width" AD 1t~ rutp.;l1ij~ t~
t~Me-P.;U I114NX~zed~t~
width"u 151*.
CiDe soiutiof//
AltlvolNjh"fAte,VMf.,Me-M;DetU DINY
fY~y
tDoljDrSD~
tIU. fYDb~ fAte,~
stiUfoItdit~MM;juMtWf1,A(x)
tU MOlfe,~dtw
slvow-tMt t~jrtJ..f41 A ~ P.;
I114NX~lUU/
fD~t or stMe- th"M t~ I114NX~1NI*
vrdu,e., mU
OWUA'~f1,A'(x) = o.
~
aoc to dra"w-t~ jrtJ..f41
M~YOINY
1
A(x) ~
foItdi
ike. WC4tWf1, tke. I114NX~1NI*
fD~t.
Texas Instruments
Casio
DL-------------------------~C
Find the width AD of the rectangle that gives it the
maximum area and justify your answer.
Be prepared
• Setting the derivative equal to zero does not guarantee
that a function will have a maximum or minimum value.
You should be able to:
Example
• understand that integration is the reverse of the process of
differentiation
Iff'(x)
• integrate algebraic functions containing sums and
differences of kx", n e. (jJ)
• integrate the functions sinx, cosx, ~ and
x
e, and
= -2, findf(x).
= cosx, andf(~)
We,a,ye,3~ M'r> UfUA<tio~iAwo~
tVde,yWa"t~
I'(x) = cosx, MtJ.,tVbolMUiMy condaio«;
= -2.
We-a,ye,Mladto fof1A;tVFtio"'j{x).
So,
f(~)
f(x)
=
J VOfXttX = sle»: +
C
composites of these functions with the linear function
Su.bfttt~tiAtj tke,bo~ry
ax+ b
• integrate f'(x) to find a function f(x) and use a given point
(called a boundary condition) to find the constant term.
f(~) = f?",~+ c =
twefore-c
-2,
vonA.itio", tojUdC,
so
(.{J'e,/uvv-e,
1 + C = -2,
= -3 tVf1A;j(x) = 5~X
-
3
You should know:
• the basic integration rules are obtained by simply looking
at integration as the "reverse process" of differentiation
Be prepared
• Write a function in an appropriate form before trying to
3
x"+1
•
J kx" dx = k n + 1 + C, n
-cosx +
Jsinxdx
=
JCOSXdX
= sinx + C
;t
find the integral. For example, Jx ;
-1
3 dx is best written
J (x + 3[2) dx using the laws of exponents and is then
easily integrated as ±
3x- + C.
as
C
X2 -
1
• Include the constant of integration C when integrating to
find a function f(x)·
J ~dx
= Inx
Jedx=
e+
+C
C
• the solution to an equation that contains a derivative such
as f'(x)
= 3x
2
or ~~
= sin (2x -
1) is a function of the form
y = f(x)+C. A boundary condition can then be substituted
to find a specific value of C.
121
You should be able to: -
y
1
• evaluate a definite integral
A
• write and evaluate an expression to find the area under a
curve
• find the area between two curves
• use a GDCto evaluate definite integrals and areas.
-1
(a) Write down an expression for the area of A.
You should know:
• to evaluate a definite integral, use the following rule:
ff(X) dx = F(b) - F(a), where F is the anti-derivative of f(x)
'flu, y~io~ UUlA'WA UM VD~teiy
aboo» t4 X-PvXU
fr0UfJX =
to x = 27C. 'fIu,yefoy~ t4delv~ite,iM~ytd
3;
1
2"
VDf
• the properties of the definite integral are best explained
using an example. Suppose that I:f(X) dx
1
6f(x)dx = -3 and
2
= 5,
x d« YqwU~tf t4M~of
A.
37</2
(b) Calculate the area of A.
1
AiUUyttud
2g(x)dx = 7 then
~
fottdion-
• ff(x) dx = - ff(x) dx = -(-3) = 3
IM~YM~
t4 exfWM!io~ ~A Mt,d,m~ tke: Y~fOY
~
PvtUfurkte,~t~Ya4 unfuwe,
• I>(X) dx = i:f(x) dx +ff(x) dx = 5 + (-3)
1
2"
cos x d«
=2
= [,i~XJ::/2
3,,/2
= f~27C
• I:[3f(x)
+ 2g(x)]dx = 3 i:f(x)dx
=0 =1
= 3(5) + 2(7) = 29
• if a function is continuous and positive on an interval
a :s; x s b, then the area under the curve is given by the
definite integral ff(x)dX
• if a function lies below the x-axis on an interval a :s; x s b,
a
tf(X) dx
or
a
if(x)dx
• the area between two curvesf(x)
f
,UtJ!!2
+2 J)(x)dx
then the area under the curve is given by -
-
(-1)
4DCso(.,u,tion-
EMe,y~ tlu,exfWM!io~fr()UfJ~
/uvv-e-;
Texas Instruments
fnlnt(cos(X),X,3
(Pv)~to the. ate, un
Casio
f(cos
X,3n.;.2,2n)
:rr/2,2Jt)
and g(x) is given by
[j(x) - g(x)]dx where f(x) > g(x) on an interval a :s; x s b.
The lower and upper limits of integration a and bare
often but not always the x-coordinates where the two
graphs intersect.
Texas Instruments
Casio
Example
The following diagram shows part of the graph of
y = cosx for 0 s x s 21t. Regions A and B are shaded.
122
\,
L\)(,.lER=q. 1123
f~;';= 1
/
..
UPf'Et,=6.2831
- ~ =-~--~~
Area and the
d;fi.;it;in!~gralj(conthiue(l)
-.::::--~.-:'....
-~;;..
(ii) Hence, find the total area of the shaded
·)1
C c) C
G'l~en th at sm
. 41t = -13
2 fin d th e area 0f
3
region B.
"R.~io~8 lies bdourt~ x-axic. We-a.nMWfor ike. tdfat
Me-atso we-rUfuMe-tke. Me-atof r~io~ 8 to be-tofUWe-. 1ItM
ce« h~v-eJ-by
fwit~
t~Umktf on.ou« k~te4"/3
1
1
Uvt~rat. So)
cos x dx rqwe-rmts t~Me-atof r~io~ 8.
3,,/2
cos x dx
4"13
J3"/2
r.'
4n
3
,3n]
- SIA1r2
{3
= - - - (-1)
2
=
"Wfxdx +
2
3"12
by
14"/3cos x d«
3,,/2
= 1+ 1--{3
2
{3
'lMqDC mAJ' be-tUUM t~(b) tofo~t~e-tdf~4
tMXut
Me-atofl.13.
::;:;LflAtX
r,
1
u t~3Wut,
2
37[/2
= lf~
The. tdftda.nat
::;:;2--
4,,/3
Are-atof 8 =
regions.
Be prepared
• Use more than one expression to find the area under a
curve that lies partially above and below the x-axis.
• Enter the expression for area into your GDCcorrectly.
• Indicate the required area on a sketch.
{3
1-2
• Find a limit of integration if you are given the area.
123
You should be able to:
Texas Instruments
• calculate the volume of a solid of revolution formed when
a region under a graph is rotated 3600 or 2n radians
around the x-axis
F·1.;.t1 F'l,)t;:: F'1(,t;:
",\' 1 El,::····,,..
"''''-1 ') +~/ (
2X-i)
'._.,
" - "
....1-.'2=
···.v 3: =
• use the integral features of your GDCto calculate the
volume.
···.\l~=
....V~=
:rrfnlnt(V1 z, i'~' 1,1
.5)
•
1(14.76538
.•••1•••16=
You should know:
• when you fully rotate the area beneath the curve around
the x-axis, a solid of positive volume is formed
Casio
:rrJ'( "'1z , 1 , 1.5)
104.76538
• to find the volume of this solid, use the formula
V=
f nl
dx, where a and b are the x-intercepts of the
function.
1ft,e,yefore-,V ~
Example
Consider the functionf(x)
=e
2x
- 1
+ 2x 5_ l ' x
'* ~.
The
region between the curve and the x-axis between x = 1
and x = 1.5 is rotated through 360 about the x-axis.
Let V be the volume formed.
SiJlt;{;~M
a;/l'a;tuoj
105
correct
to three.
fojU.iYM.
Be prepared .
0
(a) Write down an expression to represent V.
• Don't forget to square the function when entering the data
into the GDC.
Ufi1tj tu forUf.,~ for oolaece, t-n kaoe:
V
=
n fS(e-2X -
1
+
Y
• Find an intercept/limit
2x ~ 1 d»:
(b) Hence, write down the value of V.
1ft,e,umu4fAlftdtuUf., "toricedoton" f~jMts
t~t
t-nM
bu,t yMur a;(jDC. Fj;yst
ence« tu fU.if1Ntwl/;mto tu UJU.!a;tWI/;
editor. Thee: 1Me- tu
mt~yalfe-a;tU.iYe- to ()bt~ tke: correct snstoe«
Mf
1Me- MI;MfA.lLytkal anwOM~
• First set up an expression that represents the volume even
if the question does not ask for it.
if a volume is given.
You should be able to:
Example
• obtain the velocity and acceleration functions, v(t) and
ott), from the displacement function s(t)
A car starts moving from a fixed point A. Its velocity v
metres per second after t seconds is given by
v = 4t + S - Se-I. Let d be the displacement from A
when t= 4.
• obtain the velocity and the displacement functions from
the acceleration function
• interpret the area under a velocity-time
travelled.
graph as distance
You should know:
• displacement measures the change of position of an
object as a function of time
• velocity is the derivative of the displacement function,
that is, v(t)
= ~~
• acceleration is the derivative of velocity with respect to
time or the second derivative of position, that is
dv d2s
ott) = dt = dt2
• if v(t) = 0, the object is not moving and has reached its
maximum or minimum height or distance
• if ott) = 0, the object's velocity is constant and the object
may have reached its maximum or minimum velocity
• jO(t)dt
= v(t) + C and jv(t)dt = s(t) + C
(a) Write down an integral that represents d.
1Iu- tniAJaty ~tiDlt;
if fOfUiAJ-e,
MU£ UtNreMiAtj over
tu ford 4 fW(HtM. 'flU" C4Vpt,
be, de,te,y~iAud by t(){)i<-i>1tj
at P.!jrP.f# of o-orrwoj~z-iAtj t~ tU t increase«, so
doe« /J: H~
tu Mea" tMUI..e¥tu caroe. if Ufu,a,tto tu
dUftMMIUMfr0~A.
1Iu-reforf0
d = 1(4t + 5 - s~-t)dt
(b) Find the value of d.
AH/alfttuda..-ptwotJU:Jt,
IM~ratiAtj, we-/uw-e,
d
= [2t + st + 5~-tl;
2
= (32
+ 20 + S~-1)
- (5)
= 47 + S~-4
aoc a..-ptw0a.vn,
UsiAtj tu~t~ra.l.fu,ftVtiDpt,~jr~
14W~
we-luvv-e-:
• if the graph of v(t) is continuous and positive on an
interval 0 :s;t s; b, then the area under the curve is given by
the definite integral fV(t)dt and represents the distance
Texas Instruments
Casio
----_.
travelled by an object during that interval of time.
lO(·.IEB=D
U~~ER=q
fdx=Q1. D9151B 19
9.
[Maximum mark: 14J
Let !(x)=e2xcosx,
(a) ~w
~!/(X)
-1S:xS:2.
= e2X(2cosx-sinx).
[3 marksJ
Let the line L be the normal to the curve of ! at x = 0 .
equation of L.
(b) ~he
[5 marks]
The graph of ! and the line L intersect at the point (0, 1) and at a second point P.
(c)
(i) @hex-coordinate
(ii) Gi~the
ofP.
area of the region enclosed by the graph of ! and the line L.
[6 marksJ
[Taken from paper 2, November 2008J
(a)
f(x)
is the product of two functions so use the product
(c) Using your GDCyou can find the x-coordinate of the
rule. Also remember the chain rule when differentiating
intersection of two functions. Make sure you write down
eZx• This is a "show that" question so be very clear in your
the equation used to find this value, not just an answer
wording.
with no working.
(b) Th'e derivative
function gives the gradient of a
tangent at any x-value
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for which it exists. You can
find this gradient in the GDC, or you can usej'(O) as
the derivative
normal
is given in part (a). Remember that L is
to the curve.
Plot1
Plot:::
Plot:::
'Y1e~A(2X)cos(X)
....•.•.•
z El-.5>:;+ 1.
···.Vs=
•••• 1',.1 Lt
=
....1.•.1&=
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Casio
l('1_e/'(2~~~
~-.-
-, t('1i
Casio
\
It
~
:'::0
d!,.-'dx=:::.I)(lo)(I(I(I:::
\
~~·....
d:~:2
\':1
\
Func
"IGr'':'F'h
1Se"'"(2:X:) cos
\
I,
:,:::.-c...-'-..~~
"
,i.
Ir,t.;;n,,·:tiMI
\
:l=1.!=Ii11:::B't ~'=.;:19't~:!=B:::
..",.,
-» /.'.'~.---'\""
__ ~-l~-
=
--•....
..•.•.•...•.
,...-.....
: '1-
>~
[-J
'1'2=-.5:0;+11
'1'3:
[-J
'1'4:
[-J
'1'5:
[-J
't'6:
[-J
IImI•• ElllU IDI
Once you have the limits of integration, you should write a correct integral
expression for the area between curves that involves the subtraction of integrals.
Then use your GDCto calculate the area. Don't forget to write your answer to
three significant figures!
• Derivative interpreted as
gradient function
• Finding equations of tangents
and norma Is
• Derivative of eX and cosx
Texas Instruments
fnlnt.(V1-V2, >(, 0,
1. 5613)
3. 275222E.58
Casio
• The chain rule for composite
functions
1(~1-~2,0,1.5613)
3.275222658
•
• The product rule
• Areas between curves
• Solutions of equations graphically
(a) /(x)
J:.
= &''' cor
x, -1
sx s2
/,(x) = 2&''' (s-sin.x)
"1----:
= 2&'''(cor
(b)
~~'----'
i sin.x)
x -
= &'''(2 cor
A1 A1 MO
x - sin.x)
AG
= 1
A1
oJ/(x)
.'. tohen:x
»
.'. /'(0)
=&,(O)(2corO-rUf,0)
M1
= 1(2 - 0)
/,(0)
A1
= 2
IWYuwJ,
L = .l: = 1x
111/
2
/(0)
then.
lx + c- =
=
+ c..1----...,
AO
1
1
2
1(0)+c..=1
2
J
c..=1
AO
121
/(x)
= e-
j'(x)
= 2e-
2x
cos x
2x
•
(cos x) + e-2x (- siscx)
M1 A1 A1
= e-2x (2 cos x - sin.x)
(b)
ne.T""!JmC
=
t(O)
(2
CbS 0 - f~01
AG
M1
.I
= 1(2 - 0) = 2
A1
1
= =z::". =-2
~NOYma.£
1
A1
T""!JmC
L....-:---t
y
y
=/(0)
= ~X
= t(O) CbS 0 =
1
(OJ
1)
A1
+b
1 = (- ~) (0) + b
=:>
b
= 1
A1
(e)
m
P(1.56, 0.219)
Area-
=
'.56
1
(e-2x cos x)clx -
o
4.23 - 0.952 = 3.28
.
;
N1
1,·56
(- !.X + Ok
0
M1 A2
2
A1
,
.
I1
..
.,'
I •
129