1 MATH HL - OPTION CALCULUS REVISION (one example for each case) by Christos Nikolaidis Example Remarks 1. L’ Hôpital Solution It works only for fractions 0 0 or ∞ ∞ 2. Continuity - Differentiability Solution • f is continuous at x=2, so 2 x lim → (x 2 +4)= 2 x lim → ax = f(2) Thus, 8=2a=8 ⇒ a=4 • f is continuous at x=3, so 3 x lim → ax= 3 x lim → (b-cx 2 ) =f(3) Thus 12=b-9c=12 ⇒ b-9c=12 (*) • f must be differentiable at x=2. Indeed, for x=2 = ′ − (2) f (x 2 +4)΄ =2x=4, = ′ + (2) f (ax)΄=a=4 • f must be differentiable at x=3. That is, for x=3 = ′ − (3) f (ax)΄=a=4, = ′ + (3) f (b-cx 2 )΄=-2cx=-6c so that 4=-6c (**) (*) and (**) give c=-2/3 and b=6 Therefore a=4, b=6, c=-2/3 Continuity at x=a means − →a x lim f = + →a x lim f = f(a) Differentiability at x=a means (a) f (a) f + − ′ = ′ i.e. the derivatives before and after point a are equal for x=a Find 3 2 x 0.5x x 1 x e 0 x lim − − − → 0 0 Let > − ≤ < ≤ + = 3 x , cx b 3 x 2 ax, 2 x 4, x f(x) 2 2 Find the values of a, b, c given that the function is continuous and differentiable.
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1
MATH HL - OPTION CALCULUS REVISION
(one example for each case) by Christos Nikolaidis
Example Remarks
1. L’ Hôpital
Solution
It works only for fractions
00 or
∞
∞
2. Continuity - Differentiability
Solution
• f is continuous at x=2, so 2x
lim→
(x2+4)=2x
lim→
ax = f(2)
Thus, 8=2a=8 ⇒ a=4 • f is continuous at x=3, so
3xlim
→ax=
3xlim
→(b-cx2) =f(3)
Thus 12=b-9c=12 ⇒ b-9c=12 (*)
• f must be differentiable at x=2. Indeed, for x=2 =′− (2)f (x2+4)΄ =2x=4, =′+ (2)f (ax)΄=a=4
• f must be differentiable at x=3. That is, for x=3 =′− (3)f (ax)΄=a=4, =′+ (3)f (b-cx2)΄=-2cx=-6c so that 4=-6c (**)
(*) and (**) give c=-2/3 and b=6
Therefore a=4, b=6, c=-2/3
Continuity at x=a means
−→axlim f =
+→axlim f = f(a)
Differentiability at x=a means (a)f(a)f +− ′=′
i.e. the derivatives before and after point a are equal for x=a
Find 3
2
x0.5xx1xe
0xlim
−−−
→
00
Let
>−
≤<
≤+
=
3x,cxb3x2ax,
2x4,xf(x)
2
2
Find the values of a, b, c given that the function is continuous and differentiable.
2
3. Rolle Theorem – Mean Value Theorem (MVT)
Solution
(a) sinxf(x) = is continuous and differentiable in [0,π] and )f(π0f(0) ==
Thus there is some value c∈]0,π[ s.t. 0(c)f =′ Indeed, xc(x)f os=′ and cosc=0 for c=π/2
(b) f(x) is still continuous/differentiable in [0.1,π-0.1] )0.1-f(πf(0.1) =
Again c=π/2
(c) f(x) is continuous and differentiable in [0,π/2] Thus there is some value c∈]0,π/2[ such that
0π/2f(0)-)/2f(π(c)f
−=′ , i.e. cosc=
π2
Indeed the GDC gives c=0.881
The minimum requirement is: - Continuous in [a,b] - Differentiable in ]a,b[ - f(a)=f(b)
Then 0(c)f =′ for some c∈]a,b[ The minimum requirement is: - Continuous in [a,b] - Differentiable in ]a,b[
Then abf(a)-f(b)(c)f
−=′
for some c∈[a,b]
Solution
-1f(1) = <0 and f(2)=11>0 Thus there is at least one root between 1 and 2 Suppose that there are two roots a and b
Then - f(x) is continuous and differentiable in [a,b] - )f(b0f(a) ==
Thus, by Rolle, there is a value c∈[a,b] such that 0(c)f =′
But, 53x(x)f 2 +=′ which has no roots which contradicts the existence of c
Therefore there is only one root.
A polynomial is continuous and differentiable everywhere In general, if f(x) has n distinct roots, then (x)f ′ must have at least n-1 roots (apply Rolle between any two consecutive roots)
Let sinxf(x) = (a) Explain why Rolle’s Thm holds in the interval
[0,π] and find the corresponding value of c (b) Does it hold in the interval [0.1 , π-0.1]? (c) Apply the Mean Value Theorem in [0 , π/2]
and find the corresponding value of c
Let 75xxf(x) 3 −+= Use Rolle’s Thm to show that f(x) has only one root
3
Solution
Consider f(x)=ex
It is continuous and differentiable everywhere. We apply the theorem in the interval [0, x] where x>0
0xe-e(c)f
0x
−=′ for come c∈[0,x]
That is 1ex
1-e cx
>= ⇒ ex-1 > x ⇒ ex > x +1
We also apply in the interval [x, 0] where x<0
0xe-e(c)f
0x
−=′ for come c∈[x, 0]
That is 1ex1-e c
x<= ⇒ ex-1 > x ⇒ ex > x +1
Finally, for x=0, ex = x +1
Therefore, in any case ex ≥ x +1
since x is positive
since x is negative
4. Fundamental Theorem of Calculus
Solution
(a) ∫′
++
−+1
02
2dx
1xx2x3x =
1
01xx2x3x
2
2
++
−+=
32)(
32 8
=−−
(b) ∫x
2
3dt(lnt)dxd = 3(lnx)
(c) ∫2x
2
3dt(lnt)dxd = 32 )2x(lnx
(d) ∫2x
xdt(lnt)
dxd 3 =
dxd
− ∫∫ x
0
3
0
3 dt(lnt)dt(lnt)2x
= 32 )2x(lnx - 3(lnx)
• [ ] f(a)f(b)f(x)(x)dxf ba
a
b
−==′∫
• ∫x
af(t)dt
dxd = f(x)
• ∫g(x)
af(t)dt
dxd = f(g(x)) (x)g′
use =∫g(x)
h(x)f(t)dt ∫∫ −
h(x)g(x)
aaf(t)dtf(t)dt
Use the MVT for f(x)=ex to show that 1xex +≥
Find
(a) ∫′
++
−+1
02
2dx
1xx2x3x (b) ∫
x
2
3dt(lnt)dxd
(c) ∫2x
2
3dt(lnt)dxd (d) ∫
2x
xdt(lnt)
dxd 3
4
5. Riemann Sum
Solution
(a) Subintervals [0,4π ] [
4π ,
2π ] [
2π ,
43π ] [
43π ,π]
(i) R.S. = f(0)4π + f(
4π )
4π + f(
43π )
4π +f(π)
4π = 1.11
(ii) R.S. = f(4π )
4π + f(
2π )
4π + f(
2π )
4π +f(
43π )
4π = 2.68
(b) Subintervals [0,2π ] [
2π ,π]
R.S. = f(4π )
2π + f(
43π )
2π = 2.22
(c) 2sinxdxπ
0=∫
So the approximation in (b) is much better despite the larger subintervals
Riemann sum = ∑i
ii ∆x)f(x
6. Improper Integrals
Solution
(a) dxx1lim
1b ∫∞→
b
[ ]b1blnxlim
∞→= [ ]0lnblim
b−=
∞→+∞=
(b) dxx1lim
12b ∫∞→
b b
1b x1lim
−=∞→
+−=∞→
1b1lim
b1=
(c) Find the indefinite integral first
dxexx∫ c+=+== ∫∫ xx
x-x-x-
e1-
ex-dxe-xedxxe
Thus,
dxex
0x∫
+∞ b
0xb e1x-lim
+=
∞→
+
+=
∞→1
e1b-lim bb
1=
It diverges
It converges to 1
(in the last limit use l’Hôpital)
It converges to 1
Consider ∫π
0sinxdx . Find the Riemman sum
(a) if you consider 4 subintervals of equal length and as representative values xi (i) the minima (ii) the maxima
(b) if you consider 2 subintervals of equal length and the midpoints as representative values
(c) Compare with the actual value of the integral
Find (a) dxx1
1∫+∞
, (b) dxx1
12∫
+∞
, (c) dxex
0x∫
+∞
5
7. Differential Equations (D.E.)
Solution
xxdy2y
d=+1
∫∫ =+
⇒ xx2y
dyd
1
c2xarctany2
+=⇒
For x=0, y=1 c4π
=⇒
Hence ,
4π
2xarctany2
+=
Notice: If they ask to express y in terms of x:
General solution: c)2xtany2
+= ( ,
Particcular solution: )4π
2xtany2
+= (
Separable
as we can separate x’s from y’s and get
∫∫ = g(x)dxf(y)dy
This is the general solution Using the initial condition
This is the particular solution
Solution
Homogeneous
as it can take the form
)xy
F(dxdy
=
[this line is always the same!] Now it is a D.E. of separable variables [this is always the case!] General solution Using the initial condition Particular solution
Solve 2
22
xy13xy36x
dxdy ++
= , if y=1 when x=1.
Solve 1)x(ydxdy 2 += , if y=1 when x=0.
6
Solution
With Integrating Factor as it can take the form
Q(x)P(x)ydxdy
=+ [see booklet]
Spot P(x), Q(x)
[see booklet for I= ∫P(x)dxe ] Don’t put (+c) here You have to remember that
Iy= ∫ IQdx
Solution
ytanxcosxedxdy 3x −=
0.2xx n1n +=+
0.2y n1n +=+ y )tanxycosx(e nnn3xn −
Euler
as it says so! ☺
Always in the form y)F(x,dxdy
=
[see booklet for formulas]
The GDC does it all!
Recursion 0.2aa n1n +=+
0.2bb n1n +=+ nnn
3a tanabcosa(e n − )
Set 0a0 = , 1b0 =
Notice: the smaller step h, the better approximation!
Solve cosxtanxyedxdy
e 3x3x =+ −− , if y=1 when x=0.
Let cosxtanxyedxdy
e 3x3x =+ −− , if 1y = when 0x =
Use Euler with step h=0.2 to estimate y for 1x =
7
8. Isoclines – Slope Fields
Solution (a) The isoclines are the curves cyx =+ .
That is cxy +−=
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
x
y
(b) We sketch the slope on each point
(c) We draw a curve passing through (0,0)
The grid is usually given!
If y)f(x,dxdy
=
the isoclines are cy)f(x, =
← that is all lines of slope -1
Slope 0 means Slope 1 means Slope -1 means
For example, at (0,1) the slope is
1yxdxdy
=+= , that is
Notice that on each isocline the slopes are equal. E.g. on y=-x+1 all slopes are 1. Just look at the point (0,0) and “follow the stream”. Indeed, if you solve the d.e. (using integrating factor) you will find the particular solution y=ex-x-1 and the graph looks like that!
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
x
y
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
x
y
Consider yxdxdy
+= . In the following grid
(a) Sketch the isoclines (b) Draw the Slope Field (c) Sketch the partic. solution passing through (0,0)
8
9. Series
Solution
It diverges since 032
43n12nlim
n≠=
++
∞→
Divergence Test
[rarely used] if the limit is not 0, we stop there! The series diverges! If it is 0, we use another test.
Solution
We check the corresponding integral dxx(lnx)
1I2
2∫+∞
=
=∫ dxx(lnx)
12 =∫ dx
x(lnx)-2
c+−lnx1
Thus ∞→
=blimI
b
lnx1
2
−∞→
=blim
+−ln21
lnb1
ln21
=
The integral converges so the series converges.
Integral Test
2x(lnx)1
f(x) = must be positive,
continuous and decreasing Usually
• for p-series ∑∞
=1npn
1 → ∫∞
1px
dx
• if you see lnn • if you recognize a known
integral, e.g.∑∞
=1nnen → ∫
∞−
1
xdxxe
Solution
Since 22 n
143n
sinn≤
+ and ∑
∞
=1n2n
1 converges
our series converges as well.
Comparison Test
[rarely used! they usually mention it in the question] larger converges ⇒ smaller too smaller diverges ⇒ larger too
Solution
na looks like 2n n
1b = and ∑∞
=1n2n
1 converges
n
nn b
alim∞→ 1
n43n52nlim
2
3n +
+=
∞→0const
32
43n5n2nlim 3
23
n>==
+
+=
∞→
Our series converges as well.
Limit Comparison Test
[Popular!] We usually compare with a
p-series ∑∞
=1npn
1
1pifdiv1pifconv
≤
>
←any positive constant will do
[we know the result beforehand]
Solution
n
1nn a
alim +
∞→
( )( )
( )( )2n
21n
n !n2!2n
!22n!1)(n2lim
+
+=
+
∞→
( ) 2)(2n12n
1)n2lim2
n ++
+=
∞→
( 121
<=
Our series converges.
Ratio Test
[Popular!]
Use if you see n!, an or similar expressions e.g. 1×3×…………×(2n+1)