2e3×+3x)(mayaj/142s19wir9completed.pdfa model rocket has upward velocity v(t)=42t2 ft/s,t seconds...
TRANSCRIPT
Math 142 Week-in-Review # 9 (Substitution Method & Estimating Distance Traveled )
Brief Overview of Section 6.2:
• u-Substitution: We use u-subcaption when our integral contains the result (or close to it) of the chain rule.
• How to Choose “u”: The following are just a few cases:
– If you want to evaluateZ
messn ·mess0 dx, choose u = mess ) du = mess0dx
– If you want to evaluateZ
emess ·mess0 dx, choose u = mess ) du = mess0dx
– If the integral has an lnx, then choose u = lnx.
• Take the derivative of u usingdudx
, or if u = f (x), then du = f 0(x)dx.
• Solve for dx by dividing, meaning dx =du
f 0(x).
• Substitute to replace all x-terms with u-terms in the integral.
• Integrate with respect to u. (Not Done!)
• Replace u-terms back with x’s.
1. Evaluate the following integrals.
(a)Z x5
px6 +6
dx
--
-
- -
--
T - d÷=f4×)
--
- du=f4x)dx- -
=
⇐write
1×51×6+65"2d×,
u=meas=x6t6, du - 6x5dx- u
mess" 2
solve for dx ( divide ) :
dx -
- dy6×5
Replace x - terms with a 's :
W
New integral in terms of ie "
-I i"
. }÷ -
- f #' ⇒i'"
du -
- f 's . i' "du
-
- In to ¥91:Lions '
'Ic. 'zlx6+b-
Not done !
Math 142 WIR, c�Maya Johnson, Spring 2019
(b)Z(x3 +3) 3
q(6x4 +72x+2)5 dx
(c)Z e�2x
(e�2x �4)5 dx
(d)Z(2x4 �8x)e2x5�20x2�10 dx
2
ategrewI J(×3 + 3) ( 6×4+72×+2 )%dx
,Uz 6×4+72×+2
IfU
w
mess du : I 24×3+72 )dx§ L x3t3 )
13
du Solve for dx L divide ) :¥⇒=fx¥¥¥, is "du Idx=¥¥E5÷I
-
- Shau'
"du=zy.uI÷tCT.tw
-_÷4l6x4tRxtz5
" ""
Se- "
(
e-zx-4-jdx.u-e-44.du-ze-wd.sc.
- - -
emesis - 5
• messSolve for dx ( divide ) :
-
mess's
wins out !dx=du-2eSe- "
u-s.dz#..=SeX:fza.u-5du--fEu-5du=-tzu.I+c3botne
:
= 'gle-445I
za4=2×5-20×110 , du -
- ( 10×4-40×74" '
{Solve fordo (
divide)
dx -
-du
fl2xE8×)e" )Toi-_sE¥x )
s.de#sj-fzxYxis**edu=f'zeudu--tseutChE
=¥ek×Ezo×z,
Math 142 WIR, c�Maya Johnson, Spring 2019
(e)Z 5 4
p(lnx)x
dx
(f)Z 2e3x +3x
2e3x +4.5x2 �4dx
(g)Z ✓
4x+ x3(x4 +2)3
◆dx
3
✓ ,U -
-lux
,du -
- Ixdxcontains -
lux"
Solve for dx L divide ) :
dx -
- defy=xd154¥.
xrdu-f-suikdu-suIY.ie?=s.4zunx5sHtc--4llnx5HtCJ-
Not done
ite
,fewer
'
€§2e3×+3x)(
21×+4.5×2-45"dx,u=2t4. 5×24,
du-f6e3×-w t9x)dx
mess' '
solve for dx ( divide ) :
" ""
" " "'
a÷±x5Ske⇒;¥¥⇒.EDU/dx--:ExIx--iEeEsx5--ftz.tudu--LzlnlultC
jog =tzH2e"t4.5x34.
.
i %ff.
=f¥dxtSx3(x4t2Pdx
,U=x4t2 ,
du -
-4x3dx
. - ifSolve for dx ( divide ) :
^Do need substitution DX =dupont
need substitution 4×3
= 5¥
dxtfxfu.dz#=f4.zdxtf'qu3du4lnlxlttyuItC
⇐ ,=4hlx-ltfzlx4+zg
Math 142 WIR, c�Maya Johnson, Spring 2019
(h)Z 30e�6/x2
x3 dx
(i)Z
x(3x�1)2 dx
2. Find f (x) if f (0) = 10 and f 0(x) = 6x5(x6 �12)2.
4
EW'
30EI.tn?dwx.u---6'x-3du=l2x-3dxemos Solve for dx ( divide ) :
dx -
- rd÷=xYgu_.
-
1304¥. #duzifso.tzeudu-ftzeudu-IE.FI#..ifse-6xIc-- Ee"
et-
i U -
- 3×-1,
du =3 dx
y mess-
solve for dx ( divide )LDoes not cancel . dx = dy
Solve for x in terms of u :3
+9--3×1,
⇒ uh =3 × ⇒ Utf = ⇒ Utfwfluty
. uh . day --Sgtcufhudu-ffzy 't4 du
atianya.dz#=tisx-iYtti3x-BtIfz
-
Find C
Hx ) -
- Sf 'c⇒d× -16×51×6-125 dx , U -
-XI 12
,du -=6x5dx
⇒ dx -- due
6×5
fWx5uZd÷y=Su2du=u÷tC=l×
'
t C
ft ) --
10¥25 + C =p ⇒ -576 tc = to ⇒ C =
10-1576=586fLx)=tx÷%-
Math 142 WIR, c�Maya Johnson, Spring 2019
Brief Overview of Section 6.3:
• The area under the curve of a velocity function represents the total distance traveled.
• The area under the curve of a positive function f (x) on the interval [a,b] can be estimated by partitioning the area
into n rectangles of equal width. The width of each rectangle is Dx =b�a
nand the height of each rectangle can
be determined by using any point on each subinterval but we typically use the left endpoint (left-hand sum), rightendpoint (right-hand sum), or midpoint.
• The exact area under the curve of the positive function f (x) can be found by taking the limit as n ! • (i.e. increasingthe number of rectangles without bound) of the sum of the areas of the rectangles.
3. The table below gives the velocity in miles per hour (mi/h) of a vehicle for the first ten hours of the drive.
t(h) 0 1 2 3 4 5 6 7 8 9 10v(mi/h) 0 25 40 46 52 65 66.5 67 69.2 73.4 75
(a) Estimate the total distance traveled by the vehicle on the interval [2,10] using a left-hand sum with n = 4 rectangles.Is this an upper or lower estimate?
(b) Estimate the total distance traveled by the vehicle on the interval [1,7] using a right-hand sum with n = 3 rectangles.Is this an upper or lower estimate?
5
-
-
-
-
T → a → a → →
-- -
On [ 4103,
n -
- 4. width D×=b=l0 =LLeft - endptso.xo-z.xc-4.xz-6.xz-u.se ^
"
Ly zDx(v( 2) tV( 4) tvlb )tVl8 ) )4 6 8 to
=2(
401-52+66.5+69.2)
" " " "
221227.7 ) - -455.4mi LowerEstimateT
- - -
[ 1,7 ],
n -
- 3. width DX -_b = =3
Right - endpts :X, -31×2=51×3--7
↳" ÷iE÷"
= 211787=356 miles UpperEstimateJ
Math 142 WIR, c�Maya Johnson, Spring 2019
(c) Estimate the total distance traveled by the vehicle on the interval [1,9] using midpoints with n = 4 rectangles. Is thisan upper or lower estimate?
4. A model rocket has upward velocity v(t) =�42t2 ft/s, t seconds after launch.
(a) Estimate the total distance traveled by the rocket on the interval [1,11] using a left-hand sum with n = 5 rectangles. Isthis an upper or lower estimate?
(b) Estimate the total distance traveled by the rocket on the interval [4,12] using a right-hand sum with n = 4 rectangles.Is this an upper or lower estimate?
6
-
- -
on [ 1,93,
n -
- 4,
width Dx -
- b=9 = Z
Midpoints : Me -
-
Xotxi == Iz =3 ; Mz -
-
Xitzxz =3t=4;mz=xztz=5tIz=6; my = x3tz=7tz9 -- 8-
My = DX ( v ( 27 TV (4) tv 16 ) TV ( 8 ) )= 2140 t 52+66.5+69.2 ) / es abbey:&
-
- z luv . 7) - -455.4mi
caunotdete.myMenkar
C should be positive
-
- -
on Cl,113
, n -
-5 ;
the width DX -
- bene = =3 n
Left - end pts : Xo -
- I,
X ,=3
,X 2=5 ,
X 3=7 ,X 4=9
↳ -
- ox has . us , msn.ru , + u , , ! overestimates
= 2(42115+42135+421551-42171742195) a y , y ,
= 2 ( 6930 ) = 13,860ft
-- -
on [ 4,123,
. n -
- 4,
the width Dx -
- but = K -
- Zµ ,
Right - end pts : * , = 6, X 2=8 , X 3=10 ,
X 4=12
R -4 -
- Bx ( HH t r (8) tulio , + veryUpperIstimateIi.::c:::::i÷
" " '
a T T A
Math 142 WIR, c�Maya Johnson, Spring 2019
(c) Estimate the total distance traveled by the rocket on the interval [0,12] using midpoints with n = 6 rectangles. Is thisan upper or lower estimate?
5. Estimate the area bounded between the graph of f (x) =�20ex+2 and the x-axis on the interval [�2,4] by
(a) using n = 3 rectangles of equal width and left-endpoints. What is the area of the second rectangle?
(b) using n = 4 rectangles of equal width and right-endpoints. What is the area of the third rectangle?
7
- -
On [ 0,12 ] with n -
-
6, the width DX =b =
I =3
Midpt : Mi =Xot# = # =L;Mz=Xitzxz=2tz4-
- Z;Mz=XztzX3=4zt6=5;Mu=XstX4=6tzI-
- I;ms=X4t= =L
M6=Xsz -10+127=11
Mb -
- DX ( VCI )tV( 3) t VC 5) tv ( 7) tv ( 9) tvlll ) )
=2142451-42135+4.2151742173742195+421115) qggfgtm)" "
=
2112012) =2402 ^€¥I→
i. ; 's ; i i.
O 2 4 6 8 10 12
--- ( Round to two decimal places ?
on C- 2,43 with n =3,
width DX -- 4-(3--2)=3 -
2nd rectangleLeft - eudpts : Xo = - 2 , Xi
-
- O,
×z=2
↳ = DX ( ft - 2) tf Lo ) tf (2) )=ztoe*zzoeo+¥zo"
Area of 2nd
Rectangle:
=2( - 1259.74 ) = - 2519.49
. DX#031=2/1-20E) f- 48.86J
- - -
on f- 2,4 ] with n=4,
the width DX -_b =
4-4-22=6-4--1.5Right - end pts :X ,= -
2+1.5=2.5, Xz = - it 1.5=1j X3--1+1.5=2.5 , X 4--2.5+1.5=1
^
3rd rectangleRy -
- DX ( ft.5) tfc 1) tf ( 2. 5) tf 14 ) )
= 1.5/-205.5+2+-20 e' +2+-202.5+2+-20 .e4t2) ↳¥→- 2 - is I 2.5 4
= 1.5/-10360.26) = -15540.39J Area of 3rd Rectangle :
DX
If( 2. 5)
1=1.5/-202-5+21
-_270
Math 142 WIR, c�Maya Johnson, Spring 2019
(c) using n = 6 rectangles of equal width and midpoints. What is the area of the fourth rectangle?
8
- - -
on E- 2,4 ] with n-
-6
,width DX =b=4-{ =L
Midpoints : Mi -
- xotz=-2tz = -1.5; Mz=Xit ==
jMz=Xzt == .Ijmy=XztX4=.lt#=1.5jMs=X4tzX5--ZtzI=
M6=Xst=3tz4=Mb = DX ( ft - I. 5) tfc -
. 5) tf C. 5) tf ( I. 5) tf ( 2. 5) tf C 3.5 ))
= I f- 20 e- " 5+2+-20 I 5+2+-2065+7 - Zoe
" 5+7-20<25+2+-20 e3-5 t
2)
= I ( - 7722.75 ) = -7722.75J a
→ 4threctangle
II. s'
.'s '
.'s ' i' s
'a's
' 's s'
- 2 - I 0 I 2 3 4
Area of 4th Rectangle :
Dxffll. 5) 1=1/-20 e' . 5+4=662.310