Post on 25-Aug-2020
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Final Review - Chapter 5(Antiderivatives and Applications of Anti-Differentiation)
Example 1: Find the most general antiderivative of the function.
a) g(x) =1
x+
1
x2 + 1b) f(x) =
x2 +
px
x
Example 2: Given f00(x) = 5x3 + 6x2 + 2, f(0) = 3, f(1) = �2, find f(x).
Example 3: A particle is moving with v(t) = 2t � 1/(1 + t2) and s(0) = 1. Fin the position of the
particle.
1
" k= Xtx
G(x)= Inlxltarctanxtc FCx)=£x2+ 2×42+0
fkxl '=f(5×3+6×72)dx=I,×4 +2×3+2×+0
f G) = ff 'Cx)dx=f(§,xll+2×3+2×+44=14×5+1×4 + x2+Cx + D
3 = flo ) =D . So D =3 . and :
- 2=fa)=÷ ,+ £+1 + C +3 fC×)=t4x5ttzx4tx2 - 2¥×+z
C= - 6314=-27140
vCH=2t - ¥2slttfvcadttffnt-,t⇒dt=t2 - arctant to
1=54--02 - avctano to =c .
So CH
So answer : SCtf=t2 . arotant +1
Example 4: Estimate the area under the curve y = x2 + 2 on the interval [0, 8] using 4 sub-intervals
and the method given below.
a) left endpoints. b) midpoints.
Example 5: Evaluate the following definite integrals.
a)Z ⇡/4
0
sec2 t
tan t+ 1dt b)
Z 4
1
x� 2px
dx
2
30
+62%
40
¥¥¥¥t+¥l÷¥EtmIYIYIItIg→⇒
L4=2(flo)tH2)tf(4) tf( 6 )) =2( 94=184
=2(
2-+6+18+39)=Z(
641=128
=ln( tutti )/%=f
,
44"2-2×42)d×
o= 5×312-4×42 ] ?
=ln(tan¥+Dtn(tano+D
=lh2 - lnklnz
-- (2314512-4442)-(3-4)
.
= 113-8-23+4=143- y4" G
- ÷
Example 6: Find the most general anti-derivaatives.
a)Z ✓
secx tanx+2p
1� x2
◆dx b)
Zx
(x� 2)3dx
Example 7: Find the most general anti-derivatives.
a)Z
sin(1/x)
x2dx b)
Zcos�1
xp1� x2
dx
Example 8: Find the derivative of the following functions.
a) F (x) =
Z x3
2
p1 + t4dt b) H(x) =
Z x2
exsec tdt
3
= f(u+2)ui3du=fu2+2u3da
= secx + Zarcsmxtc let U=X -2 = . ui'
-uttc
du=dx
×=ut2=
- ( × -251 - ( x-252 + c
= .
fsinudu= |udu=lzu2+c
letn=¥=x'
= cosutcletwarccosx 2
du= 1-=tz@rccosx )
due - Ehdx = cos (E) + C asd×
. du=
j2dx-dxx2Ftx7-1i@y4M.z
,
={Isectdttf
.SI?tdt=3x2,+#= - {¥ectdt+f×2seaµ.
Now HtA= - ( seed )°e×+(secx2)2×=
- e×sec( d) + Zxsecx'
.
Example 9: A particle moves along a line with velocity function v(t) = t2 � 2t, where v is measured
in meters per second.
(a) Find the displacement over the time interval [0, 3]
(b) Find the total distance traveled during the time interval [0, 3]
Example 10: A bacteria population is 4000 at time t = 0 and its rate of growth is 1000 ⇥ 2t bacteriaper hour after t hours. What is the population after one hour?
4
Estado
a.
displacement = fobCost dt = Sint ] ! = since - Sino = sin 6.
a.
Ilitsatanu = foblcostldt = Must at -
f3y%cost# + feast #
3% 3% 6tie,
= snttotk - (snttq+ ( smttg
,62
= @net . and-fin (E) - smEH]+[sm6 - sin #= 1
- [ - 1 - 1 ] + sin 6-+1
= 1 + 2 + sin 6+1
= 4 + Since
Let PLH be the population .at timet .
Thus P(•)=4000 and Ptt ) = 1000 . 2£.
1
So PCD = 4000 + follow . Edt = 4000+1000 ;÷z . 240
= 4000 + Ynozt ( 2'
- i ) =4ooo +1000
NTTnz
bacteria
2-1=1