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