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the Fundamental Theorem of Calculus PDF

Added on - 15 Nov 2021

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Solution Task 1:
(a)
Consider12lim,1Rt
tLteR

Since1R, let’s consider a particular case
3
2R, then



12
3122
1
22
2
lim
lim
limlim
1limlim
Rt
t
t
t
t
tt
t
tt
Lte
te
te
et











Simplify further,
12lim
1
0
Rt
t
Lte
e







Hence12limRt
tte

(b)
Consider the limit23limRtt
tLttee

Since0e
So,



23
2
2
lim
0
0
Rtt
t
Rt
Rt
Lttee
ttee
tte







Hence23lim0Rtt
tttee

Solution Task 2:
(a)
Consider the definite integral
0
1
F
aFxsxKeedxwhere,saandFare positive
constants.
Now,




0
0
0
0
1
1
1
F
aFxsx
F
sxaFax
F
sxaFax
F
saxaFsx
Keedx
eeedx
eeedx
eeedx












Simplify further,





00
00
1
111
111
FF
saxaFsx
aFFFsaxsx
aFsaFsF
sFaFsF
Keedxedx
eeesas
eeesas
eeesas









Simplify further,
sFseK

aFsFsese



11
sF
sFaF
aesa
ssa
aese
ssa





Hence


11sFaFaeseKssa

(b)
To show11sFaFssaKaese
Last result from part (a)



11
11
sFaF
sFaF
aeseKssa
ssaKaese




This completes the proof.
Solution Task 3:
Consider the system of equation,
1233
21 14
0112
11 0
x
y
za

(a)
The augmented matrix is,
1233
21 14
0112
11 0a

Now let’s perform elementary row operations.
Perform2213312,RRRRRR
1233
05510
0112
0333a


2
25
RR
1233
0112
0112
0333a


332442,3RRRRRR
12 33
0 1 12
0 0 00
0 0 03a


(a)
Since system has four equations and three unknown that is system has either infinitely
many solutions or no solution.
Now,
If30athat is for3asystem has infinitely many solutions and if3a0that is
ifa3then the system is inconsistent
(b)
If3athen system has infinitely many solutions that is
233
2
xyz
yz


Supposeztthen
1,2andxtytztWheretR
Solution Task 4:
Given the matrices,
0 04341
3 0 0 and121
0 123
tsAB
ts

(a)
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