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Edurev123
5. Infinite and Improper Integrals
5.1 Examine if the improper integral ?
?? ?? ?
?? ?????? (?? -?? ?? )
?? /?? exists.
Solution:
(2017 : 10 Marks)
Given integrand,
2?? (1-?? 2
)
2/3
is undefined at ?? =1.
Hence, we split the limit at ?? =1.
? ?
3
0
?
2?? (1-?? 2
)
2/3
???? =? ?
1
0
?
2?? (1-?? 2
)
2/3
???? +? ?
3
1
?
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
0
?? ?
?? 0
?
2?? (1-?? 2
)
2/3
???? + lim
?? ???
?? ?
?[
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
-
?[-3(-1-?? 2
)
1/3
]
0
?? + lim
?? ?1
+
?[-3(1-?? 2
)
1/3
]
?? 3
=[0-(-3)]+[-3(-2)-0]
=3+6=9
5.2 Find the maximum and minimum values of ?? ?? -?? ?? ?? +?? on the interval [?? ,?? ].
(2018 : 13 marks)
Solution:
?????? ?? (?? ) =?? 4
-5?? 2
+4 (given)
Now, ?? (?? ) =4?? 3
-10?? +4
?? '
(?? )=0 at ?? 1
=1.32,?? 2
=-1.75,0.43
Also, ?? '
(?? )=12?? 2
-10
In the interval [2, 3], ?? (?? ) is monotonous increasing function as ?? '
(?? )>0 for ?? >1.32.
Therefore, in the interval [2,3], minimum value occurs at ?? =2 and maximum value
occurs at ?? =3.
?? (2)=0= minimum value
?? (3)=40= maximum value
Thus, ?? (?? ) has minimum value of 0 and maximum value of 40 in the interval [2,3].
Edurev123
Page 2
Edurev123
5. Infinite and Improper Integrals
5.1 Examine if the improper integral ?
?? ?? ?
?? ?????? (?? -?? ?? )
?? /?? exists.
Solution:
(2017 : 10 Marks)
Given integrand,
2?? (1-?? 2
)
2/3
is undefined at ?? =1.
Hence, we split the limit at ?? =1.
? ?
3
0
?
2?? (1-?? 2
)
2/3
???? =? ?
1
0
?
2?? (1-?? 2
)
2/3
???? +? ?
3
1
?
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
0
?? ?
?? 0
?
2?? (1-?? 2
)
2/3
???? + lim
?? ???
?? ?
?[
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
-
?[-3(-1-?? 2
)
1/3
]
0
?? + lim
?? ?1
+
?[-3(1-?? 2
)
1/3
]
?? 3
=[0-(-3)]+[-3(-2)-0]
=3+6=9
5.2 Find the maximum and minimum values of ?? ?? -?? ?? ?? +?? on the interval [?? ,?? ].
(2018 : 13 marks)
Solution:
?????? ?? (?? ) =?? 4
-5?? 2
+4 (given)
Now, ?? (?? ) =4?? 3
-10?? +4
?? '
(?? )=0 at ?? 1
=1.32,?? 2
=-1.75,0.43
Also, ?? '
(?? )=12?? 2
-10
In the interval [2, 3], ?? (?? ) is monotonous increasing function as ?? '
(?? )>0 for ?? >1.32.
Therefore, in the interval [2,3], minimum value occurs at ?? =2 and maximum value
occurs at ?? =3.
?? (2)=0= minimum value
?? (3)=40= maximum value
Thus, ?? (?? ) has minimum value of 0 and maximum value of 40 in the interval [2,3].
Edurev123
6. Double and Triple Integrals
6.1 Let ?? be the region determined by the inequalities ?? >?? ,?? >?? ,?? <?? and ?? >
?? ?? +?? ?? . Compute ?
?? ??? ??????????????
Solution:
(2010 : 20 Marks)
The given region is ?? ,?? >0 and ?? <8.
?? >?? 2
+?? 2
? Integral, I=??
?? ?2?? ????????????
? ?? =? ?
8
?? =?? 2
+?? 2
?? ?
8
2
?2??????????????
=? ?
?
?
?[?? ]
?? 2
+?? 2
8
·2?????????? =? (8-?? 2
-?? 2
)·2??????????
=2? (8?? -?? 3
-?? 2
?? )????????
Let ?? =?? cos ?? ,?? =?? sin ??
?? varies from 0 to
?? 2
as ?? ,?? >0 in given region.
Page 3
Edurev123
5. Infinite and Improper Integrals
5.1 Examine if the improper integral ?
?? ?? ?
?? ?????? (?? -?? ?? )
?? /?? exists.
Solution:
(2017 : 10 Marks)
Given integrand,
2?? (1-?? 2
)
2/3
is undefined at ?? =1.
Hence, we split the limit at ?? =1.
? ?
3
0
?
2?? (1-?? 2
)
2/3
???? =? ?
1
0
?
2?? (1-?? 2
)
2/3
???? +? ?
3
1
?
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
0
?? ?
?? 0
?
2?? (1-?? 2
)
2/3
???? + lim
?? ???
?? ?
?[
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
-
?[-3(-1-?? 2
)
1/3
]
0
?? + lim
?? ?1
+
?[-3(1-?? 2
)
1/3
]
?? 3
=[0-(-3)]+[-3(-2)-0]
=3+6=9
5.2 Find the maximum and minimum values of ?? ?? -?? ?? ?? +?? on the interval [?? ,?? ].
(2018 : 13 marks)
Solution:
?????? ?? (?? ) =?? 4
-5?? 2
+4 (given)
Now, ?? (?? ) =4?? 3
-10?? +4
?? '
(?? )=0 at ?? 1
=1.32,?? 2
=-1.75,0.43
Also, ?? '
(?? )=12?? 2
-10
In the interval [2, 3], ?? (?? ) is monotonous increasing function as ?? '
(?? )>0 for ?? >1.32.
Therefore, in the interval [2,3], minimum value occurs at ?? =2 and maximum value
occurs at ?? =3.
?? (2)=0= minimum value
?? (3)=40= maximum value
Thus, ?? (?? ) has minimum value of 0 and maximum value of 40 in the interval [2,3].
Edurev123
6. Double and Triple Integrals
6.1 Let ?? be the region determined by the inequalities ?? >?? ,?? >?? ,?? <?? and ?? >
?? ?? +?? ?? . Compute ?
?? ??? ??????????????
Solution:
(2010 : 20 Marks)
The given region is ?? ,?? >0 and ?? <8.
?? >?? 2
+?? 2
? Integral, I=??
?? ?2?? ????????????
? ?? =? ?
8
?? =?? 2
+?? 2
?? ?
8
2
?2??????????????
=? ?
?
?
?[?? ]
?? 2
+?? 2
8
·2?????????? =? (8-?? 2
-?? 2
)·2??????????
=2? (8?? -?? 3
-?? 2
?? )????????
Let ?? =?? cos ?? ,?? =?? sin ??
?? varies from 0 to
?? 2
as ?? ,?? >0 in given region.
Also, ?? varies from 0 to 2v2 as ?? 2
+?? 2
<?? .
?? =2? (8·?? cos ?? -?? 3
cos
3
?? -?? 3
sin
2
?? ·cos ?? )?????????? =2? {8?? cos ?? -?? 3
cos ?? (cos
2
?? +sin
2
?? )}?????????? }
=2? ?
?? /2
?? =0
?? ?
2v2
?? =0
?(8?? 2
cos ?? -?? 4
cos ?? )???????? =2[8[
?? 3
3
]
0
2v2
[sin ?? ]
0
?? /2
-[
?? 5
5
]
0
2v2
[sin ?? ]
0
?? /2
]
=2[8×
16v2
3
×1-
128v2
5
×1]=
512v2
15
6.2 Evaluate ?
?? ????????? where ?? is the region bounded by the line ?? =?? -?? and the
parabola ?? ?? =?? ?? +?? .
(2013 : 15 Marks)
Solution:
The required areas is shown shaded in the diagram.
The points of intersection of the curves are
(?? -1)
2
=2?? +6??? 2
-4?? -5=0
?? =5,-1
i.e., (-1,-2) and (5,4)
Page 4
Edurev123
5. Infinite and Improper Integrals
5.1 Examine if the improper integral ?
?? ?? ?
?? ?????? (?? -?? ?? )
?? /?? exists.
Solution:
(2017 : 10 Marks)
Given integrand,
2?? (1-?? 2
)
2/3
is undefined at ?? =1.
Hence, we split the limit at ?? =1.
? ?
3
0
?
2?? (1-?? 2
)
2/3
???? =? ?
1
0
?
2?? (1-?? 2
)
2/3
???? +? ?
3
1
?
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
0
?? ?
?? 0
?
2?? (1-?? 2
)
2/3
???? + lim
?? ???
?? ?
?[
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
-
?[-3(-1-?? 2
)
1/3
]
0
?? + lim
?? ?1
+
?[-3(1-?? 2
)
1/3
]
?? 3
=[0-(-3)]+[-3(-2)-0]
=3+6=9
5.2 Find the maximum and minimum values of ?? ?? -?? ?? ?? +?? on the interval [?? ,?? ].
(2018 : 13 marks)
Solution:
?????? ?? (?? ) =?? 4
-5?? 2
+4 (given)
Now, ?? (?? ) =4?? 3
-10?? +4
?? '
(?? )=0 at ?? 1
=1.32,?? 2
=-1.75,0.43
Also, ?? '
(?? )=12?? 2
-10
In the interval [2, 3], ?? (?? ) is monotonous increasing function as ?? '
(?? )>0 for ?? >1.32.
Therefore, in the interval [2,3], minimum value occurs at ?? =2 and maximum value
occurs at ?? =3.
?? (2)=0= minimum value
?? (3)=40= maximum value
Thus, ?? (?? ) has minimum value of 0 and maximum value of 40 in the interval [2,3].
Edurev123
6. Double and Triple Integrals
6.1 Let ?? be the region determined by the inequalities ?? >?? ,?? >?? ,?? <?? and ?? >
?? ?? +?? ?? . Compute ?
?? ??? ??????????????
Solution:
(2010 : 20 Marks)
The given region is ?? ,?? >0 and ?? <8.
?? >?? 2
+?? 2
? Integral, I=??
?? ?2?? ????????????
? ?? =? ?
8
?? =?? 2
+?? 2
?? ?
8
2
?2??????????????
=? ?
?
?
?[?? ]
?? 2
+?? 2
8
·2?????????? =? (8-?? 2
-?? 2
)·2??????????
=2? (8?? -?? 3
-?? 2
?? )????????
Let ?? =?? cos ?? ,?? =?? sin ??
?? varies from 0 to
?? 2
as ?? ,?? >0 in given region.
Also, ?? varies from 0 to 2v2 as ?? 2
+?? 2
<?? .
?? =2? (8·?? cos ?? -?? 3
cos
3
?? -?? 3
sin
2
?? ·cos ?? )?????????? =2? {8?? cos ?? -?? 3
cos ?? (cos
2
?? +sin
2
?? )}?????????? }
=2? ?
?? /2
?? =0
?? ?
2v2
?? =0
?(8?? 2
cos ?? -?? 4
cos ?? )???????? =2[8[
?? 3
3
]
0
2v2
[sin ?? ]
0
?? /2
-[
?? 5
5
]
0
2v2
[sin ?? ]
0
?? /2
]
=2[8×
16v2
3
×1-
128v2
5
×1]=
512v2
15
6.2 Evaluate ?
?? ????????? where ?? is the region bounded by the line ?? =?? -?? and the
parabola ?? ?? =?? ?? +?? .
(2013 : 15 Marks)
Solution:
The required areas is shown shaded in the diagram.
The points of intersection of the curves are
(?? -1)
2
=2?? +6??? 2
-4?? -5=0
?? =5,-1
i.e., (-1,-2) and (5,4)
?? =??
?? ????????? =? ?
?? 1
????????? +? ?
?? 2
????????? =? ?
-1
?? =-3
?? ?
v2?? +6
-v2?? +6
????????????? +? ?
?? =5
?? =-1
?? ?
v2?? +6
?? =?? -1
????????????? =? ?
-1
?? =-3
?[
?? ?? 2
2
]
?? =-v2?? +6
v2?? +6
???? +? ?
5
?? =-1
??? [
?? 2
2
]
?? -1
v2?? +6
????
=? ?
-1
?? =-3
?
?? 2
[(2?? +6)-(2?? +6)]???? +? ?
5
?? =-1
?
?? 2
[(2?? +6)-(?? -1)
2
]????
=0+? ?
5
?? =-1
?
?? 2
(4?? +5-?? 2
)?? '
?? =? ?
5
?? =-1
?(2?? 2
+
5
2
?? -
?? 3
2
)????
=[
2?? 3
3
+
5
4
?? 2
-
?? 4
8
]
?? =-1
5
=36
6.3 By using the transformation ?? +?? =?? ,?? =???? , evaluate the integral ?{???? (?? -
?? -?? )}
?? /?? ???????? taken over the area enciosed by the straight lines ?? =?? ,?? =?? and
?? +?? =?? .
(2014 : 15 Marks)
Solution:
?????????? : ?? +?? =?? ,?? =????
? ?? =?? -????
?? =?? (1-?? )
??????
?(?? ,?? )
?(?? ,?? )
=|
??? ??? ??? ??? ??? ??? ??? ??? |+|
-?? -?? ?? ?? |
=?? -???? +????
=??
?????? ???????? =
?(?? ,?? )
?(?? ,?? )
???????? ???????? =??????????
vxy(1-?? -?? )=v?? (1-?? )???? (1-?? )=?? ?? 1/2
v(1-?? )(1-?? )
Clearly, the region of integration is OAB.
The integration formulae are ?? +?? :=?? ,?? =???? =(?? +?? )??
? ?? =
?? 1-?? ??
Page 5
Edurev123
5. Infinite and Improper Integrals
5.1 Examine if the improper integral ?
?? ?? ?
?? ?????? (?? -?? ?? )
?? /?? exists.
Solution:
(2017 : 10 Marks)
Given integrand,
2?? (1-?? 2
)
2/3
is undefined at ?? =1.
Hence, we split the limit at ?? =1.
? ?
3
0
?
2?? (1-?? 2
)
2/3
???? =? ?
1
0
?
2?? (1-?? 2
)
2/3
???? +? ?
3
1
?
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
0
?? ?
?? 0
?
2?? (1-?? 2
)
2/3
???? + lim
?? ???
?? ?
?[
2?? (1-?? 2
)
2/3
????
= lim
?? ?1
-
?[-3(-1-?? 2
)
1/3
]
0
?? + lim
?? ?1
+
?[-3(1-?? 2
)
1/3
]
?? 3
=[0-(-3)]+[-3(-2)-0]
=3+6=9
5.2 Find the maximum and minimum values of ?? ?? -?? ?? ?? +?? on the interval [?? ,?? ].
(2018 : 13 marks)
Solution:
?????? ?? (?? ) =?? 4
-5?? 2
+4 (given)
Now, ?? (?? ) =4?? 3
-10?? +4
?? '
(?? )=0 at ?? 1
=1.32,?? 2
=-1.75,0.43
Also, ?? '
(?? )=12?? 2
-10
In the interval [2, 3], ?? (?? ) is monotonous increasing function as ?? '
(?? )>0 for ?? >1.32.
Therefore, in the interval [2,3], minimum value occurs at ?? =2 and maximum value
occurs at ?? =3.
?? (2)=0= minimum value
?? (3)=40= maximum value
Thus, ?? (?? ) has minimum value of 0 and maximum value of 40 in the interval [2,3].
Edurev123
6. Double and Triple Integrals
6.1 Let ?? be the region determined by the inequalities ?? >?? ,?? >?? ,?? <?? and ?? >
?? ?? +?? ?? . Compute ?
?? ??? ??????????????
Solution:
(2010 : 20 Marks)
The given region is ?? ,?? >0 and ?? <8.
?? >?? 2
+?? 2
? Integral, I=??
?? ?2?? ????????????
? ?? =? ?
8
?? =?? 2
+?? 2
?? ?
8
2
?2??????????????
=? ?
?
?
?[?? ]
?? 2
+?? 2
8
·2?????????? =? (8-?? 2
-?? 2
)·2??????????
=2? (8?? -?? 3
-?? 2
?? )????????
Let ?? =?? cos ?? ,?? =?? sin ??
?? varies from 0 to
?? 2
as ?? ,?? >0 in given region.
Also, ?? varies from 0 to 2v2 as ?? 2
+?? 2
<?? .
?? =2? (8·?? cos ?? -?? 3
cos
3
?? -?? 3
sin
2
?? ·cos ?? )?????????? =2? {8?? cos ?? -?? 3
cos ?? (cos
2
?? +sin
2
?? )}?????????? }
=2? ?
?? /2
?? =0
?? ?
2v2
?? =0
?(8?? 2
cos ?? -?? 4
cos ?? )???????? =2[8[
?? 3
3
]
0
2v2
[sin ?? ]
0
?? /2
-[
?? 5
5
]
0
2v2
[sin ?? ]
0
?? /2
]
=2[8×
16v2
3
×1-
128v2
5
×1]=
512v2
15
6.2 Evaluate ?
?? ????????? where ?? is the region bounded by the line ?? =?? -?? and the
parabola ?? ?? =?? ?? +?? .
(2013 : 15 Marks)
Solution:
The required areas is shown shaded in the diagram.
The points of intersection of the curves are
(?? -1)
2
=2?? +6??? 2
-4?? -5=0
?? =5,-1
i.e., (-1,-2) and (5,4)
?? =??
?? ????????? =? ?
?? 1
????????? +? ?
?? 2
????????? =? ?
-1
?? =-3
?? ?
v2?? +6
-v2?? +6
????????????? +? ?
?? =5
?? =-1
?? ?
v2?? +6
?? =?? -1
????????????? =? ?
-1
?? =-3
?[
?? ?? 2
2
]
?? =-v2?? +6
v2?? +6
???? +? ?
5
?? =-1
??? [
?? 2
2
]
?? -1
v2?? +6
????
=? ?
-1
?? =-3
?
?? 2
[(2?? +6)-(2?? +6)]???? +? ?
5
?? =-1
?
?? 2
[(2?? +6)-(?? -1)
2
]????
=0+? ?
5
?? =-1
?
?? 2
(4?? +5-?? 2
)?? '
?? =? ?
5
?? =-1
?(2?? 2
+
5
2
?? -
?? 3
2
)????
=[
2?? 3
3
+
5
4
?? 2
-
?? 4
8
]
?? =-1
5
=36
6.3 By using the transformation ?? +?? =?? ,?? =???? , evaluate the integral ?{???? (?? -
?? -?? )}
?? /?? ???????? taken over the area enciosed by the straight lines ?? =?? ,?? =?? and
?? +?? =?? .
(2014 : 15 Marks)
Solution:
?????????? : ?? +?? =?? ,?? =????
? ?? =?? -????
?? =?? (1-?? )
??????
?(?? ,?? )
?(?? ,?? )
=|
??? ??? ??? ??? ??? ??? ??? ??? |+|
-?? -?? ?? ?? |
=?? -???? +????
=??
?????? ???????? =
?(?? ,?? )
?(?? ,?? )
???????? ???????? =??????????
vxy(1-?? -?? )=v?? (1-?? )???? (1-?? )=?? ?? 1/2
v(1-?? )(1-?? )
Clearly, the region of integration is OAB.
The integration formulae are ?? +?? :=?? ,?? =???? =(?? +?? )??
? ?? =
?? 1-?? ??
i.e., clearly the area for new variable is to be divided by the lines parallel to ?? +?? =1
and by lines ?? =
?? ?? -?? ?? :
i.e.,
?? =?? tan ??
where
tan ?? =
?? 1-??
where ?? varies from 0 to
?? 2
and so ?? varies from 0 to 1 and ?? +?? +?? varies from 0 to 1
i.e., limits of ?? are 0 to 1 .
Hence, the given integral
=? ?
1
0
? ?
1
0
?? ?? 1/2
v(1-?? )(1-?? )??????????
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Infinite and Improper Integrals UPSC Questions
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