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Page 1
Edurev123
3. Homan's Definition of Definite Integrals
3.1 Show that the function
?? (?? )=[?? ?? ]+|?? -?? |
is a Riemann-integrable in the interval [?? ,?? ], where [?? ] denotes the greatest
integer less than or equal to ?? . Can you give an example of a function that is not
Riemann integrable on [?? ,?? ] ? Compute ?
?? ?? ??? (?? )???? , where ?? (?? ) is as above.
(2010: 12 Marks)
Solution:
Given:
?? (?? )=
{
1-?? if 0=?? <1
?? if 1=?? <v2
?? +1 v2=?? <v3
?? +2 v3=?? <2
Now, ?? (?? ) is discontinuous at x=1,v2,v3, i.e., finite number of discontinuity.
??? (?? ) is Riemann integrable in [0,2].
Now,
? ?
2
0
??? (?? )???? =? ?
1
0
?(1-?? )???? +? ?
v2
1
??????? +? ?
v3
v2
?(?? +1)???? +? ?
2
v3
?(?? +2)????
=[?? ]
0
1
-[
?? 2
2
]
0
1
+[
?? 2
2
]
1
v2
+[
?? 2
2
]
v2
v3
+[?? ]
v2
v3
+[
?? 2
2
]
v3
2
+[2?? ]
v3
2
=1-
1
2
+
2
2
-
1
2
+
3
2
-
2
2
+v3-v2+
4
2
-
3
2
+2(2-v3)
=6-v2-v3
3.2 Evaluate: ?
?? ?? ????? ??????
(2011: 12 Marks)
Solution:
Page 2
Edurev123
3. Homan's Definition of Definite Integrals
3.1 Show that the function
?? (?? )=[?? ?? ]+|?? -?? |
is a Riemann-integrable in the interval [?? ,?? ], where [?? ] denotes the greatest
integer less than or equal to ?? . Can you give an example of a function that is not
Riemann integrable on [?? ,?? ] ? Compute ?
?? ?? ??? (?? )???? , where ?? (?? ) is as above.
(2010: 12 Marks)
Solution:
Given:
?? (?? )=
{
1-?? if 0=?? <1
?? if 1=?? <v2
?? +1 v2=?? <v3
?? +2 v3=?? <2
Now, ?? (?? ) is discontinuous at x=1,v2,v3, i.e., finite number of discontinuity.
??? (?? ) is Riemann integrable in [0,2].
Now,
? ?
2
0
??? (?? )???? =? ?
1
0
?(1-?? )???? +? ?
v2
1
??????? +? ?
v3
v2
?(?? +1)???? +? ?
2
v3
?(?? +2)????
=[?? ]
0
1
-[
?? 2
2
]
0
1
+[
?? 2
2
]
1
v2
+[
?? 2
2
]
v2
v3
+[?? ]
v2
v3
+[
?? 2
2
]
v3
2
+[2?? ]
v3
2
=1-
1
2
+
2
2
-
1
2
+
3
2
-
2
2
+v3-v2+
4
2
-
3
2
+2(2-v3)
=6-v2-v3
3.2 Evaluate: ?
?? ?? ????? ??????
(2011: 12 Marks)
Solution:
We know that ? ???????? =?? ? ?????? -? (
????
????
·? ???? ?? )????
? ? ln (?? )???? =ln ?? ? 1???? -? ((
?? ????
ln (?? ))? ?????? )????
=ln ?? ·?? -?
1
?? ·?????? =?? -ln ?? -? 1????
=?? in ?? -?? +?? , where ?? is the constant of integration
3.3 Evaluate: ?
?? ?? ?(?? ?? ??????
?? ?? -??????
?? ?? )????
(2013 : 10 Marks)
Solution:
Given integral ?
0
1
?(2?? sin
1
?? -cos
1
?? )????
Let
1
?? =?? ??? =
1
?? ????? =-
1
?? 2
????
when ?? =0,?? =8;?? =1,?? =1.
? ?? =? ?
1
8
?(
2
?? sin ?? -cos ?? )(-
1
?? 2
???? )
=? ?
1
8
?(
2
?? 3
sin ?? -
1
???
2
cos ?? )????
Using integration by parts on 2nd term
=? ?
8
1
?
2
?? 3
sin ?? -{[
1
?? 2
sin ?? ]
1
8
-? ?
8
1
?-
2
?? 3
sin ?? }
=-[
1
?? 2
sin ?? ]
1
8
=sin 1
3.4 Evaluate: ?
?? ?? ?
???? ?? ?? (?? +?? )
?? +?? ?? ????
(2014: 10 marks)
Solution:
Let
Page 3
Edurev123
3. Homan's Definition of Definite Integrals
3.1 Show that the function
?? (?? )=[?? ?? ]+|?? -?? |
is a Riemann-integrable in the interval [?? ,?? ], where [?? ] denotes the greatest
integer less than or equal to ?? . Can you give an example of a function that is not
Riemann integrable on [?? ,?? ] ? Compute ?
?? ?? ??? (?? )???? , where ?? (?? ) is as above.
(2010: 12 Marks)
Solution:
Given:
?? (?? )=
{
1-?? if 0=?? <1
?? if 1=?? <v2
?? +1 v2=?? <v3
?? +2 v3=?? <2
Now, ?? (?? ) is discontinuous at x=1,v2,v3, i.e., finite number of discontinuity.
??? (?? ) is Riemann integrable in [0,2].
Now,
? ?
2
0
??? (?? )???? =? ?
1
0
?(1-?? )???? +? ?
v2
1
??????? +? ?
v3
v2
?(?? +1)???? +? ?
2
v3
?(?? +2)????
=[?? ]
0
1
-[
?? 2
2
]
0
1
+[
?? 2
2
]
1
v2
+[
?? 2
2
]
v2
v3
+[?? ]
v2
v3
+[
?? 2
2
]
v3
2
+[2?? ]
v3
2
=1-
1
2
+
2
2
-
1
2
+
3
2
-
2
2
+v3-v2+
4
2
-
3
2
+2(2-v3)
=6-v2-v3
3.2 Evaluate: ?
?? ?? ????? ??????
(2011: 12 Marks)
Solution:
We know that ? ???????? =?? ? ?????? -? (
????
????
·? ???? ?? )????
? ? ln (?? )???? =ln ?? ? 1???? -? ((
?? ????
ln (?? ))? ?????? )????
=ln ?? ·?? -?
1
?? ·?????? =?? -ln ?? -? 1????
=?? in ?? -?? +?? , where ?? is the constant of integration
3.3 Evaluate: ?
?? ?? ?(?? ?? ??????
?? ?? -??????
?? ?? )????
(2013 : 10 Marks)
Solution:
Given integral ?
0
1
?(2?? sin
1
?? -cos
1
?? )????
Let
1
?? =?? ??? =
1
?? ????? =-
1
?? 2
????
when ?? =0,?? =8;?? =1,?? =1.
? ?? =? ?
1
8
?(
2
?? sin ?? -cos ?? )(-
1
?? 2
???? )
=? ?
1
8
?(
2
?? 3
sin ?? -
1
???
2
cos ?? )????
Using integration by parts on 2nd term
=? ?
8
1
?
2
?? 3
sin ?? -{[
1
?? 2
sin ?? ]
1
8
-? ?
8
1
?-
2
?? 3
sin ?? }
=-[
1
?? 2
sin ?? ]
1
8
=sin 1
3.4 Evaluate: ?
?? ?? ?
???? ?? ?? (?? +?? )
?? +?? ?? ????
(2014: 10 marks)
Solution:
Let
?? =? ?
1
0
log (1+?? )
1+?? 2
????
Put ?? =tan ?? ? ???? =sec
2
??????
when ?? =0,?? =0 and when ?? =1,?? =
?? ?? 1
?? =? ?
?? /4
0
?
log (1+tan ?? )
1+tan
2
?? ·sec
2
?????? =? ?
?? /4
0
?log(1+tan?? )???? …(??)
=? ?
?? /4
0
?log [1+tan (
?? 4
-?? )]( by the property ? ?
?? 0
??? '
?? )???? =? ?
?? 0
??? (?? -?? ))
=? ?
?? /4
0
?log (1+
1-tan ?? 1+tan ?? )????
=? ?
?? /4
0
?log (
2
1+tan ?? )????
=? ?
?? /4
0
?{log 2-log (1+tan ?? }????
=log 2? ?
?? /4
0
????? -? ?
?? /4
0
?log (1+tan ?? )????
?? =log 2·[?? ]
0
?? 4
-?? ???????? (??)
? 2?? =
?? 4
log 2
? ?? =
?? 8
log 2
? ? ?
?? /4
0
?
log (1+?? )
1+?? 2
???? =
?? 8
log 2
3.5 Evaluate the following integral: ?
?? /?? ?? /?? ?
v?????? ?? ?? v?????? ?? ?? +v?????? ?? ?? ????
(2015: 10 Marks)
Solution:
Page 4
Edurev123
3. Homan's Definition of Definite Integrals
3.1 Show that the function
?? (?? )=[?? ?? ]+|?? -?? |
is a Riemann-integrable in the interval [?? ,?? ], where [?? ] denotes the greatest
integer less than or equal to ?? . Can you give an example of a function that is not
Riemann integrable on [?? ,?? ] ? Compute ?
?? ?? ??? (?? )???? , where ?? (?? ) is as above.
(2010: 12 Marks)
Solution:
Given:
?? (?? )=
{
1-?? if 0=?? <1
?? if 1=?? <v2
?? +1 v2=?? <v3
?? +2 v3=?? <2
Now, ?? (?? ) is discontinuous at x=1,v2,v3, i.e., finite number of discontinuity.
??? (?? ) is Riemann integrable in [0,2].
Now,
? ?
2
0
??? (?? )???? =? ?
1
0
?(1-?? )???? +? ?
v2
1
??????? +? ?
v3
v2
?(?? +1)???? +? ?
2
v3
?(?? +2)????
=[?? ]
0
1
-[
?? 2
2
]
0
1
+[
?? 2
2
]
1
v2
+[
?? 2
2
]
v2
v3
+[?? ]
v2
v3
+[
?? 2
2
]
v3
2
+[2?? ]
v3
2
=1-
1
2
+
2
2
-
1
2
+
3
2
-
2
2
+v3-v2+
4
2
-
3
2
+2(2-v3)
=6-v2-v3
3.2 Evaluate: ?
?? ?? ????? ??????
(2011: 12 Marks)
Solution:
We know that ? ???????? =?? ? ?????? -? (
????
????
·? ???? ?? )????
? ? ln (?? )???? =ln ?? ? 1???? -? ((
?? ????
ln (?? ))? ?????? )????
=ln ?? ·?? -?
1
?? ·?????? =?? -ln ?? -? 1????
=?? in ?? -?? +?? , where ?? is the constant of integration
3.3 Evaluate: ?
?? ?? ?(?? ?? ??????
?? ?? -??????
?? ?? )????
(2013 : 10 Marks)
Solution:
Given integral ?
0
1
?(2?? sin
1
?? -cos
1
?? )????
Let
1
?? =?? ??? =
1
?? ????? =-
1
?? 2
????
when ?? =0,?? =8;?? =1,?? =1.
? ?? =? ?
1
8
?(
2
?? sin ?? -cos ?? )(-
1
?? 2
???? )
=? ?
1
8
?(
2
?? 3
sin ?? -
1
???
2
cos ?? )????
Using integration by parts on 2nd term
=? ?
8
1
?
2
?? 3
sin ?? -{[
1
?? 2
sin ?? ]
1
8
-? ?
8
1
?-
2
?? 3
sin ?? }
=-[
1
?? 2
sin ?? ]
1
8
=sin 1
3.4 Evaluate: ?
?? ?? ?
???? ?? ?? (?? +?? )
?? +?? ?? ????
(2014: 10 marks)
Solution:
Let
?? =? ?
1
0
log (1+?? )
1+?? 2
????
Put ?? =tan ?? ? ???? =sec
2
??????
when ?? =0,?? =0 and when ?? =1,?? =
?? ?? 1
?? =? ?
?? /4
0
?
log (1+tan ?? )
1+tan
2
?? ·sec
2
?????? =? ?
?? /4
0
?log(1+tan?? )???? …(??)
=? ?
?? /4
0
?log [1+tan (
?? 4
-?? )]( by the property ? ?
?? 0
??? '
?? )???? =? ?
?? 0
??? (?? -?? ))
=? ?
?? /4
0
?log (1+
1-tan ?? 1+tan ?? )????
=? ?
?? /4
0
?log (
2
1+tan ?? )????
=? ?
?? /4
0
?{log 2-log (1+tan ?? }????
=log 2? ?
?? /4
0
????? -? ?
?? /4
0
?log (1+tan ?? )????
?? =log 2·[?? ]
0
?? 4
-?? ???????? (??)
? 2?? =
?? 4
log 2
? ?? =
?? 8
log 2
? ? ?
?? /4
0
?
log (1+?? )
1+?? 2
???? =
?? 8
log 2
3.5 Evaluate the following integral: ?
?? /?? ?? /?? ?
v?????? ?? ?? v?????? ?? ?? +v?????? ?? ?? ????
(2015: 10 Marks)
Solution:
?? =? ?
?? /3
?? /6
?
(sin ?? )
1/3
(sin ?? )
1/3
+(cos ?? )
1/3
?? ?? (??)
=? ?
?? /3
?? /6
?
[sin (
?? 6
+
?? 3
-?? )]
1/3
????
sin (
?? 6
+
?? 3
-?? )]
1/3
+[cos (
?? 6
+
?? 3
-?? )]
1/3
?? =? ?
?? /3
?? /6
?
(cos ?? )
1/3
(cos ?? )
1/3
+(sin ?? )
1/3
???? (???? )
=[? ?
?? ?? ??? (?? )???? =? ?
?? ?? ??? (?? +?? -?? )???? ]
Adding (i) and (ii)
2?? =? ?
?? /3
?? /6
?
(sin ?? )
1/3
+(cos)
1/3
(sin ?? )
1/3
+(cos ?? )
1/3
???? =? ?
?? /3
?? /6
????? =[?? ]
?? /6
?? /3
=[
?? 3
-
?? 6
]
? ?? =
1
2
×
?? 6
=
?? 12
3.6 Evaluate: ?? =?
?? ?? ??? v?? ?????? (
?? ?? )???? .
(2016 : 10 Marks)
Solution:
Improper integral as integrand becomes undefined at lower limit, i.e., ?? =0.
Let log
1
?? =?? ? ?? =?? -?? ????? =-?? -?? ????
???????????? , ?? =0
+
??? ?8
x=1??? =0
?? =-? ?
0
8
?(?? -?? ?? )
1/3
·?? -?? ???? =? ?
8
0
??? 1/3
·?? -
4
3
?? ????
Putting
4
3
?? =?? ? ???? =
3
4
????
?? =? ?
8
0
?(
3
4
?? )
1/3
?? -?? ·
3
4
???? =
3
4
·(
3
4
)
1/3
? ?
8
0
??? 1/3
·?? -?? ????
Page 5
Edurev123
3. Homan's Definition of Definite Integrals
3.1 Show that the function
?? (?? )=[?? ?? ]+|?? -?? |
is a Riemann-integrable in the interval [?? ,?? ], where [?? ] denotes the greatest
integer less than or equal to ?? . Can you give an example of a function that is not
Riemann integrable on [?? ,?? ] ? Compute ?
?? ?? ??? (?? )???? , where ?? (?? ) is as above.
(2010: 12 Marks)
Solution:
Given:
?? (?? )=
{
1-?? if 0=?? <1
?? if 1=?? <v2
?? +1 v2=?? <v3
?? +2 v3=?? <2
Now, ?? (?? ) is discontinuous at x=1,v2,v3, i.e., finite number of discontinuity.
??? (?? ) is Riemann integrable in [0,2].
Now,
? ?
2
0
??? (?? )???? =? ?
1
0
?(1-?? )???? +? ?
v2
1
??????? +? ?
v3
v2
?(?? +1)???? +? ?
2
v3
?(?? +2)????
=[?? ]
0
1
-[
?? 2
2
]
0
1
+[
?? 2
2
]
1
v2
+[
?? 2
2
]
v2
v3
+[?? ]
v2
v3
+[
?? 2
2
]
v3
2
+[2?? ]
v3
2
=1-
1
2
+
2
2
-
1
2
+
3
2
-
2
2
+v3-v2+
4
2
-
3
2
+2(2-v3)
=6-v2-v3
3.2 Evaluate: ?
?? ?? ????? ??????
(2011: 12 Marks)
Solution:
We know that ? ???????? =?? ? ?????? -? (
????
????
·? ???? ?? )????
? ? ln (?? )???? =ln ?? ? 1???? -? ((
?? ????
ln (?? ))? ?????? )????
=ln ?? ·?? -?
1
?? ·?????? =?? -ln ?? -? 1????
=?? in ?? -?? +?? , where ?? is the constant of integration
3.3 Evaluate: ?
?? ?? ?(?? ?? ??????
?? ?? -??????
?? ?? )????
(2013 : 10 Marks)
Solution:
Given integral ?
0
1
?(2?? sin
1
?? -cos
1
?? )????
Let
1
?? =?? ??? =
1
?? ????? =-
1
?? 2
????
when ?? =0,?? =8;?? =1,?? =1.
? ?? =? ?
1
8
?(
2
?? sin ?? -cos ?? )(-
1
?? 2
???? )
=? ?
1
8
?(
2
?? 3
sin ?? -
1
???
2
cos ?? )????
Using integration by parts on 2nd term
=? ?
8
1
?
2
?? 3
sin ?? -{[
1
?? 2
sin ?? ]
1
8
-? ?
8
1
?-
2
?? 3
sin ?? }
=-[
1
?? 2
sin ?? ]
1
8
=sin 1
3.4 Evaluate: ?
?? ?? ?
???? ?? ?? (?? +?? )
?? +?? ?? ????
(2014: 10 marks)
Solution:
Let
?? =? ?
1
0
log (1+?? )
1+?? 2
????
Put ?? =tan ?? ? ???? =sec
2
??????
when ?? =0,?? =0 and when ?? =1,?? =
?? ?? 1
?? =? ?
?? /4
0
?
log (1+tan ?? )
1+tan
2
?? ·sec
2
?????? =? ?
?? /4
0
?log(1+tan?? )???? …(??)
=? ?
?? /4
0
?log [1+tan (
?? 4
-?? )]( by the property ? ?
?? 0
??? '
?? )???? =? ?
?? 0
??? (?? -?? ))
=? ?
?? /4
0
?log (1+
1-tan ?? 1+tan ?? )????
=? ?
?? /4
0
?log (
2
1+tan ?? )????
=? ?
?? /4
0
?{log 2-log (1+tan ?? }????
=log 2? ?
?? /4
0
????? -? ?
?? /4
0
?log (1+tan ?? )????
?? =log 2·[?? ]
0
?? 4
-?? ???????? (??)
? 2?? =
?? 4
log 2
? ?? =
?? 8
log 2
? ? ?
?? /4
0
?
log (1+?? )
1+?? 2
???? =
?? 8
log 2
3.5 Evaluate the following integral: ?
?? /?? ?? /?? ?
v?????? ?? ?? v?????? ?? ?? +v?????? ?? ?? ????
(2015: 10 Marks)
Solution:
?? =? ?
?? /3
?? /6
?
(sin ?? )
1/3
(sin ?? )
1/3
+(cos ?? )
1/3
?? ?? (??)
=? ?
?? /3
?? /6
?
[sin (
?? 6
+
?? 3
-?? )]
1/3
????
sin (
?? 6
+
?? 3
-?? )]
1/3
+[cos (
?? 6
+
?? 3
-?? )]
1/3
?? =? ?
?? /3
?? /6
?
(cos ?? )
1/3
(cos ?? )
1/3
+(sin ?? )
1/3
???? (???? )
=[? ?
?? ?? ??? (?? )???? =? ?
?? ?? ??? (?? +?? -?? )???? ]
Adding (i) and (ii)
2?? =? ?
?? /3
?? /6
?
(sin ?? )
1/3
+(cos)
1/3
(sin ?? )
1/3
+(cos ?? )
1/3
???? =? ?
?? /3
?? /6
????? =[?? ]
?? /6
?? /3
=[
?? 3
-
?? 6
]
? ?? =
1
2
×
?? 6
=
?? 12
3.6 Evaluate: ?? =?
?? ?? ??? v?? ?????? (
?? ?? )???? .
(2016 : 10 Marks)
Solution:
Improper integral as integrand becomes undefined at lower limit, i.e., ?? =0.
Let log
1
?? =?? ? ?? =?? -?? ????? =-?? -?? ????
???????????? , ?? =0
+
??? ?8
x=1??? =0
?? =-? ?
0
8
?(?? -?? ?? )
1/3
·?? -?? ???? =? ?
8
0
??? 1/3
·?? -
4
3
?? ????
Putting
4
3
?? =?? ? ???? =
3
4
????
?? =? ?
8
0
?(
3
4
?? )
1/3
?? -?? ·
3
4
???? =
3
4
·(
3
4
)
1/3
? ?
8
0
??? 1/3
·?? -?? ????
=(
3
4
)
4/3
·? ?
8
0
??? 4
3
-1
·?? -?? ???? =(
3
4
)
4
3
·
v
(
4
3
)
=(
3
4
)
4
3
·
v
(1+
1
3
)=
1
3
(
3
4
)
4
3
·[
1
3
[G(?? )=? ?
8
0
??? -?? ·?? ?? -1
???? ;G(??? +1)=?? G(?? )]
3.7 Evaluate: ?
?? ?? ??????? -?? (?? -
?? ?? )????
(2020: 15 marks)
Solution:
Let
?? =? ?
1
0
?tan
-1
(1-
1
?? )
=? ?
1
0
?tan
-1
(
?? -1
?? ) …(??)
?? =? ?
1
0
?tan
-1
(
1-?? -1
1-?? )???? (? ?
?? ?? ??? (?? )???? =? ?
?? ?? ??? (?? +?? -?? )???? )
=? ?
1
0
?tan
-1
(
-?? 1-?? )????
=? ?
1
0
?tan
-1
(
?? 1-?? )???? …(???? )
Adding (i) and (ii), we get
2?? =? ?
1
0
?tan
-1
(
?? -1
?? )???? +? ?
1
0
?tan
-1
(
?? ?? -1
)????
=? ?
1
0
?[tan
-1
(
?? -1
?? )+tan
-1
(
?? ?? -1
)]???? ..(?????? )
(?tan
-1
?? +tan
-1
1
?? =(
?? 2
if ?? >0
-
?? 2
if ?? <0
)
Given ?? ?(0,1)
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