JEE Exam > JEE Videos > Mathematics (Maths) Class 12 > Examples : Solving using Properties of Definite integrals
Examples : Solving using Properties of Definite integrals Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Solving using Properties of Definite integrals Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What are the properties of definite integrals? |  |
Ans. The properties of definite integrals include linearity, additivity, and the property of the definite integral over an interval being equal to the negative of the integral over the same interval in the opposite direction.
| 2. How can linearity be applied to definite integrals? | |
Ans. Linearity allows us to split a definite integral of a sum or difference of functions into the sum or difference of the definite integrals of the individual functions.
| 3. Can you explain the additivity property of definite integrals? | |
Ans. The additivity property states that if we have two intervals [a, b] and [b, c], then the definite integral over the combined interval [a, c] is equal to the sum of the definite integrals over the individual intervals [a, b] and [b, c].
| 4. How can we use the property of the definite integral over an interval being equal to the negative of the integral over the same interval in the opposite direction? | |
Ans. This property allows us to change the direction of integration by negating the limits of integration. For example, if we have the definite integral from a to b of a function f(x), we can rewrite it as the negative of the definite integral from b to a of f(x).
| 5. What are some practical applications of the properties of definite integrals? | |
Ans. The properties of definite integrals are frequently used in various fields, such as physics, engineering, and economics, to solve problems involving areas, volumes, work, and accumulation of quantities over time. These properties provide a powerful tool for simplifying calculations and solving complex problems.
Text Transcript from Video
hello friends this video on integrals
part 44 is brought to you by exam fear
calm no more fear from exam before
watching this video please make sure
that you have watched part 1 - part 43
we can do some more examples integral
sine root sine inverse X by root sign
words this would cost here also if you
see even if you say this guy's PI by 2
minus X this guy comes Cossacks this
guy's cost X sine X which we means seen
actually correct you can add these two
that becomes one just by looking and see
this way alright same way
sign up PI by 2 minus X by minus X
that's root of cos of Pi by 2 minus X
correct and this guy is nothing but 0 to
PI by 2 cos Phi cos x by cos x plus this
way ie this is also I if you add these 2
by 2 I is nothing but dog sine X by root
of cos x plus negative sine X plus root
of this guy cos x by u cos x plus root
of sine X this is nothing if you see
into your log root of sine X plus root
of cos x value of sine X plus root over
cos x DX and this guy is 1 so now what I
am left with is if you put the upper
limit in lower element 2 y is nothing
but X and X that is from 0 to PI by 2
because 1 dot DX is X so what I get here
is 2 y is equal to PI by 2 correct
here also I is nothing but PI so if you
see here also just by using the
properties we didn't exactly Ponte
indigo of sine X of cos x just by using
properties the condom receive Amazon see
kind of question actually you can't
solve the doubt properties please pay
attention every critical question here
because it's more than X plus 2 and the
rain is changing from minus 5 to 30 if
you see mod of X plus 2 grabbed this is
something like this X equal to minus 2 y
becomes 0 you increase there are no
effects because like this we may be
degrees it goes like this
correct hexagon doing breaks you take X
go to 0 it is to accessibility if you
see - you see there are so it is so X is
equal to minus 2 is the breaking point
okay so and minus 2 is part of this guy
minus 5 to 5 you see it is minus 5 to 5
to integrate minus 2 is part of this so
my function is breaking so I can't use a
single function because not continuous
so as I told for integrating this it has
to be continuous right so what I'll do
is I'll break this guide two functions
from here to here is continuous from
here to there also continuous so break
this guided to function and I know that
X plus 2 mod is equal to positive X plus
2 for X is there is no minus 2 and X
plus 2 more is equal to minus of X plus
2 if my X is less than this thing I'm so
this guy will break from minus PI to
minus 2 mod of X plus 2 DX plus minus 2
to 5 mod of X plus T DS correct now if
you see as I don't since it is minus 5
to minus 2 this guy here the value has
to be minus of X plus 2 so let's do this
so this becomes minus of X plus 2 DX
plus this becomes this guy's through -
to fight this guy is X plus 2 n this is
where becomes X plus 2 X
oh my industry now if you see now on
both this integrals are continuous hair
it was not continuous right so we broke
into two continuous integrals and now I
can just solve it so this becomes minus
off this becomes X square by 2 plus two
x from minus 5 to minus 2 plus this
becomes again X square by n plus 2 X
from minus 2 to 5 this is a minus here
so you saw this this guy's nothing minus
of minus 2 square by 2 plus 2 into minus
2 minus minus Phi minus Phi square by 2
correct minus plus 2 into minus Phi this
whole thing plus v square by 2 plus into
5 minus because minus 2 minus 2 square
by 2 plus 2 into so we saw that this is
nothing but minus 4 by 2 that is 2 minus
of 2 plus this becomes minus 4 minus 4
this becomes minus this square is class
so the minus of 25 by 2 again minus
minus plus plus 10 plus this is 25 by 2
plus 2 into 5 10 minus square is plus
minus 2 and this becomes minus minus
plus so we solve this further you get 29
and the reserve answer so the crooks
part was this guy was not continuous
over x over the interval
so what we did was we broke this guy
into two intervals and in these two he
traversed from here to here this guy is
continuous from here to here the guys
companies and we found by itself so
please keep in mind this kind of pushes
can be tricky because it's already all
already told that if you want to
integrate then that in that particular
interval the integrand has to be
continuous this is not a property
question but I want to solve this
because the general question we get so
many equation of this form X by 1 minus
X to the power N and we have solved some
question but this is the general form so
I took this example in this case if any
of X 1 minus X or some K minus X to the
power n form it is difficult to expand
this so better take the whole thing St
so you take 1 minus X st you get minus
DX as DT right and you get X is nothing
but 1 minus T so this equation turns out
to be x becomes 1 minus T and minus X is
T DX becomes minus TD right and this guy
also changes because from 0 to 1 so when
X is going to 0 all right X is equal to
0 T is equal to 1 minus 0 that is 1 when
X is equal to 1 T is equal to 1 minus 1
that is 0 so it changed from 0 to 1 to 1
to 0 here correct this will be e to the
power n yeah now if you see this guy is
simple to solve this becomes T to the
power n minus 3 to the power n plus 1 DT
sorry 1 to 0 that is nothing but T to
the power n plus 1 by n plus 1 minus T
to the power n plus u by n plus 2 on 1
dozen so if you solve this you put 0
this becomes 0 minus 0 right 0 minus 0
minus if you put 1 1 by n plus 1 minus 1
by n plus
in the - here so this becomes 1 by n
plus 2 minus 1 by n minus 1 so you can
solve this becomes n plus 2 into n plus
1 and this you get as n plus 1 and minus
n minus T so this was minus here so well
this - I'm missing here - here - here so
this is one more - they'll become class
and this becomes minus so minus of this
that is is gone
- of this guy that is 1 by n plus T I
miss will - here know there is a minus
here actually this is minus of me right
so minus minus hey this guy also will
try to solve it in properties what
should we do not getting any clue let's
try first thing let's make it tan of Pi
therefore minus X because then you can
approach just try yeah we don't have any
other option you do so this becomes 0 to
PI by 4 log of 1 plus tan out PI by 4
minus 6 just buy it in Thailand way
because I don't have any other options 1
plus tan PI by 4 minus X if you know is
nothing but n PI by 4 minus 3 X by 1
plus n by 2 X what is the value of tan
PI by 4 1 1 minus X by 1 this guy
because one 10 power 4 is 1 into 10 X
that is X so you get 1 plus tan X by 1
minus tan X this guy is nothing but 0 to
PI by 4 log of this guy's eye is log of
1 plus 1 minus tan X by 1 plus 10 X days
let's add these two so this becomes 0 to
PI
to log of 1 plus 10x minus 10x so you
get 2 by 1 plus 10x correct if you
observe carefully this is nothing but 0
to PI by 2 log of 2 minus 0 to PI by 2
log of 1 plus tan X yes you see this guy
is same as this guy so I can write
nothing but I is nothing but 0 to PI by
2 log 2 D takes minus 5 correct this is
nothing but 2 pi is equal to this guy
and what's the value of this X log 2 and
the value of exchange no thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks and sorry
part 44 is brought to you by exam fear
calm no more fear from exam before
watching this video please make sure
that you have watched part 1 - part 43
we can do some more examples integral
sine root sine inverse X by root sign
words this would cost here also if you
see even if you say this guy's PI by 2
minus X this guy comes Cossacks this
guy's cost X sine X which we means seen
actually correct you can add these two
that becomes one just by looking and see
this way alright same way
sign up PI by 2 minus X by minus X
that's root of cos of Pi by 2 minus X
correct and this guy is nothing but 0 to
PI by 2 cos Phi cos x by cos x plus this
way ie this is also I if you add these 2
by 2 I is nothing but dog sine X by root
of cos x plus negative sine X plus root
of this guy cos x by u cos x plus root
of sine X this is nothing if you see
into your log root of sine X plus root
of cos x value of sine X plus root over
cos x DX and this guy is 1 so now what I
am left with is if you put the upper
limit in lower element 2 y is nothing
but X and X that is from 0 to PI by 2
because 1 dot DX is X so what I get here
is 2 y is equal to PI by 2 correct
here also I is nothing but PI so if you
see here also just by using the
properties we didn't exactly Ponte
indigo of sine X of cos x just by using
properties the condom receive Amazon see
kind of question actually you can't
solve the doubt properties please pay
attention every critical question here
because it's more than X plus 2 and the
rain is changing from minus 5 to 30 if
you see mod of X plus 2 grabbed this is
something like this X equal to minus 2 y
becomes 0 you increase there are no
effects because like this we may be
degrees it goes like this
correct hexagon doing breaks you take X
go to 0 it is to accessibility if you
see - you see there are so it is so X is
equal to minus 2 is the breaking point
okay so and minus 2 is part of this guy
minus 5 to 5 you see it is minus 5 to 5
to integrate minus 2 is part of this so
my function is breaking so I can't use a
single function because not continuous
so as I told for integrating this it has
to be continuous right so what I'll do
is I'll break this guide two functions
from here to here is continuous from
here to there also continuous so break
this guided to function and I know that
X plus 2 mod is equal to positive X plus
2 for X is there is no minus 2 and X
plus 2 more is equal to minus of X plus
2 if my X is less than this thing I'm so
this guy will break from minus PI to
minus 2 mod of X plus 2 DX plus minus 2
to 5 mod of X plus T DS correct now if
you see as I don't since it is minus 5
to minus 2 this guy here the value has
to be minus of X plus 2 so let's do this
so this becomes minus of X plus 2 DX
plus this becomes this guy's through -
to fight this guy is X plus 2 n this is
where becomes X plus 2 X
oh my industry now if you see now on
both this integrals are continuous hair
it was not continuous right so we broke
into two continuous integrals and now I
can just solve it so this becomes minus
off this becomes X square by 2 plus two
x from minus 5 to minus 2 plus this
becomes again X square by n plus 2 X
from minus 2 to 5 this is a minus here
so you saw this this guy's nothing minus
of minus 2 square by 2 plus 2 into minus
2 minus minus Phi minus Phi square by 2
correct minus plus 2 into minus Phi this
whole thing plus v square by 2 plus into
5 minus because minus 2 minus 2 square
by 2 plus 2 into so we saw that this is
nothing but minus 4 by 2 that is 2 minus
of 2 plus this becomes minus 4 minus 4
this becomes minus this square is class
so the minus of 25 by 2 again minus
minus plus plus 10 plus this is 25 by 2
plus 2 into 5 10 minus square is plus
minus 2 and this becomes minus minus
plus so we solve this further you get 29
and the reserve answer so the crooks
part was this guy was not continuous
over x over the interval
so what we did was we broke this guy
into two intervals and in these two he
traversed from here to here this guy is
continuous from here to here the guys
companies and we found by itself so
please keep in mind this kind of pushes
can be tricky because it's already all
already told that if you want to
integrate then that in that particular
interval the integrand has to be
continuous this is not a property
question but I want to solve this
because the general question we get so
many equation of this form X by 1 minus
X to the power N and we have solved some
question but this is the general form so
I took this example in this case if any
of X 1 minus X or some K minus X to the
power n form it is difficult to expand
this so better take the whole thing St
so you take 1 minus X st you get minus
DX as DT right and you get X is nothing
but 1 minus T so this equation turns out
to be x becomes 1 minus T and minus X is
T DX becomes minus TD right and this guy
also changes because from 0 to 1 so when
X is going to 0 all right X is equal to
0 T is equal to 1 minus 0 that is 1 when
X is equal to 1 T is equal to 1 minus 1
that is 0 so it changed from 0 to 1 to 1
to 0 here correct this will be e to the
power n yeah now if you see this guy is
simple to solve this becomes T to the
power n minus 3 to the power n plus 1 DT
sorry 1 to 0 that is nothing but T to
the power n plus 1 by n plus 1 minus T
to the power n plus u by n plus 2 on 1
dozen so if you solve this you put 0
this becomes 0 minus 0 right 0 minus 0
minus if you put 1 1 by n plus 1 minus 1
by n plus
in the - here so this becomes 1 by n
plus 2 minus 1 by n minus 1 so you can
solve this becomes n plus 2 into n plus
1 and this you get as n plus 1 and minus
n minus T so this was minus here so well
this - I'm missing here - here - here so
this is one more - they'll become class
and this becomes minus so minus of this
that is is gone
- of this guy that is 1 by n plus T I
miss will - here know there is a minus
here actually this is minus of me right
so minus minus hey this guy also will
try to solve it in properties what
should we do not getting any clue let's
try first thing let's make it tan of Pi
therefore minus X because then you can
approach just try yeah we don't have any
other option you do so this becomes 0 to
PI by 4 log of 1 plus tan out PI by 4
minus 6 just buy it in Thailand way
because I don't have any other options 1
plus tan PI by 4 minus X if you know is
nothing but n PI by 4 minus 3 X by 1
plus n by 2 X what is the value of tan
PI by 4 1 1 minus X by 1 this guy
because one 10 power 4 is 1 into 10 X
that is X so you get 1 plus tan X by 1
minus tan X this guy is nothing but 0 to
PI by 4 log of this guy's eye is log of
1 plus 1 minus tan X by 1 plus 10 X days
let's add these two so this becomes 0 to
PI
to log of 1 plus 10x minus 10x so you
get 2 by 1 plus 10x correct if you
observe carefully this is nothing but 0
to PI by 2 log of 2 minus 0 to PI by 2
log of 1 plus tan X yes you see this guy
is same as this guy so I can write
nothing but I is nothing but 0 to PI by
2 log 2 D takes minus 5 correct this is
nothing but 2 pi is equal to this guy
and what's the value of this X log 2 and
the value of exchange no thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks and sorry
About this Video
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Dec 26, 2024 Last updated |
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