JEE Exam > JEE Videos > Mathematics (Maths) Class 12 > Examples : Evaluation of Definite Integrals using Substitution method
Examples : Evaluation of Definite Integrals using Substitution method Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Evaluation of Definite Integrals using Substitution method Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. How does the substitution method help in evaluating definite integrals? |  |
Ans. The substitution method is a technique used in calculus to simplify the evaluation of definite integrals. It involves substituting a new variable in place of the original variable and rewriting the integral in terms of this new variable. This substitution helps in transforming the integrand into a simpler form, making it easier to evaluate the integral.
| 2. What is the general process of evaluating definite integrals using the substitution method? | |
Ans. The general process of evaluating definite integrals using the substitution method involves the following steps:
1. Identify a suitable substitution: Look for a subexpression within the integrand that can be replaced by a new variable.
2. Substitute the new variable: Replace the identified subexpression with the new variable.
3. Calculate the differential: Determine the differential of the new variable.
4. Rewrite the integral: Express the original integral in terms of the new variable and its differential.
5. Evaluate the integral: Use the substitution to simplify the integrand and evaluate the integral.
6. Apply the limits: Replace the original limits of integration with the corresponding values in terms of the new variable.
| 3. Can you provide an example of evaluating a definite integral using the substitution method? | |
Ans. Sure! Let's consider the integral ∫(2x + 1)^2 dx with limits of integration from 0 to 1. We can use the substitution method by letting u = 2x + 1. By differentiating both sides of this equation, we get du = 2dx. Rearranging this, we have dx = du/2. Substituting these into the integral, we get ∫(2x + 1)^2 dx = ∫u^2 (du/2). Simplifying, we have (1/2)∫u^2 du. Integrating, we get (1/2) * (u^3/3) + C. Finally, substituting back u = 2x + 1 and applying the limits, we find the value of the definite integral.
| 4. Are there any specific guidelines for choosing an appropriate substitution? | |
Ans. Yes, there are some guidelines to help choose an appropriate substitution:
1. Look for a subexpression within the integrand that is easily differentiable.
2. Choose a substitution that simplifies the integrand or reduces it to a known form.
3. Avoid choosing a substitution that results in a more complicated expression or a more difficult integral.
4. Consider the limits of integration and ensure that they can be expressed in terms of the new variable.
5. If possible, choose a substitution that cancels out any square roots or trigonometric functions present in the integrand.
| 5. Can the substitution method be used for all definite integrals? | |
Ans. No, the substitution method may not be applicable to all definite integrals. It is most effective for integrals that involve algebraic functions or expressions that can be simplified using substitutions. However, for certain integrals involving transcendental functions or complex expressions, other techniques like integration by parts or trigonometric identities may be more suitable. It is important to consider the specific form of the integrand and apply the appropriate method accordingly.
Text Transcript from Video
hello friends this video on integrals
path 40 is granty by exam Viacom no more
fear from exam before watching this
video please make sure that you have
watched part 1 to part 40 these kind of
questions we have done actually where we
have X into X to the power for example X
into X plus Z to the power M kind of
thing right so what we do in this case
because my X plus K to the power M we
have which is difficult to expand do you
assume this guy as T so this becomes
nothing but X plus K if you take P X
becomes t minus K into T to the power n
this is easy to solve because T to the
power M into t minus K is nothing but Q
to the power n plus 1 minus K in the
bottom integration of this is simple
actually correct but indication of this
is difficult because it expand this so
here I will do same thing will take X
plus 2 s T we can take or we can take
T's fire also because if you take T
squared then also you can solve do you
dig tea or something us against all so
let's take T only this becomes DX is
equal to reading so what this guy X is
nothing but t minus 2 so this is nothing
but e minus 2 into e to the power 1 by 2
DT now when x is 0 when x is 0 is what x
plus 2 that is 0 plus 2 that is 2 this
becomes 2 when X is equal to two T's
word T is equal to 2 plus 2 that is so
this is my mother so this is nothing but
I will say into your Lord to the power 3
by 2 minus 2 into t to power 1 by 2 DT
from 2 to 4 this is nothing but e to the
power 5 by 2 by 5 by 2 minus 2 into 2 e
to the power 1 by 2 plus 1 the raised to
the power 3 by 2 by 3 by 2 and the value
is from 2 to 4
if you saw this this is nothing bad five
into T to the power 5 by 2 that is for
the power 5 by 2 right sorry this is 2
here by 5
- I can even take 4 2 into 2 by 3 into T
there is 4 to the power 3 by 2 this
thing - not a dick - yes - so this
becomes 2 by 5 into 2 to the power 5 by
2
- here 2 into 2 by 3 into instead of or
I'll take the s 2 2 to the power 3 right
this is what I get you can solve this
this is nothing but if you see for the
power 5 by 2 that is 2 to the power 5
that is 32 into 2 there is 64 by 5 minus
4 to the power 3 by 2 is nothing but 2
to the power 3 8 8 into 4 32 by 3 - here
2 to the power 5 by 2 into 2 that is
nothing but a 5 by 5 and this is glass
this becomes o by 3 okay and if you
solve this actually what you'll get is 1
1 16 1 5 and that is masked so not a
great question with the easy question
actually we knew that if you have this
kind of integrand we convert this into
simpler form by taking you can you don't
have taken X plus T is the root T Square
also in this case that also would have
got the same answer in this question we
have to find integration of sine X by 1
plus cos square X
if you see this is a very not bad very
difficult question because if you just
put costs X as T the equation turns out
to be 1 by 1 plus T squared really
correct so you can integrate
differentiate both sides you get minus
sine X the external DV so this guy is
sine X DX we have so this guy is nothing
but minus DT by 1 plus T squared
integral alright now we would change the
value of this because this is from the
very X very strong 0 to PI by T I have
to find the how the T values and the
equation we has cost X is going to be
cos 0 is what 1 cos PI over 2 is what 0
so my t varies from 1 to 0 that is one
reason we integrate there this is
nothing but minus of tan inverse T and
this is nothing but from 0 1 to 0 and
this is nothing but minus of x 0 minus
tan inverse 1 and the other thing will -
of time where 0 is 0 and time as 1 is 5
at 4 minus of minus PI by 4 is plus PI
also you note that when you do an
integration of definite integral you
don't have any constant part you with
the fixed value Y because if you have
any function you are doing a finite
integral of this definite integral this
part is always constant so you don't
need the constant part thank you visit
exam viacom to watch free' h-e-b should
videos try free online tests get the
best quality study materials study from
the best tutors and mentors and much
more thanks and sorry
path 40 is granty by exam Viacom no more
fear from exam before watching this
video please make sure that you have
watched part 1 to part 40 these kind of
questions we have done actually where we
have X into X to the power for example X
into X plus Z to the power M kind of
thing right so what we do in this case
because my X plus K to the power M we
have which is difficult to expand do you
assume this guy as T so this becomes
nothing but X plus K if you take P X
becomes t minus K into T to the power n
this is easy to solve because T to the
power M into t minus K is nothing but Q
to the power n plus 1 minus K in the
bottom integration of this is simple
actually correct but indication of this
is difficult because it expand this so
here I will do same thing will take X
plus 2 s T we can take or we can take
T's fire also because if you take T
squared then also you can solve do you
dig tea or something us against all so
let's take T only this becomes DX is
equal to reading so what this guy X is
nothing but t minus 2 so this is nothing
but e minus 2 into e to the power 1 by 2
DT now when x is 0 when x is 0 is what x
plus 2 that is 0 plus 2 that is 2 this
becomes 2 when X is equal to two T's
word T is equal to 2 plus 2 that is so
this is my mother so this is nothing but
I will say into your Lord to the power 3
by 2 minus 2 into t to power 1 by 2 DT
from 2 to 4 this is nothing but e to the
power 5 by 2 by 5 by 2 minus 2 into 2 e
to the power 1 by 2 plus 1 the raised to
the power 3 by 2 by 3 by 2 and the value
is from 2 to 4
if you saw this this is nothing bad five
into T to the power 5 by 2 that is for
the power 5 by 2 right sorry this is 2
here by 5
- I can even take 4 2 into 2 by 3 into T
there is 4 to the power 3 by 2 this
thing - not a dick - yes - so this
becomes 2 by 5 into 2 to the power 5 by
2
- here 2 into 2 by 3 into instead of or
I'll take the s 2 2 to the power 3 right
this is what I get you can solve this
this is nothing but if you see for the
power 5 by 2 that is 2 to the power 5
that is 32 into 2 there is 64 by 5 minus
4 to the power 3 by 2 is nothing but 2
to the power 3 8 8 into 4 32 by 3 - here
2 to the power 5 by 2 into 2 that is
nothing but a 5 by 5 and this is glass
this becomes o by 3 okay and if you
solve this actually what you'll get is 1
1 16 1 5 and that is masked so not a
great question with the easy question
actually we knew that if you have this
kind of integrand we convert this into
simpler form by taking you can you don't
have taken X plus T is the root T Square
also in this case that also would have
got the same answer in this question we
have to find integration of sine X by 1
plus cos square X
if you see this is a very not bad very
difficult question because if you just
put costs X as T the equation turns out
to be 1 by 1 plus T squared really
correct so you can integrate
differentiate both sides you get minus
sine X the external DV so this guy is
sine X DX we have so this guy is nothing
but minus DT by 1 plus T squared
integral alright now we would change the
value of this because this is from the
very X very strong 0 to PI by T I have
to find the how the T values and the
equation we has cost X is going to be
cos 0 is what 1 cos PI over 2 is what 0
so my t varies from 1 to 0 that is one
reason we integrate there this is
nothing but minus of tan inverse T and
this is nothing but from 0 1 to 0 and
this is nothing but minus of x 0 minus
tan inverse 1 and the other thing will -
of time where 0 is 0 and time as 1 is 5
at 4 minus of minus PI by 4 is plus PI
also you note that when you do an
integration of definite integral you
don't have any constant part you with
the fixed value Y because if you have
any function you are doing a finite
integral of this definite integral this
part is always constant so you don't
need the constant part thank you visit
exam viacom to watch free' h-e-b should
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 18, 2024 Last updated |
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