JEE > Mathematics (Maths) Class 12 > Example : Area under Simple Curves
Example : Area under Simple Curves Video Lecture - Mathematics (Maths) Class 12 - JEE
FAQs on Example : Area under Simple Curves Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is the formula to find the area under a simple curve? |  |
Ans. The formula to find the area under a simple curve is given by integrating the function that represents the curve over the desired interval. For example, if the curve is represented by the function f(x), the formula would be ∫f(x)dx.
| 2. How can I determine the limits of integration for finding the area under a curve? | |
Ans. The limits of integration for finding the area under a curve depend on the given problem or the specific interval of interest. They are usually provided in the problem statement or can be identified based on the context of the curve. The lower limit is typically the starting point of the interval, and the upper limit is the ending point.
| 3. Is it necessary for the curve to be a simple curve to find its area? | |
Ans. No, it is not necessary for the curve to be a simple curve to find its area. The concept of finding the area under a curve can be extended to more complex curves as well. However, the method of finding the area may differ depending on the complexity of the curve.
| 4. Can we find the area under a curve using numerical methods instead of integration? | |
Ans. Yes, it is possible to find the area under a curve using numerical methods instead of integration. Numerical methods, such as the trapezoidal rule or Simpson's rule, can be used to approximate the area under a curve when it is not possible or practical to find the exact solution using integration.
| 5. What are some real-life applications of finding the area under simple curves? | |
Ans. Finding the area under simple curves has various real-life applications. It is used in physics to calculate the work done by a force, in economics to determine the total revenue or profit, in statistics to calculate probabilities, and in engineering to analyze the performance of systems or structures. Additionally, it is also used in fields such as finance, biology, and environmental science to solve various problems and make informed decisions.
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208 videos|243 docs|139 tests
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Text Transcript from Video
hello friends this video an application
of integrals part 2 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 let's
do the same thing with the Eclipse so
you have to find the area bounded by
this ellipse X square by a square plus y
square B Square is no Google I am
solving this for a general equation so
in exams if you get for some value of a
and B again then so on so here is my
lips here also after I al goodbye both
approach the first approaches and take
this small cotton and I will say area of
the lips is nothing but 4 into area of
shaded region why because by symmetric
their into four paths so and then in
this case also I take one horizontal
rectangle and I say this guy is nothing
but area you integration of Y dot DX
where my Y is very sorry D X so X is
varying from 0 to you know here it is 0
0 and it is easy so X is varying from 0
to it so you say 0 to start but I have
to write Y in terms of X so Y if you
solve if you gain from the same equation
y square by B Square is nothing but 1
minus X square by a square + y square
something but B Square into a square
minus X square by a square and Y is
nothing but V by a root a square - it's
good
plus - any any period take the observed
values of plasma not doesn't matter for
us so this becomes 4 into R all right
here only for into degradation of B by a
2 a square minus X square DX a virus 1 X
that is 1 degree so let us say V by a
common so this become 4 B by a root of a
square minus X square DX
correct DX 0 to a so we have to
integrate this we know how to integrate
this same same formula slide right here
over B by a this becomes X by 2 mu a
square minus X square plus a square by 2
sine inverse X by a and this a var X bar
is from 0 to a you put the values you
get for B by a who a is but a by 2 a
square minus a square plus a square by 2
sine inverse a by a times 1 minus 0
minus a square by 2 sine inverse 0 so
this becomes this is 0 sine 4 0 0 this
is also 0
so what you get for B by a into a square
by 2 and sine inverse 1 is PI by 2 we
know since we solve or cancel a is
square cancel you get PI into E and that
is we know that area of a ellipse is PI
ap for the ellipse equation X square by
a square plus y square by B Square by 2
ma and there's max so what we have done
we have this bigger area and I didn't
want to find the area of all this
quadrant and my novi symmetric or are
equal I found the area of 1 and say the
area of ellipse is nothing but 4 into
area of the shaded region I'll do the
same thing same worship in the other
window where I will vary Y same thing I
am doing the same questions in different
ways
you can get comfortable in the concept
that you can choose any other option you
can either say X dot dy or say why don't
hear what will give you same answer so
same thing here we will say area of
ellipse is nothing but or e into area of
red regime and this guy is nothing but X
dot dy integration of text
Bui now if you see Y is varying from
zero to B please note earlier my X is
varying from zero to a by the Y is
varying from zero to B only y is varying
from zero to D correct not right X in
terms of Y so same thing into X square
by n square nothing but one minus y
square v square or X square is nothing
but a square mm v square into v square
minus y square or X is nothing but V by
V u v square minus plus minus is there
but I can ignore that because I am
talking about the absolute values so
this becomes more V by V right
integration of d square minus y square
dy and y varies from 0 to B so this guy
is nothing but we know same thing for a
be anyways constant outside becomes Y by
2 v square minus y square plus v square
by 2 sine inverse Y by so if you're
having difficulty in this formula please
watch my previous videos where we have
explained how this integration works and
we also given the proof of this
integration so and this varies from 0 to
B so let us put the values H so this
becomes for a by B Y becomes B by 2
minus v square minus B square plus v
square by 2 sine versus B by a D by B
minus y becomes 0 0 minus v square by 2
my inverse Z so this guy is 0 sine of 0
is 0 and this guy of 0 v square minus v
square is 0 so this I am left with for a
B into v square by 2 and sine inverse 1
is 5 by 2 this is nothing but pi a B and
that is nice so if you see in both case
we got the same answer 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
thank
of integrals part 2 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 let's
do the same thing with the Eclipse so
you have to find the area bounded by
this ellipse X square by a square plus y
square B Square is no Google I am
solving this for a general equation so
in exams if you get for some value of a
and B again then so on so here is my
lips here also after I al goodbye both
approach the first approaches and take
this small cotton and I will say area of
the lips is nothing but 4 into area of
shaded region why because by symmetric
their into four paths so and then in
this case also I take one horizontal
rectangle and I say this guy is nothing
but area you integration of Y dot DX
where my Y is very sorry D X so X is
varying from 0 to you know here it is 0
0 and it is easy so X is varying from 0
to it so you say 0 to start but I have
to write Y in terms of X so Y if you
solve if you gain from the same equation
y square by B Square is nothing but 1
minus X square by a square + y square
something but B Square into a square
minus X square by a square and Y is
nothing but V by a root a square - it's
good
plus - any any period take the observed
values of plasma not doesn't matter for
us so this becomes 4 into R all right
here only for into degradation of B by a
2 a square minus X square DX a virus 1 X
that is 1 degree so let us say V by a
common so this become 4 B by a root of a
square minus X square DX
correct DX 0 to a so we have to
integrate this we know how to integrate
this same same formula slide right here
over B by a this becomes X by 2 mu a
square minus X square plus a square by 2
sine inverse X by a and this a var X bar
is from 0 to a you put the values you
get for B by a who a is but a by 2 a
square minus a square plus a square by 2
sine inverse a by a times 1 minus 0
minus a square by 2 sine inverse 0 so
this becomes this is 0 sine 4 0 0 this
is also 0
so what you get for B by a into a square
by 2 and sine inverse 1 is PI by 2 we
know since we solve or cancel a is
square cancel you get PI into E and that
is we know that area of a ellipse is PI
ap for the ellipse equation X square by
a square plus y square by B Square by 2
ma and there's max so what we have done
we have this bigger area and I didn't
want to find the area of all this
quadrant and my novi symmetric or are
equal I found the area of 1 and say the
area of ellipse is nothing but 4 into
area of the shaded region I'll do the
same thing same worship in the other
window where I will vary Y same thing I
am doing the same questions in different
ways
you can get comfortable in the concept
that you can choose any other option you
can either say X dot dy or say why don't
hear what will give you same answer so
same thing here we will say area of
ellipse is nothing but or e into area of
red regime and this guy is nothing but X
dot dy integration of text
Bui now if you see Y is varying from
zero to B please note earlier my X is
varying from zero to a by the Y is
varying from zero to B only y is varying
from zero to D correct not right X in
terms of Y so same thing into X square
by n square nothing but one minus y
square v square or X square is nothing
but a square mm v square into v square
minus y square or X is nothing but V by
V u v square minus plus minus is there
but I can ignore that because I am
talking about the absolute values so
this becomes more V by V right
integration of d square minus y square
dy and y varies from 0 to B so this guy
is nothing but we know same thing for a
be anyways constant outside becomes Y by
2 v square minus y square plus v square
by 2 sine inverse Y by so if you're
having difficulty in this formula please
watch my previous videos where we have
explained how this integration works and
we also given the proof of this
integration so and this varies from 0 to
B so let us put the values H so this
becomes for a by B Y becomes B by 2
minus v square minus B square plus v
square by 2 sine versus B by a D by B
minus y becomes 0 0 minus v square by 2
my inverse Z so this guy is 0 sine of 0
is 0 and this guy of 0 v square minus v
square is 0 so this I am left with for a
B into v square by 2 and sine inverse 1
is 5 by 2 this is nothing but pi a B and
that is nice so if you see in both case
we got the same answer 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
thank
About this Video
|
4.86/5 Rating |
|
Tarun Kaushik |
Video Description: Example : Area under Simple Curves for JEE 2023 is part of Mathematics (Maths) Class 12 preparation.
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