Finding Areas Under Simple Curves Video Lecture | Mathematics (Maths) for JEE Main & Advanced

area under simple curves in the previous
lessons we have learned that definite
integral represents area bounded by the
curve y is equal to FX x is equal to e x
is equal to b and x axis we can think of
complete area PQRS consists of large
number of very thin vertical strips let
us consider an arbitrary strip of height
Y and width DX the area of element is
flip T a is equal to Y into DX where Y
is a function of X this area is called
the elementary area which is located at
arbitrary position the total area of the
region between y is equal to FX
ordinates X is equal to a X is equal to
B and the x-axis is sum of such
elementary areas of thin strips from X
is equal to e 2 X is equal to B so a is
equal to definite integral of FX DX from
X is equal to e 2 X is equal to B
now let us apply similar method to find
area bound by curve X is equal to Gy
y is equal to C y is equal to D and Y
axis let us consider a small thin strip
and horizontal directions the elementary
area of flip is X into D Y the total
area of region bound by the curve X is
equal to Gy yxs and line y is equal to C
and Y is equal to D is given by sum of
all such elementary areas ay is equal to
sum of elementary areas or a is equal to
definite integral of DA from y is equal
to C 2 y is equal to D or a is equal to
definite integral of dy dy from y is
equal to C 2 y is equal to D
when the curve under consideration is
below x-axis the function FX will have
negative values so the area bound by
curve x axis and ordinates x is equal to
a and X is equal to B will come out to
the negative so when area is negative we
will take its absolute value sometime it
may happen that some portion of the
curve is above x-axis and some is below
the x-axis the portion below x-axis will
give negative area value so while
finding total area we will take some of
even and absolute of a 2 using these
methods let us find the area and close
by the circle x squared plus y squared
is equal to a square because circle is
symmetric in all four quadrants we will
find area of circle in first quadrant
and then multiply it by for area of
circle is equal to 4 times the area of
region A or B let us find the area of
region e
Opa bounded by the curve x is equal to 0
X is equal to E and x-axis
let us consider a very thin vertical
strip inside the circle of high dry and
wet the X the area of elementary strip
da is equal to Y into DX area of circle
in first quadrant f1 is equal to
definite integral of Y DX from X is
equal to 0 to X is equal to a as X
square plus y square is equal to e
square we can write it in the form of Y
as a function of X as Y is equal to
square root of a square minus X square
substituting in area formula we will get
given is equal to definite integral of
square root of a square minus X square
from X is equal to 0 to X is equal to a
integral of square root of a square
minus X square is X by 2 square root of
e square minus X square plus a square by
2 sine inverse X by a so a 1 is equal to
X by 2 square root of a square minus X
square plus a square by 2 sine inverse X
by a from 0 to a substituting X is equal
to e and X is equal to 0 we will get a 1
is equal to a square by 2 sine inverse 1
or a 1 is equal to a square PI by 4 the
total area of circle is 4 times a 1 so
area of circle is equal to PI a square
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