Courses

# Area Under The Curve JEE Notes | EduRev

## JEE : Area Under The Curve JEE Notes | EduRev

``` Page 1

E
1
J E E - M a t h e m a t i c s
1 . AREA UNDER THE CURVES :
( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and
O x=a x=b dx
y=ƒ (x)
x
y

x = b is given by A =
b
a
y dx
?
, where y = ƒ (x) lies above the x-axis
and b > a. Here vertical strip of thickness dx is considered at distance x.
(b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider
O
x
y
a b
the magnitude only, i.e. A =
b
a
y dx
?
(c) If curve crosses the x-axis at x = c, then A =
c b
a c
y dx ydx ?
? ?
O
x
y
x=a
x=b
c
(d ) Sometimes integration w.r.t. y is very useful (horizontal strip) :

O
x
y
y=b
dy
y=a
Area bounded by the curve, y-axis and the two abscissae at
y = a & y = b is written as
b
a
A xdy ?
?
.
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one
symmetric portion).
Illustration 1 : Find the area bounded by y = sec
2
x, x =
6
?
, x =
3
?
& x-axis
Solution : Area bounded =
/ 3
/ 6
ydx
?
?
?
=
/ 3
2
/ 6
sec xdx
?
?
?
=
/ 3
/ 6
[tan x]
?
?
= tan
3
?
– tan
6
?
=
3
–
1
3
=
2
3
sq.units.
Illustration 2 : Find the area in the first quadrant bounded by y = 4x
2
,  x = 0, y = 1 and y = 4.
Solution : Required area =
4
1
x
?
dy =
4
1
y
2
?
dy  =
4
3 / 2
1
1 2
y
2 3
? ?
? ?
? ?

y=1
y=4
x=0
Y
O
X
=
1
3
[4
3/2
– 1] =
1
3
[8 – 1]
=
7
3
= 2
3
?
sq.units.
Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x =
4
?
and
3
x
4
?
?
Solution : Required area =
/ 2 3 / 4
/ 4 / 2
sin 2xdx sin 2xdx
? ?
? ?
?
? ?
=
/ 2 3 / 4
/ 4 / 2
cos2x cos2x
2 2
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?/4
3 /4 ? ?/2
=
1 1
[ 1 0] (0 ( 1))
2 2
? ? ? ? ? ?
= 1 sq. unit
AREA UNDER THE CURVE
JEEMAIN.GURU
Page 2

E
1
J E E - M a t h e m a t i c s
1 . AREA UNDER THE CURVES :
( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and
O x=a x=b dx
y=ƒ (x)
x
y

x = b is given by A =
b
a
y dx
?
, where y = ƒ (x) lies above the x-axis
and b > a. Here vertical strip of thickness dx is considered at distance x.
(b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider
O
x
y
a b
the magnitude only, i.e. A =
b
a
y dx
?
(c) If curve crosses the x-axis at x = c, then A =
c b
a c
y dx ydx ?
? ?
O
x
y
x=a
x=b
c
(d ) Sometimes integration w.r.t. y is very useful (horizontal strip) :

O
x
y
y=b
dy
y=a
Area bounded by the curve, y-axis and the two abscissae at
y = a & y = b is written as
b
a
A xdy ?
?
.
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one
symmetric portion).
Illustration 1 : Find the area bounded by y = sec
2
x, x =
6
?
, x =
3
?
& x-axis
Solution : Area bounded =
/ 3
/ 6
ydx
?
?
?
=
/ 3
2
/ 6
sec xdx
?
?
?
=
/ 3
/ 6
[tan x]
?
?
= tan
3
?
– tan
6
?
=
3
–
1
3
=
2
3
sq.units.
Illustration 2 : Find the area in the first quadrant bounded by y = 4x
2
,  x = 0, y = 1 and y = 4.
Solution : Required area =
4
1
x
?
dy =
4
1
y
2
?
dy  =
4
3 / 2
1
1 2
y
2 3
? ?
? ?
? ?

y=1
y=4
x=0
Y
O
X
=
1
3
[4
3/2
– 1] =
1
3
[8 – 1]
=
7
3
= 2
3
?
sq.units.
Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x =
4
?
and
3
x
4
?
?
Solution : Required area =
/ 2 3 / 4
/ 4 / 2
sin 2xdx sin 2xdx
? ?
? ?
?
? ?
=
/ 2 3 / 4
/ 4 / 2
cos2x cos2x
2 2
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?/4
3 /4 ? ?/2
=
1 1
[ 1 0] (0 ( 1))
2 2
? ? ? ? ? ?
= 1 sq. unit
AREA UNDER THE CURVE
JEEMAIN.GURU
2
E
J E E - M a t h e m a t i c s
Do yourself - 1 :
(i) Find the area bounded by y = x
2
+ 2 above x-axis between x = 2 &  x = 3.
(ii) Using integration, find the area of the curve
2
y 1 x ? ?
with co-ordinate axes bounded in first quadrant.
(iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?.
(iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x =
1
2
? and x=1.
2 . AREA ENCLOSED BETWEEN TWO CURVES :
O
y
y =ƒ (x)
1
y =g(x)
2
x
1
x
2
x
( a ) Area bounded by two curves y = ƒ (x) & y = g(x)
such that ƒ (x) > g(x) is
2
1
x
1 2
x
A (y y )dy ? ?
?
2
1
x
x
A [ƒ(x) g(x)]dx ? ?
?
(b ) In case horizontal strip is taken we have
y
x =g(y)
2
x =ƒ (y)
1
x
y
2
y
1
2
1
y
1 2
y
A (x x )dy ? ?
?
2
1
y
y
A [ƒ(y) g(y)]dy ? ?
?
(c) If the curves y
1
= ƒ (x) and y
2
= g(x) intersect at x = c, then required area
x=a
x=b
c
y
x
y =g(x)
2
y =ƒ (x)
1
A =
c b
a c
(g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ?
? ?
=
b
a
ƒ(x) g(x) dx ?
?
Note : Required area must have all the boundaries indicated in the problem.
Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and
x = 3 and x-axis
Solution : To determine the sign, we follow the usual rule of change of sign.
y = +ve for x > 3
y = –ve for 2 < x < 3
y = +ve for 1 < x < 2
y = –ve for x < 1.
3
0
| y| dx
?
=
1
0
| y| dx
?
+
2
1
| y| dx
?
+
3
2
| y| dx
?

1 2 3
E
D F
C
B
O
Y
A
(0, –6)
X
=
1
0
– y dx
?
+
2
1
y dx
?
+
3
2
–y dx
?
Now let F(x)  =
?
(x – 1) (x – 2) (x – 3) dx =
?
(x
3
– 6x
2
+ 11x – 6) dx =
1
4
x
4
– 2x
3
+
11
2
x
2
– 6x.
? F(0) = 0, F(1) = –
9
4
, F(2) = –2, F(3) = –
9
4
.
Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2
3
4
sq.units.
JEEMAIN.GURU
Page 3

E
1
J E E - M a t h e m a t i c s
1 . AREA UNDER THE CURVES :
( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and
O x=a x=b dx
y=ƒ (x)
x
y

x = b is given by A =
b
a
y dx
?
, where y = ƒ (x) lies above the x-axis
and b > a. Here vertical strip of thickness dx is considered at distance x.
(b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider
O
x
y
a b
the magnitude only, i.e. A =
b
a
y dx
?
(c) If curve crosses the x-axis at x = c, then A =
c b
a c
y dx ydx ?
? ?
O
x
y
x=a
x=b
c
(d ) Sometimes integration w.r.t. y is very useful (horizontal strip) :

O
x
y
y=b
dy
y=a
Area bounded by the curve, y-axis and the two abscissae at
y = a & y = b is written as
b
a
A xdy ?
?
.
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one
symmetric portion).
Illustration 1 : Find the area bounded by y = sec
2
x, x =
6
?
, x =
3
?
& x-axis
Solution : Area bounded =
/ 3
/ 6
ydx
?
?
?
=
/ 3
2
/ 6
sec xdx
?
?
?
=
/ 3
/ 6
[tan x]
?
?
= tan
3
?
– tan
6
?
=
3
–
1
3
=
2
3
sq.units.
Illustration 2 : Find the area in the first quadrant bounded by y = 4x
2
,  x = 0, y = 1 and y = 4.
Solution : Required area =
4
1
x
?
dy =
4
1
y
2
?
dy  =
4
3 / 2
1
1 2
y
2 3
? ?
? ?
? ?

y=1
y=4
x=0
Y
O
X
=
1
3
[4
3/2
– 1] =
1
3
[8 – 1]
=
7
3
= 2
3
?
sq.units.
Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x =
4
?
and
3
x
4
?
?
Solution : Required area =
/ 2 3 / 4
/ 4 / 2
sin 2xdx sin 2xdx
? ?
? ?
?
? ?
=
/ 2 3 / 4
/ 4 / 2
cos2x cos2x
2 2
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?/4
3 /4 ? ?/2
=
1 1
[ 1 0] (0 ( 1))
2 2
? ? ? ? ? ?
= 1 sq. unit
AREA UNDER THE CURVE
JEEMAIN.GURU
2
E
J E E - M a t h e m a t i c s
Do yourself - 1 :
(i) Find the area bounded by y = x
2
+ 2 above x-axis between x = 2 &  x = 3.
(ii) Using integration, find the area of the curve
2
y 1 x ? ?
with co-ordinate axes bounded in first quadrant.
(iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?.
(iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x =
1
2
? and x=1.
2 . AREA ENCLOSED BETWEEN TWO CURVES :
O
y
y =ƒ (x)
1
y =g(x)
2
x
1
x
2
x
( a ) Area bounded by two curves y = ƒ (x) & y = g(x)
such that ƒ (x) > g(x) is
2
1
x
1 2
x
A (y y )dy ? ?
?
2
1
x
x
A [ƒ(x) g(x)]dx ? ?
?
(b ) In case horizontal strip is taken we have
y
x =g(y)
2
x =ƒ (y)
1
x
y
2
y
1
2
1
y
1 2
y
A (x x )dy ? ?
?
2
1
y
y
A [ƒ(y) g(y)]dy ? ?
?
(c) If the curves y
1
= ƒ (x) and y
2
= g(x) intersect at x = c, then required area
x=a
x=b
c
y
x
y =g(x)
2
y =ƒ (x)
1
A =
c b
a c
(g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ?
? ?
=
b
a
ƒ(x) g(x) dx ?
?
Note : Required area must have all the boundaries indicated in the problem.
Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and
x = 3 and x-axis
Solution : To determine the sign, we follow the usual rule of change of sign.
y = +ve for x > 3
y = –ve for 2 < x < 3
y = +ve for 1 < x < 2
y = –ve for x < 1.
3
0
| y| dx
?
=
1
0
| y| dx
?
+
2
1
| y| dx
?
+
3
2
| y| dx
?

1 2 3
E
D F
C
B
O
Y
A
(0, –6)
X
=
1
0
– y dx
?
+
2
1
y dx
?
+
3
2
–y dx
?
Now let F(x)  =
?
(x – 1) (x – 2) (x – 3) dx =
?
(x
3
– 6x
2
+ 11x – 6) dx =
1
4
x
4
– 2x
3
+
11
2
x
2
– 6x.
? F(0) = 0, F(1) = –
9
4
, F(2) = –2, F(3) = –
9
4
.
Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2
3
4
sq.units.
JEEMAIN.GURU
E
3
J E E - M a t h e m a t i c s
Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves
y = 2
x
, y = 2x – x
2
.
Solution : Figure is self-explanatory  y = 2
x
, (x – 1)
2
= – (y – 1)
y=2x–x
2
x=2
R(2,4)
M(2,0)
y=2
x
Q
(0,1)
O
The required area =
2
1 2
0
(y y ) ?
?
dx
where y
1
= 2
x
and y
2
= 2x – x
2
=
2
x 2
0
(2 2x x )dx ? ?
?
=
2
x
2 3
0
2 1
x x
ln 2 3
? ?
? ?
? ?
? ?
=
4 8
4
ln 2 3
? ?
? ?
? ?
? ?
–
1
ln 2
=
3 4
–
ln 2 3
sq.units.
Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y
2
, x = 1 – 3y
2
.
Solution : Solving the equations x = –2y
2
, x = 1 – 3y
2
, we find that ordinates of the points of intersection of
the two curves as y
1
= – 1, y
2
= 1.
The points are (–2, –1) and (–2, 1).
1
Y
x=1–3y
2
P
1
1
X
–1
–2 –1
O
P
2
x=–2y
2
(–2, 1)
(–2, –1)
The required area
2
1
1 2
0
(x x ) ?
?
dy = 2
1
2 2
0
[(1 3y ) (–2y )]dy ? ?
?
=2
1
2
0
(1 y )dy ?
?
= 2
1
3
0
y
y
3
? ?
?
? ?
? ?
=
4
3
sq.units.
Do yourself - 2 :
(i) Find the area bounded by y x ? and y = x.
(ii) Find the area bounded by the curves x = y
2
and x = 3 – 2y
2
.
(iii) Find the area of the region bounded by the curves x =
1
2
, x = 2, y = logx and y = 2
x
.
3 . CURVE TRACING :
The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be
extremely useful to quickly and correctly evaluate the area under the curves.
( a ) Symmetry : The symmetry of the curve is judged as follows :
(i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x.
(ii) If all the powers of x are even, the curve is symmetrical about the axis of y.
(iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y.
(iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is
(v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is
(b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents.
(c) Find the points where the curve crosses the x ?axis & also the y ?axis.
(d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when
x ? ? or ? ?.
Illustration 7 : Find the area of a loop as well as the whole area of the curve a
2
y
2
= x
2
(a
2
– x
2
).
Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0)
Area of a loop = 2
a
0
y dx
?
= 2
a
2
0
x
a x dx
a
?
?
?
Y
O
A
(a,0)
X X'
A'
(–a,0)
= –
a
2 2
0
1
a x (–2x)dx
a
?
? = –
a
2 2 3 / 2
0
1 2 2
(a x )
a 3 3
? ?
? ?
? ?
? ?
a
2
Total area = 2 ×
2
3
a
2
=
4
3
a
2
sq.units. s.
JEEMAIN.GURU
Page 4

E
1
J E E - M a t h e m a t i c s
1 . AREA UNDER THE CURVES :
( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and
O x=a x=b dx
y=ƒ (x)
x
y

x = b is given by A =
b
a
y dx
?
, where y = ƒ (x) lies above the x-axis
and b > a. Here vertical strip of thickness dx is considered at distance x.
(b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider
O
x
y
a b
the magnitude only, i.e. A =
b
a
y dx
?
(c) If curve crosses the x-axis at x = c, then A =
c b
a c
y dx ydx ?
? ?
O
x
y
x=a
x=b
c
(d ) Sometimes integration w.r.t. y is very useful (horizontal strip) :

O
x
y
y=b
dy
y=a
Area bounded by the curve, y-axis and the two abscissae at
y = a & y = b is written as
b
a
A xdy ?
?
.
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one
symmetric portion).
Illustration 1 : Find the area bounded by y = sec
2
x, x =
6
?
, x =
3
?
& x-axis
Solution : Area bounded =
/ 3
/ 6
ydx
?
?
?
=
/ 3
2
/ 6
sec xdx
?
?
?
=
/ 3
/ 6
[tan x]
?
?
= tan
3
?
– tan
6
?
=
3
–
1
3
=
2
3
sq.units.
Illustration 2 : Find the area in the first quadrant bounded by y = 4x
2
,  x = 0, y = 1 and y = 4.
Solution : Required area =
4
1
x
?
dy =
4
1
y
2
?
dy  =
4
3 / 2
1
1 2
y
2 3
? ?
? ?
? ?

y=1
y=4
x=0
Y
O
X
=
1
3
[4
3/2
– 1] =
1
3
[8 – 1]
=
7
3
= 2
3
?
sq.units.
Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x =
4
?
and
3
x
4
?
?
Solution : Required area =
/ 2 3 / 4
/ 4 / 2
sin 2xdx sin 2xdx
? ?
? ?
?
? ?
=
/ 2 3 / 4
/ 4 / 2
cos2x cos2x
2 2
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?/4
3 /4 ? ?/2
=
1 1
[ 1 0] (0 ( 1))
2 2
? ? ? ? ? ?
= 1 sq. unit
AREA UNDER THE CURVE
JEEMAIN.GURU
2
E
J E E - M a t h e m a t i c s
Do yourself - 1 :
(i) Find the area bounded by y = x
2
+ 2 above x-axis between x = 2 &  x = 3.
(ii) Using integration, find the area of the curve
2
y 1 x ? ?
with co-ordinate axes bounded in first quadrant.
(iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?.
(iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x =
1
2
? and x=1.
2 . AREA ENCLOSED BETWEEN TWO CURVES :
O
y
y =ƒ (x)
1
y =g(x)
2
x
1
x
2
x
( a ) Area bounded by two curves y = ƒ (x) & y = g(x)
such that ƒ (x) > g(x) is
2
1
x
1 2
x
A (y y )dy ? ?
?
2
1
x
x
A [ƒ(x) g(x)]dx ? ?
?
(b ) In case horizontal strip is taken we have
y
x =g(y)
2
x =ƒ (y)
1
x
y
2
y
1
2
1
y
1 2
y
A (x x )dy ? ?
?
2
1
y
y
A [ƒ(y) g(y)]dy ? ?
?
(c) If the curves y
1
= ƒ (x) and y
2
= g(x) intersect at x = c, then required area
x=a
x=b
c
y
x
y =g(x)
2
y =ƒ (x)
1
A =
c b
a c
(g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ?
? ?
=
b
a
ƒ(x) g(x) dx ?
?
Note : Required area must have all the boundaries indicated in the problem.
Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and
x = 3 and x-axis
Solution : To determine the sign, we follow the usual rule of change of sign.
y = +ve for x > 3
y = –ve for 2 < x < 3
y = +ve for 1 < x < 2
y = –ve for x < 1.
3
0
| y| dx
?
=
1
0
| y| dx
?
+
2
1
| y| dx
?
+
3
2
| y| dx
?

1 2 3
E
D F
C
B
O
Y
A
(0, –6)
X
=
1
0
– y dx
?
+
2
1
y dx
?
+
3
2
–y dx
?
Now let F(x)  =
?
(x – 1) (x – 2) (x – 3) dx =
?
(x
3
– 6x
2
+ 11x – 6) dx =
1
4
x
4
– 2x
3
+
11
2
x
2
– 6x.
? F(0) = 0, F(1) = –
9
4
, F(2) = –2, F(3) = –
9
4
.
Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2
3
4
sq.units.
JEEMAIN.GURU
E
3
J E E - M a t h e m a t i c s
Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves
y = 2
x
, y = 2x – x
2
.
Solution : Figure is self-explanatory  y = 2
x
, (x – 1)
2
= – (y – 1)
y=2x–x
2
x=2
R(2,4)
M(2,0)
y=2
x
Q
(0,1)
O
The required area =
2
1 2
0
(y y ) ?
?
dx
where y
1
= 2
x
and y
2
= 2x – x
2
=
2
x 2
0
(2 2x x )dx ? ?
?
=
2
x
2 3
0
2 1
x x
ln 2 3
? ?
? ?
? ?
? ?
=
4 8
4
ln 2 3
? ?
? ?
? ?
? ?
–
1
ln 2
=
3 4
–
ln 2 3
sq.units.
Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y
2
, x = 1 – 3y
2
.
Solution : Solving the equations x = –2y
2
, x = 1 – 3y
2
, we find that ordinates of the points of intersection of
the two curves as y
1
= – 1, y
2
= 1.
The points are (–2, –1) and (–2, 1).
1
Y
x=1–3y
2
P
1
1
X
–1
–2 –1
O
P
2
x=–2y
2
(–2, 1)
(–2, –1)
The required area
2
1
1 2
0
(x x ) ?
?
dy = 2
1
2 2
0
[(1 3y ) (–2y )]dy ? ?
?
=2
1
2
0
(1 y )dy ?
?
= 2
1
3
0
y
y
3
? ?
?
? ?
? ?
=
4
3
sq.units.
Do yourself - 2 :
(i) Find the area bounded by y x ? and y = x.
(ii) Find the area bounded by the curves x = y
2
and x = 3 – 2y
2
.
(iii) Find the area of the region bounded by the curves x =
1
2
, x = 2, y = logx and y = 2
x
.
3 . CURVE TRACING :
The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be
extremely useful to quickly and correctly evaluate the area under the curves.
( a ) Symmetry : The symmetry of the curve is judged as follows :
(i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x.
(ii) If all the powers of x are even, the curve is symmetrical about the axis of y.
(iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y.
(iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is
(v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is
(b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents.
(c) Find the points where the curve crosses the x ?axis & also the y ?axis.
(d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when
x ? ? or ? ?.
Illustration 7 : Find the area of a loop as well as the whole area of the curve a
2
y
2
= x
2
(a
2
– x
2
).
Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0)
Area of a loop = 2
a
0
y dx
?
= 2
a
2
0
x
a x dx
a
?
?
?
Y
O
A
(a,0)
X X'
A'
(–a,0)
= –
a
2 2
0
1
a x (–2x)dx
a
?
? = –
a
2 2 3 / 2
0
1 2 2
(a x )
a 3 3
? ?
? ?
? ?
? ?
a
2
Total area = 2 ×
2
3
a
2
=
4
3
a
2
sq.units. s.
JEEMAIN.GURU
4
E
J E E - M a t h e m a t i c s
Illustration 8 : Find the whole area induded between the curve x
2
y
2
= a
2
(y
2
– x
2
) and its asymptotes.
Solution : (i) The curve is symmetric about both the axes (even powers of x & y)
(ii) Asymptotes are x = ± a
a
0
A 4 ydx ?
?
y
x=–a x=a
dx
x
a
2 2
0
ax
4 dx
a x
?
?
?
a
2 2
0
4a a x ? ? ?
= 4a
2
Illustration 9 : Find the area bounded by the curve xy
2
= 4a
2
(2a–x) and its asymptote.
Solution : (i) The curve is symmetrical about the x-axis as it contains even powers of y.
(ii) It passes through (2a,0).
(iii) Its asymptote is x = 0, i.e., y-axis.
y
x
(2a,0)
2a 2a
0 0
2a x
A 2 ydx 2 2a dx
x
?
? ?
? ?
Put x = 2a sin
2
?
A = 16a
2
/ 2
2
0
cos d
?
? ?
?
= 4 ?a
2
4 . IMPORTANT POINTS :
( a ) Since area remains invariant even if the co-ordinate axes are shifted, hence shifting of origin in many
cases proves to be very convenient in computing the area.
Illustration 10 : Find the area enclosed by |x – 1| + |y + 1| = 1.
Solution : Shift the origin to (1, –1).
(0,1)
(0,–1)
(1,0) (–1,0)
2
X = x – 1 Y = y + 1
|X| + |Y| = 1
Area = 2 2 ? = 2 sq. units
Illustration 11 : Find the area of the region common to the circle x
2
+ y
2
+ 4x + 6y – 3 = 0 and the parabola
x
2
+ 4x = 6y + 14.
Solution : Circle is x
2
+ y
2
+ 4x + 6y – 3 = 0
? (x + 2)
2
+ (y + 3)
2
= 16
x
(0,4)
(–2 3,2) ? (2 3,2) ?
Shifting origin to (–2,–3).
X
2
+ Y
2
= 16
equation of parabola ? ? (x + 2)
2
= 6(y + 3)
? X
2
= 6Y
Solving circle & parabola, we get X = ± 2 3
Hence they intersect at
? ?
2 3,2 ? &
? ?
2 3,2
2 4
2
0 2
A 2 6Y dY 16 Y dY
? ?
? ? ?
? ?
? ?
? ?
4
2
3 / 2 2 1
0
2
2 1 16 Y
2 6 Y Y 16 Y sin
3 2 2 4
?
? ?
? ?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ?
4 3 16
3 3
? ?
?
? ? ? ?
? ?
? ?
sq. units
JEEMAIN.GURU
Page 5

E
1
J E E - M a t h e m a t i c s
1 . AREA UNDER THE CURVES :
( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and
O x=a x=b dx
y=ƒ (x)
x
y

x = b is given by A =
b
a
y dx
?
, where y = ƒ (x) lies above the x-axis
and b > a. Here vertical strip of thickness dx is considered at distance x.
(b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider
O
x
y
a b
the magnitude only, i.e. A =
b
a
y dx
?
(c) If curve crosses the x-axis at x = c, then A =
c b
a c
y dx ydx ?
? ?
O
x
y
x=a
x=b
c
(d ) Sometimes integration w.r.t. y is very useful (horizontal strip) :

O
x
y
y=b
dy
y=a
Area bounded by the curve, y-axis and the two abscissae at
y = a & y = b is written as
b
a
A xdy ?
?
.
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one
symmetric portion).
Illustration 1 : Find the area bounded by y = sec
2
x, x =
6
?
, x =
3
?
& x-axis
Solution : Area bounded =
/ 3
/ 6
ydx
?
?
?
=
/ 3
2
/ 6
sec xdx
?
?
?
=
/ 3
/ 6
[tan x]
?
?
= tan
3
?
– tan
6
?
=
3
–
1
3
=
2
3
sq.units.
Illustration 2 : Find the area in the first quadrant bounded by y = 4x
2
,  x = 0, y = 1 and y = 4.
Solution : Required area =
4
1
x
?
dy =
4
1
y
2
?
dy  =
4
3 / 2
1
1 2
y
2 3
? ?
? ?
? ?

y=1
y=4
x=0
Y
O
X
=
1
3
[4
3/2
– 1] =
1
3
[8 – 1]
=
7
3
= 2
3
?
sq.units.
Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x =
4
?
and
3
x
4
?
?
Solution : Required area =
/ 2 3 / 4
/ 4 / 2
sin 2xdx sin 2xdx
? ?
? ?
?
? ?
=
/ 2 3 / 4
/ 4 / 2
cos2x cos2x
2 2
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?/4
3 /4 ? ?/2
=
1 1
[ 1 0] (0 ( 1))
2 2
? ? ? ? ? ?
= 1 sq. unit
AREA UNDER THE CURVE
JEEMAIN.GURU
2
E
J E E - M a t h e m a t i c s
Do yourself - 1 :
(i) Find the area bounded by y = x
2
+ 2 above x-axis between x = 2 &  x = 3.
(ii) Using integration, find the area of the curve
2
y 1 x ? ?
with co-ordinate axes bounded in first quadrant.
(iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?.
(iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x =
1
2
? and x=1.
2 . AREA ENCLOSED BETWEEN TWO CURVES :
O
y
y =ƒ (x)
1
y =g(x)
2
x
1
x
2
x
( a ) Area bounded by two curves y = ƒ (x) & y = g(x)
such that ƒ (x) > g(x) is
2
1
x
1 2
x
A (y y )dy ? ?
?
2
1
x
x
A [ƒ(x) g(x)]dx ? ?
?
(b ) In case horizontal strip is taken we have
y
x =g(y)
2
x =ƒ (y)
1
x
y
2
y
1
2
1
y
1 2
y
A (x x )dy ? ?
?
2
1
y
y
A [ƒ(y) g(y)]dy ? ?
?
(c) If the curves y
1
= ƒ (x) and y
2
= g(x) intersect at x = c, then required area
x=a
x=b
c
y
x
y =g(x)
2
y =ƒ (x)
1
A =
c b
a c
(g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ?
? ?
=
b
a
ƒ(x) g(x) dx ?
?
Note : Required area must have all the boundaries indicated in the problem.
Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and
x = 3 and x-axis
Solution : To determine the sign, we follow the usual rule of change of sign.
y = +ve for x > 3
y = –ve for 2 < x < 3
y = +ve for 1 < x < 2
y = –ve for x < 1.
3
0
| y| dx
?
=
1
0
| y| dx
?
+
2
1
| y| dx
?
+
3
2
| y| dx
?

1 2 3
E
D F
C
B
O
Y
A
(0, –6)
X
=
1
0
– y dx
?
+
2
1
y dx
?
+
3
2
–y dx
?
Now let F(x)  =
?
(x – 1) (x – 2) (x – 3) dx =
?
(x
3
– 6x
2
+ 11x – 6) dx =
1
4
x
4
– 2x
3
+
11
2
x
2
– 6x.
? F(0) = 0, F(1) = –
9
4
, F(2) = –2, F(3) = –
9
4
.
Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2
3
4
sq.units.
JEEMAIN.GURU
E
3
J E E - M a t h e m a t i c s
Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves
y = 2
x
, y = 2x – x
2
.
Solution : Figure is self-explanatory  y = 2
x
, (x – 1)
2
= – (y – 1)
y=2x–x
2
x=2
R(2,4)
M(2,0)
y=2
x
Q
(0,1)
O
The required area =
2
1 2
0
(y y ) ?
?
dx
where y
1
= 2
x
and y
2
= 2x – x
2
=
2
x 2
0
(2 2x x )dx ? ?
?
=
2
x
2 3
0
2 1
x x
ln 2 3
? ?
? ?
? ?
? ?
=
4 8
4
ln 2 3
? ?
? ?
? ?
? ?
–
1
ln 2
=
3 4
–
ln 2 3
sq.units.
Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y
2
, x = 1 – 3y
2
.
Solution : Solving the equations x = –2y
2
, x = 1 – 3y
2
, we find that ordinates of the points of intersection of
the two curves as y
1
= – 1, y
2
= 1.
The points are (–2, –1) and (–2, 1).
1
Y
x=1–3y
2
P
1
1
X
–1
–2 –1
O
P
2
x=–2y
2
(–2, 1)
(–2, –1)
The required area
2
1
1 2
0
(x x ) ?
?
dy = 2
1
2 2
0
[(1 3y ) (–2y )]dy ? ?
?
=2
1
2
0
(1 y )dy ?
?
= 2
1
3
0
y
y
3
? ?
?
? ?
? ?
=
4
3
sq.units.
Do yourself - 2 :
(i) Find the area bounded by y x ? and y = x.
(ii) Find the area bounded by the curves x = y
2
and x = 3 – 2y
2
.
(iii) Find the area of the region bounded by the curves x =
1
2
, x = 2, y = logx and y = 2
x
.
3 . CURVE TRACING :
The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be
extremely useful to quickly and correctly evaluate the area under the curves.
( a ) Symmetry : The symmetry of the curve is judged as follows :
(i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x.
(ii) If all the powers of x are even, the curve is symmetrical about the axis of y.
(iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y.
(iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is
(v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is
(b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents.
(c) Find the points where the curve crosses the x ?axis & also the y ?axis.
(d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when
x ? ? or ? ?.
Illustration 7 : Find the area of a loop as well as the whole area of the curve a
2
y
2
= x
2
(a
2
– x
2
).
Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0)
Area of a loop = 2
a
0
y dx
?
= 2
a
2
0
x
a x dx
a
?
?
?
Y
O
A
(a,0)
X X'
A'
(–a,0)
= –
a
2 2
0
1
a x (–2x)dx
a
?
? = –
a
2 2 3 / 2
0
1 2 2
(a x )
a 3 3
? ?
? ?
? ?
? ?
a
2
Total area = 2 ×
2
3
a
2
=
4
3
a
2
sq.units. s.
JEEMAIN.GURU
4
E
J E E - M a t h e m a t i c s
Illustration 8 : Find the whole area induded between the curve x
2
y
2
= a
2
(y
2
– x
2
) and its asymptotes.
Solution : (i) The curve is symmetric about both the axes (even powers of x & y)
(ii) Asymptotes are x = ± a
a
0
A 4 ydx ?
?
y
x=–a x=a
dx
x
a
2 2
0
ax
4 dx
a x
?
?
?
a
2 2
0
4a a x ? ? ?
= 4a
2
Illustration 9 : Find the area bounded by the curve xy
2
= 4a
2
(2a–x) and its asymptote.
Solution : (i) The curve is symmetrical about the x-axis as it contains even powers of y.
(ii) It passes through (2a,0).
(iii) Its asymptote is x = 0, i.e., y-axis.
y
x
(2a,0)
2a 2a
0 0
2a x
A 2 ydx 2 2a dx
x
?
? ?
? ?
Put x = 2a sin
2
?
A = 16a
2
/ 2
2
0
cos d
?
? ?
?
= 4 ?a
2
4 . IMPORTANT POINTS :
( a ) Since area remains invariant even if the co-ordinate axes are shifted, hence shifting of origin in many
cases proves to be very convenient in computing the area.
Illustration 10 : Find the area enclosed by |x – 1| + |y + 1| = 1.
Solution : Shift the origin to (1, –1).
(0,1)
(0,–1)
(1,0) (–1,0)
2
X = x – 1 Y = y + 1
|X| + |Y| = 1
Area = 2 2 ? = 2 sq. units
Illustration 11 : Find the area of the region common to the circle x
2
+ y
2
+ 4x + 6y – 3 = 0 and the parabola
x
2
+ 4x = 6y + 14.
Solution : Circle is x
2
+ y
2
+ 4x + 6y – 3 = 0
? (x + 2)
2
+ (y + 3)
2
= 16
x
(0,4)
(–2 3,2) ? (2 3,2) ?
Shifting origin to (–2,–3).
X
2
+ Y
2
= 16
equation of parabola ? ? (x + 2)
2
= 6(y + 3)
? X
2
= 6Y
Solving circle & parabola, we get X = ± 2 3
Hence they intersect at
? ?
2 3,2 ? &
? ?
2 3,2
2 4
2
0 2
A 2 6Y dY 16 Y dY
? ?
? ? ?
? ?
? ?
? ?
4
2
3 / 2 2 1
0
2
2 1 16 Y
2 6 Y Y 16 Y sin
3 2 2 4
?
? ?
? ?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ?
4 3 16
3 3
? ?
?
? ? ? ?
? ?
? ?
sq. units
JEEMAIN.GURU
E
5
J E E - M a t h e m a t i c s
Do yourself : 3
(i) Find the area inside the circle x
2
–2x + y
2
– 4y + 1 = 0 and outside the ellipse x
2
–2x+4y
2
–16y+13= 0
(b ) If the equation of the curve is in parametric form, then
t
t
dx
A y .dt
dt
? ?
? ?
?
?
or
t
t
dy
x .dt
dt
? ?
? ?
?
, where ? ? ?? ? are
values corresponding to values of x and ? & ? are values corresponding to values of y.
Illustration 12 : Find the area bounded by x-axis and the curve given by x = asint, y = acost for  0 ? t ? ? ?.
Solution : Area =
0
dx
y .dt
dt
?
?
=
0
a cos t(a cos t)dt
?
?
=
2 2
0 0
a a sin 2t
(1 cos2t)dt t
2 2 2
?
?
? ? ?
?
=
2 2
a a
2 2
?
? ?
Alternatively,
Area  =
0
dy
x .dt
dt
?
?
=
0
a sin t( a sin t)dt
?
?
?
=
2
0
a
(cos2t 1)dt
2
?
?
?
=
2
0
a sin 2t
t
2 2
?
? ?
2
a
2
?
?
Illustration 13 : Find the area of the figure bounded by one arc of the cycloid x = a(t – sint), y = a(1 – cost) and the
x-axis.
Solution : To find the points where an arc cuts x-axis
a(1 – cost) = 0 ? t = 0, ?
Area =
2 2
0 0
dx
y dt a (1 cos t) dt
dt
? ?
? ?
? ?

2
0
3 sin 2t
a t 2 sin t
2 4
?
? ? ?
2
2
3 3 a
a
2 2
? ? ? ?
? ?
? ?
? ?
Do yourself - 4 :
(i) Find the area of the loop of the curve :
(a) x = 3t
2
, y = 3t – t
3
(b) x = t
2
– 1, y = t
3
– t
(c) If y = ƒ (x) is a monotonic function in (a, b), then the area bounded by the ordinates at x = a, x = b,
y = ƒ (x) and y = ƒ (c) [where c ? (a, b)] is minimum when
a b
c
2
?
? .
Proof : Let the function y = ƒ (x) be monotonically increasing.
y=ƒ (c)
x
y=ƒ (x)
x=a
x=c
x=b
y
O
Required area A =
c b
a c
[ƒ(c) ƒ(x)]dx [ƒ(x) ƒ(c)]dx ? ? ?
? ?
For minimum area,
dA
0
dc
?
?
[ƒ '(c).c ƒ(c) ƒ '(c)a ƒ(c)] [ ƒ(c) ƒ '(c).b ƒ '(c).c ƒ(c)] 0 ? ? ? ? ? ? ? ? ?
?
a b
ƒ '(c) c 0
2
? ? ?
? ?
? ?
? ?
? c =
a b
2
?
( ??ƒ '(c) ? 0)
JEEMAIN.GURU
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;