1

CURVE SKETCHING
There are 5 steps to draw a curve, y = f(x)
Step 1 :
CHECK FOR SYMMETRY
1. Check whether function is even.
(a) Replace x by – x
(b) If no change in f(x), then function is even
(c) Graph is symmetrical about y-axis.
2. Check if function is odd
(a) Replace x by – x
(b) If f (– x) = – f(x), then function is odd
(c) Graph is symmetric about origin
3. Replace y by – y, if no change, then the graph is symmetrical about x-axis.
4. If function is periodic, draw graph only in the period, as the remaining curve is a repetition of
this one.

Step 2 :
POINTS OF INTERSECTION
(a) Find points of intersection with x-axis & y-axis.
for x axis, put y = 0
y axis, put x = 0
(b) Special care of double & triple roots
for example, if f (x) = (x – a)2 g (x)
here x = a, is a double root of f (x) or you can also say repeated root of f (x) = 0
note in this case f (a) = 0, f´(a) = 0 but f´´ (a)  0 i.e. can be point of local maxima / local
minima

now take f (x) = (x – a)3 g (x)

 x = a is a triple root of f (x)
here, note
f (a) = 0
f´(a) = 0
f´´ (a) = 0, but
f´´´(a)  0

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 a is the point of inflection (as shown in diagram) i.e. the point where shape of the
curve changes from convex to concave & vice-versa.
 TIP : if f (x) = (x – a)n g (x)
& nth derivative is the first non zero derivative then

if n is even ⎯⎯→ x = a is a point of local maxima / minima

n is odd ⎯⎯→ x = a is a point of inflection

let us take another example

y = x (x – 1)2
we can see that x = 1 is a double root of f (x)
Step 1 : no symmetry
Step 2 : has double root at x = 1
Step 3 : put f (x) > 0
 x (x – 1)2 > 0

· + +
0 1
double root so
no sign change across it.
 for x  (0, ) y > 0
x  (–  0) y < 0
so approximate graph would be

0 1

from this we can also deduce that x = 1 will be point of local minima & not local
maxima. If it was local maxima then the curve will come in the negative half.
Step 3 :
SIGN OF GRAPH
(a) Put y > 0 & see where the graph is above x-axis & where it is below x-axis.
for example if f (x) = x (x – 1) (x – 2)
put x (x – 1) (x – 2) > 0
(using wavy curve) 0 1 2

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So corresponding to it the graph will be above x-axis where region is +ve & below x-axis
where region is represented by – ve sign.

Further x (x – 1) (x – 2) is a polynomial function, so it will be continuous & hence the

following will be the curve.

Check the domain : Find the domain (especially in cases where denominator can become
zero). Draw the curve only in its domain.

Step 4 :
ASYMPTOTE
Examples of asymptote

y=Logx

x=0 is a vertical
asymptote
here x = 0 is a vertical asymptote
& y = 0 is a horizontal asymptote

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(a) Horizontal asymptote

• If lim f (x) = c1 , where c 1 is finite then y = c1 is horizontal asymptote in right half.

x→

• if lim f (x) = c2 , where c is finite then y = c is horizontal asymptote in left half.

x → − 2 2

for example
y = tan–1 x

y = tanx y = Log x

x = /2, a vertical asymptote x = 0, a vertical asymptote

Take special care in case of

(a) when denominator can be zero. Put denominator = 0 to get the value of x for which vertical
asymptote will be formed.

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(b) Logarithmic functions

for log (f(x)) = 0, find f (x) = 1 & for those values of x it will be a vertical asymptote.

Step 5 :
• Find the points of maxima, minima.
• find the shape of curve, i.e. either concave or convex.

Concave Convex

No line segment lies above the graph No line segment lies below the graph

A twice differentiable function f defined on interval I is

(a) concave if and only if f´´(x)  0 for x  I
(b) convex if and only if f´´(x)  0 for x  I
inflection is the point where the shape of the curve changes from convex to concave or vice-versa.

Sketch of some common cur ves :

(a) Linear inequality, straight line curve
ax + by  c or ax + by  c
• convert inequality into equality, to obtain an equation of straight line.
• For deciding which region is the answer put origin in the line and check for the region
required.
e.g 3x + 5y  15

now put origin (0, 0) in the inequation.

(0, 0) satisfies the inequality

 shaded region is the required area.

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(b) Circle
general form : x2 + y2 + 2gx + 2fy + c = 0
standard form : (x – h)2 + (y – k)2 = r2
for inequality (x – h)2 + (y – k)2  r2 or  r2

(x –h)2 + (y – k)2 < r2

(region inside the circle) 2 2 2
(x –h) + (y – k) > r
(region outside the circle)

• y2  4ax ⎯⎯→ region inside the parabola

• y2 > 4ax ⎯⎯→ region outside the parabola

Ar ea of Bounded Regions
1. For a continuous function f (x) defined over [a, b], the area bounded by the curve y = f (x), the
x-axis and the ordinates x = a and x = b is given by

b b

 f (x) dx or  y dx
a a

Other types
b

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2. When f (x) is below x-axis then the value of integral will come
out to be negative but as we know area is a positive quantity.

b b

 Area = −
 f (x) dx =  f (x) dx
a a
y = f (x)

3. Area of such curves is obtained by using the above 2 types

Area = Area (I) + Area (II) + Area (III)

b c d c

=  f (x) dx −  f (x) dx +  f (x) dx

a b c

= (( f (x) − g(x)) dx

a

The result will be the same whether the curves meet in 2 points or 1.

intersecting at one point on inter section

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Sometimes it is easier to take area with y-axis as in following cases :

1. Area =
 f ( y) dy
a

Note : The function inside the integral is a function of y not x.

It is because we have to integrate with respect to y.

2. Area = −
 f (y) dy
a

=  f ( y) dy

b c

4. Here area will be

area =
(right – left)dy
a

=  f ( y) − g( y)dy
a

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see you can use any set of formula, but choose such a one which makes your calculation easier.
let us discuss some cases.

for such cases, use

( f (x) − g(x)) dx
for such cases use ( f ( y) − g( y)) dy
because to integrate wrt x is very

( f ( y) − g( y)) dy
Illustration 1

Draw the rough sketch of the curve y = and find the area under the curve above
x-axis and between x = 0 and x = 4.
Solution :

The equation y = represents the upper half of parabola y2 = 3x + 4 whose axis of

symmetry is x-axis, focus on positive direction of x-axis and vertex also on x-axis. When y = 0,
4  4 
x = − 3 , therefore vertex is at  − 3 , 0  . The parabola opens on the right, therefore it also intersects
 
y-axis. Putting x = 0, we get y =  2, therefore parabola passes through points (0, 2) and
(0, – 2) as shown. x = 0 is equation of y-axis and x = 4 is equation of line parallel to y-axis as
shown. The region whose area is to be determined is the shaded portion shown.

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The region is bounded between two parallel lines x = 0 and x = 4
 required area

3 4

=  y dx = 
0 0
dx Y

4
2 (3x + 4)3/ 2 
= 3 
3 0
X´ X
2
= [16) 3/ 2
− (4) 3/ 2
]
9

Illustration 2

Y
2
y= 9 – x2
3

X´ (3, 0) (–3, 0) X

x=3
Y´
x=O

The region is symmetric about both coordinates axes. We determine the area of region in first
quadrant and multiply it by 4 to get the required area. The region in first quadrant is bounded
between two parallel lines x = 0 and x = 3.

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 required area

= 4 0
y dx x2
9
+
y2
4
=1

y2
= 9−x
2

4 9

2
 y=
3

3
3 2 8 x 9 x
= 4 3 dx = 
3 2
+
2
sin−1 
3 0
0

8 9 9  8 9 
= 0 + sin−1 (1) + 0 − (0) = . = 6 sq. units.

3  2   
2  3 2 2

Illustration 3

Find the area bounded by the curve y2 = 4ax and the line y = 2a and y-axis.
Solution :
The equation y2 = 4ax represents a parabola whose vertex is at origin, axis of symmetry is x-axis
and it opens on right side with focus on positive direction of x-axis, y = 2a is equation of straight
line parallel to x-axis as shown. The region whose area is to be determined is the shaded
portion.
The region is bounded between two parallel lines y = 0 and y = 2a.
 required area

2a 2a
y2
=  x dy =  4a dy
0 0
Y

1  2a 
1  y3  

2a y2 dy
= 4a = 4a 3 
0
 0  X´ X

1 8a3  2a2
 −
= 4a 0 = sq. units. Y´
 3  3

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Illustration 4
Find the area of the region included between the parabola y2 = x and the line x + y = 2.
Solution :
The equation y2 = x represents a parabola whose vertex is at origin, axis of symmetry is x-axis
and it opens on the right with vertex on positive direction of x-axis.
For finding the points of intersection of y2 = x and x + y = 2, we solve them to get
(2 – x)2 = x  4 + x2 – 4x = x
 x2 – 5x + 4 = 0
 (x – 4) (x – 1) = 0
 x = 4 and x = 1
 y = – 2 and y = 1
The region whose area is to be determined is the shaded portion.

X´ X
 − 

 1  (− 8) 
 Y´
     

 3   1 8  15
2 + 6 − 3 + 3 = 2 − 3 = 2
9 sq. units.
=
   

Illustration 5

x2 y2
Find the area of smaller region bounded by the ellipse + = 1 and the straight line
a2 b2
x y
+ =1
a b

x2 y2
Solution : The equation + = 1 represents an ellipse.
a2 b2

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The region whose area is to be determined is the shaded portion.

The region is bounded between two parallel lines x = 0 and x = b.
 required area Y

a a

=  ( y of ellipse) dx −  ( y of line) dx
0 0
X
a a =1
b b
= 
0
a
dx −
 a (a − x) dx
0 Y´

  a
b x a2 −1 a  b (a − x)2 
 + x    
= a 2  
sin
2 a  − a − 2 
 0   0 

b  a2 −1  b a2 
= a 0 + 2 sin (1) − a  2 
   
ab  ab ab  
= . − =  2 − 1  sq. units.
2 2 2 2
 

Illustration 6

Sketch the region bounded by the curve y= and y = |x – 1| and find its area.

Solution :

The equation y = represents the upper half of the circle x2 + y2 = 5 whose centre is at
origin and radius 2.
The region whose area is to be determined is the shaded region.
For finding the points of intersection of Y

y= and y = |x – 1|, we solve them to get

|x – 1| =
(x – 1)2 = (5 – x2)
 x2 – 2x + 1 = 5 – x2
X´ X
 2x2 – 2x – 4 = 0
 x2 – x – 2 = 0
 (x + 1) (x – 2) = 0
 x = – 1 and x = 2 Y´

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The region is bounded between two parallel lines x = – 1 and x = 2
 required area

2 2

=  ( y of circle) dx −  ( y of modulus function) dx

−1 −1

2 2

=  dx −
 x − 1 dx
−1 −1

2 1 2 
=  


dx −  − (x − 1) dx + (x − 1) dx 

 −1 
−1  1 


 
− −  +  
− 1  − 1 1 

 
 

5
 
   

Illustration 7
Find the area of the smaller region bounded by the curves x2 + y2 = 4 and y2 = 3 (2x – 1).
Solution :
Y
The equation x2 + y2 = 4 represents a circle with
centre at origin and radius 2 and y2 = 3 (2x – 1) is
equation of a parabola whose axis of symmetry is
x-axis and it opens on the right. The vertex is on
x-axis, putting y = 0, we get x = 1/2, therefore
(1/2, 0) is the vertex. X
X´
For finding the points of intersection of + x2 y2 =4
and y2 = 3 (2x – 1), we solve them to get
x2 + 3 (2x – 1) = 4
 x2 + 6x – 7 = 0
 (x + 7) (x – 1) = 0 Y´
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 x = 1 and x = – 7 (not possible)

The region whose area is to be determined is the shaded portion.
The region is symmetric about x-axis and region in Ist quadrant can be divided into two regions
bounded between parallel lines x = 1/2 and x = 1 and x = 1 and x = 2 respectively.
 required area

1 2 
 
1/ 2

= 2  ( y of parabola) dx +  ( y of circle) dx 

1

 

1 2

= 
2


dx +
 dx


1/ 2 1

 
  +  
  


 1
  
  

 
+− =  + 
   

Illustration 8
Find the area of the region bounded by the x-axis and the curves defined by
   3
y = tan x, −  x ; y = cot x,  x
3 3 6 2
Solution :
To find the area hold by x-axis and curves
y = tan x, –/3 < x < /3 ...(1)
and y = cotx /6 < x < 3/2 ...(2)
The curves interesect at P, where tan x = cot x, which is satisfied at x = –/4 within the given
domain of x.

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The required area is shaded area
 /4  /3
A=  /6 y1 dx +  / 4 y2 dx
where
y1 = tan x
/6 /4/3/2  3/2
y2 = cot x X
 /4  /3
 A =  /6 tan xdx +  /4 cot x dx
 /4 3 /
= [log sec x] /6 + [log sin x ] / 4

 2  +  log 3 1 
− log  
=  log 3 2 − log 2

   


 

Illustration 9

The given curves are

x2 + y2 = 4 ...(1)

x2 = − y ...(2)

x= y ...(3)
It is clear from the equations of curves that eq. (1) represents a cricle with centre (0,0) and radius
2. (2) represents a downward parabola with vertex at origin.
(3) represents a straight line, through origin making as anlge of 45 with +ve direction of x-axis.
Now points of intersection of (1) and (2)–
Substituting the value of x2 from (2) in (1), we get
y2 − y−4=0

 y= 2  2 + 16 23 2
=
2 2

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4 2 −2 2
= or = 2 or −
2 2

 x2 = (− 2 )
[rejective –ve value of y as x2 can not be –ve]
 x2 = 2

 x= 

 Pts of intersection of (1) and (2) are ( 2, − 2), (− 2, − 2) .

Also for point of intersection of (1) and (3)–

Solving eq. (1) and (3), we get

2x2 = 4  x2 = 2  x =    y= 

  ( yC − yP ) dx

− − (-

0 −

0 2
x 4 x  x2   −x3 
= 2 + sin−1 −   −  

2 2 2 0  2 −  3 2 0

 2 −1  2  −2 
−
2  2 + 2sin  2   2  −
=  
  

  2 1
= 1 + 2.  − 1 − =  +
 4 3 3

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Illustration 1 0
Find the area bounded by the curves, x2 + y2 = 25, 4y = |4 – x2| and x = 0 above the x-axis.
Solution :
We have to find the area bounded by the cruves
x2 + y2 = 25
4y = |4 – x2|
x=0
above x-axis

4 − x2 if x2  4
Now, 4 y = 4 − y2 = 
x − 4 if x  4
2 2

4 − x2 if − 2  x  2
4y = 
x − 4 if x  2 or  − 2
2

Thus we have three curves

(I) Circle x2 + y2 = 25
(II) P1 = Parabola x2 = –4(y – 1), –2 < x < – 2
(III) P 2 = Parabola x2 = 4(y + 1), x > 2 or x < – 2
(I) and (II) intersect at – 4y + 4 + y2 = 25
or (y – 2)2 = 52  y – 2 = ±5
y = 7, y = –3
y = –3, 7 are rejected since. y = –3 is below x axis and y = 7 gives imaginary value of x.
(I) and (III) intersect at
4y + 4 + y2 = 25 or (y + 2)2 = 52
y + 2 = ±5
y = 3, –7
y = –7 is rejected, y = 3 gives the points above x-axis. When y = 3, x = ±4. Hence the points of
intersection of (I) and (III) are (4, 3) and (–4, 3). Thus we have the shape of the curve as given
in figure.
Required area is
= 2 4 y y dx
4
dx− 2
y dx−
0 circle 0 P1
2 p2 
 4 1 2 1 4 
= 2  25 − x2 dx −  (4 − x2 )dx −  (x2 − 4)dx
 0 
4 0 4 2

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 4
1 3 2 4
4 25 sin−1 x 1  x3 − 4x 
= 2 + − 4x − x  − 
 0 2 5 0 4  3  4  3 2
0

= 2 2 25 −1 4  − 1  8 − 8  −1 
 64  8 
 + 2 sin 5  4  3  4 3 − 16  −  − 8
3
   

 25
= 26 + sin −1 4 − 4 − 4 − 4 
 2 5 3 3 3

4 4
= 12 + 25sin−1 − 8 = 4 + 25sin−1
5 5

Illustration 1 1

( / 4, 1) ( / 4, 1)
dx
 Equation of tangent at P is
y – 1 = 2 (x – /4)
or y = 2x + 1 – /2 ...(2)
The graph of (1) and (2) are as shown in the figure.
− 2  Y
Tangent (2) meet x-axis at, L ,0
 4 
 

Now the required area = shaded area X

= Area OPMO – Ar (PLM)

/4

1 LM.PM
= y(1) dx −
0 2

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/4 1
=
0
tan xdx − 2 (OM − OL) PM

1   − 2 
[log sec x] /4 − − .1
= 0  
2 4 4 
1 1
= log 2 − Sq. units. Ans.

2  
2 

Illustration 1 2

1
Sketh the curves and identify the region bounded by x = , x = 2, y = In x and y = 2x. Find
2
the area of this region.

 
 ex 

1
 x= or x= 1
e
At x = 1/e or ex = 1,
log x = – log e = – 1, y = – 1
1 
,−1
so that   is one pt. of intersection and at x = 1, log 1 = 0  y = 0
e 

 (1, 0) is the other common pt. of the curves.

1
Now in between, i.e.,  x1
or 1 < ex < e
e

1
and log   < log x < log 1
 e

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or – 1 < log x < 0

i.e. log x is – ve, throughout.

loge x
y1 = ex loge x, y2 =
ex

1
 x1
Clearly under the condition stated above y1 < y2 both being – ve in the interval
e
The rough sketch of the two curves is as shown in fig. and shaded area is the required area.
 The required area = shaded area
1 1  log x
= 1/ e
( y1 − y2 ) dx =
 ex log x − ex  dx
1/ e  
Y

 e



X
A
 x2
  −  
 1/ e e  1/ e

 1  
= e  − 4 −  − −  e  2 
  

 1 −e 3
e − + + + +


e e2 − 5
= − = = Ans.
4e

Illustration 1 3
2
Sketh the region bounded by the curves y = x2 and y = . Find the area.
1 + x2
Solution :

The given curves are y = x2 and y = 2

2 . Here y = x2 is upward parabola with vertex at origin.
1+ x
2
Also, y = is a curve symmetrical with respect to y-axis.
1 + x2

AREA
22

dy − 4x
At x = 0, y = 2 =  0 for x > 0
dx (1 + x2 )2

 Curve is decreasing on (0, )

dy
Moreover = 0 at x = 0
dx
 At (0, 2) tangent to curve is parallel to x-axis.
As x → , y → 0
 y = 0 is asymptote of the given curve.
For the given curveÊs pt of intersection : solving their equations we get x = 1, y = 1, i.e., (1, 1).
Thus the graph of two curves is as follows :
1 2 
 The required area = 2  1 + x
0 2
− x2  dx
 

 3  
1
4 tan−1 x − 2x  = 4. − 2 x
= 
 3  4 3
0

2
= − sq. units.
3

Illustration 1 4
Find the possible values of b > 0, so that the area of the bounded region enclosed between

x2
the parabolas y = x – bx2 and y = is maximkum.
b
Solution :
The given curves are
y = x – bx2 ... (1) and y = x2/b ...(2)
2
 1   1 
=−b x−
  y − 4b   2b 
and x2 = by
  

Here clearly first curve is a downward parabola with vertex at  1 , 1  and meeting x-axis at
2b 4b
 
(0, 0) and (1/b, 0) while second is an upward parabola with vertex at (0, 0).

 b b 
Solving (1) and (2) we get the intersection points of two curves at (0, 0) and  ,
2 2
 1 + b (1 + b ) 
2

AREA
 23

Hence the graph of given curves is as below

Shaded portion represents the required area, given by

b x 
2
A =  1 + b2  x

2
− bx −
b 
dx
0 

b
 x2 bx3 x3 1 + b2
 A =  2 − 3 − 3b 
 0

b2 − b4 b2
= −
2 (1 + b2 )2 3 (1 + b2 )3 3 (1 + b2 )3

1
 [2b (1 + b2)2 – (1 + b2) . 2b . b2] = 0

2b (1 + b2) (1 + b2 – 2b2) = 0
 1 – b2 = 0
 b = 1, – 1
but given that b > 0  b =1

Illustration 1 5

 1 
Let O (0, 0), A (2, 0) and B   be the vertices of a triangle. Let R be the region consisting

of all those points P inside OAB which satisfy, where d denotes the distance from the
points to the corresponding line. Sketch the region R and find its area.
Solution :
OAB is the given . Consider OI the  bisector of BOA. We know any point on OI must be at
equal distances from OB and OA.
Thus for d (P, OA)  min [d (P, OB), d (P, AB)]
Point P must lie under OI and similarly under AI. Thus within or on OIA
 Req. area = Ar ( OIA)

AREA
24

1/ 1
Now, tan BOA = =
1 3

 BOA = 30
 IOA = 15
Similarly IAO = 15
 Eqn of OI is y = (tan 15) x ...(1)
n
Eq of IA is y = (– tan 15) (x – 2) ...(2)
with point of intersection I (1, tan 15)
 Required area

1
2

= (tan 15) xdx + 

0
1
− tan 15 (x − 2)dx
Y



  
   

A
 1 
X
  

Illustration 1 6
Let f (x) be a continuous function given by

f (x) = f (x) =  2x | x | |

 2 
 x + ax + b, |x | |


Find the area of the region in the third quardrant bounded by the curves x = 2y2 and
y = f (x) lying on the left of the line 8x + 1 = 0.
Solution :

x2 + ax + b ; x  − 1

2x ; −1 x1
 2
x + ax + b ; x  1

 f (x) is continuous at x = – 1

AREA
 25

 (– 1)2 + a (– 1) + b = – 2
and 2 = (1)2 + a.1 + b
i.e., a –b = 3
and a+b=1
On solving we get a = 2, b = – 1

x2 + 2x − 1 ; x  − 1

 f (x) = 2x ; −1 x1
 2
 x + 2x − 1 ; x  1
x = -1 x = -1/8

Given curves are

y = f(x), x = – 2y2 and 8x + 1 = 0
Solving x = – 2y2, y = x2 + 2x – 1 (x < – 1) we get x = – 2.
Also y = 2x, x = – 2y2 meet at (0, 0)

 1 − 1
y = 2x and x = – 1/8 meet at  − ,
 8 4 
The required area is the shaded region in the figure.
 Required area

   
 
1 1/ 8
 − (x2 + 2x − 1)  dx + − 2 x dx
=  
− 2   −1  

 1 2(− x)3/ 2 x3 
−1  1 2(− x)3/ 2 − 1/ 8
2
=  − − x 2
+ x +  − x 
 2 3 3 − 2  2 3 − 1

 2 1  4 8   2 1 2 
+ − 1 − 1 − + − 4 − 2 + . 1   −1
=  3 3  3 3   3 − 64 − 3 

     

 2 − 5   4 + 8 − 18   4 − 3   2 − 3 
 
=  3 −   +  192 −  3 
 3 
   

257
= square units
192

AREA
 AREAS (EAMCET)
OBJECTIVES

1. Area bounded by y = x sin x and x − axis between x = 0 and x = 2 , is

(a) 0 (b) 2 sq. unit (c)  sq. unit (d) 4 sq. unit

2. Area bounded by the parabola y = 4 x 2, y − axis and the lines y = 1, y = 4 is

7 7
(a) 3 sq. unit (b) sq. unit (c) sq. unit (d) None of these
5 3
2
3. Area bounded by the curve y = xe x , x − axis and the ordinates x = 0, x = a

e
a2
+1 a2
−1
(a) sq. unit (b) e sq. unit (c) 2
e a + 1 sq. unit (d) e a2
− 1 sq. unit
2 2

4. Area bounded by curve y = x 3 , x − axis and ordinates x = 1 and x = 4, is

127 sq. unit (d) 255
(a) 64 sq. unit (b) 27 sq. unit (c) sq. unit
4 4

5. The area of smaller part between the circle x 2 + y 2 = 4 and the line x = 1 is

4 8 4 5
(a) − 3 (b) − 3 (c) + 3 (d) + 3
3 3 3 3

6. Area under the curve y = x 2 − 4 x within the x-axis and the line x=2, is
16 16 4
(a) sq.unit (b) − sq.unit (c) sq.unit (d) Cannot be calculated
3 3 7

7. The ratio of the areas bounded by the curves y = cos x and y = cos 2x between x = 0, x=/3

and x − axis, is
(a) 2 :1 (b) 1 : 1 (c) 1 : 2 (d) None of these

8. The area bounded by the parabola y 2 = 4 ax, and two ordinates x = 4, x = 9 is

152 a
(a) 4a2 (b) 4a 2 .4 (c) 4 a 2 (9 − 4) (d)
3

9. If the ordinate x=a divides the area bounded by the curve y = 1 + 8 , x − axis and the
 2
 x 

ordinates x = 2, x=4 into two equal parts, then a =

(a) 8 (b) 2 2 (c) 2 (d) 2
10. Area bounded by the curve y = log x , x − axis and the ordinates x = 1, x = 2 is

(a) log 4 sq. unit (b) (log 4 +1) sq. unit (c) (log 4 − 1) sq. unit (d) None of these

11. Area bounded by the lines y = x, x = −1, x = 2 and x − axis is

5 3 1
(a) sq. unit (b) sq. unit (c) sq. unit (d) None of these.
2 2 2

3
12. If the area above the x-axis, bounded by the curves y = 2kx and x=0 and x = 2 is , then
ln 2

the value of k is
1
(a) (b) 1 (c) −1 (d) 2
2

13. Area bounded by parabola y2=x and straight line 2y = x is

4 2 1
(a) (b) 1 (c) (d)
3 3 3

14. The area of the region bounded by the x − axis and the curves defined by
y = tan x,(− / 3  x   / 3) is

(a) log 2 (b) − log 2 (c) 2 log2 (d) 0

15. The area bounded by the circle x 2 + y 2 = 4, line x=3y and x − axis lying in the first
quadrant, is

(a) (b)  (c) 
(d) 
2 4 3

y = tan x, tangent drawn to the curve at x=
16. Area of the region bounded by the curve and
4

the x-axis is
1 1 1
(a) (b) log 2+ (c) log 2 − (d) None of these
4 4 4

17. The area of figure bounded by y = e x , y = e −x and the straight line x = 1 is

1 1
(a) e+ (b) e− (c) e + 1 − 2 (d) e + 1 + 2
e e e e

18. The area of the region bounded by y =| x − 1| and y = 1 is

1
(a) 2 (b) 1 (c) (d) None of these
2
19. The area enclosed by the parabolas y=x2−1 and y = 1 − x 2 is

(a) 1/3 (b) 2/3 (c) 4/3 (d) 8/3

20. The part of straight line y=x+1 between x=2 and x=3 is revolved about x-axis, then the
curved surface of the solid thus generated is
(a) 37/3 (b) 7 2 (c) 37  (d) 7 / 2

21. The area bounded by y = −x 2 + 2 x + 3 and y = 0 is

32 1 1
(a) 32 (b) (c) (d)
3 32 3

22. The area bounded by the curves y2=8x and y = x is

128 32 64
(a) sq. unit (b) sq. unit (c) sq. unit (d) 32 sq. unit
3 3 3


23. The area bounded by curves y = cos x and y = sin x and ordinates x=0 and x= is
4

(a) 2 (b) 2 +1 (c) 2 −1 (d) 2 ( 2 − 1)

24. Area bounded by the parabola y 2 = 4 ax and its latus rectum is

2 4 8 3
(a) a 2 sq. unit (b) a 2 sq. unit (c) a 2 sq. unit (d) a 2 sq. unit
3 3 3 8

25. The area bounded by the curve y=4x−x 2

and the x − axis, is
30 31 32 34
(a) sq. unit (b) sq. unit (c) sq. unit (d) sq. unit
7 7 3 3

26. The area of the region bounded by the curves y=x2 and y =| x | is

(a) 1/6 (b) 1/3 (c) 5/6 (d) 5/3

27. The area enclosed between the parabolas y2=4x and x 2 = 4 y is

14 3 3 16
(a) sq. unit (b) sq. unit (c) sq. unit (d) sq. unit
3 4 16 3

28. The area between the parabola y=x2 and the line y = x is

1 1 1
(a) sq. unit (b) sq. unit (c) sq. unit (d) None of these
6 3 2

29. Area included between the two curves y 2 = 4 ax and x 2 = 4 ay,

16
(a) 32
a2 sq. unit (b) 16 sq. unit (c) 32
sq. unit (d) a2 sq. unit
3 3 3 3
 30. Area bounded by curves y=x2 and y = 2 − x 2 is

(a) 8/3 (b) 3/8 (c) 3/2 (d) None of these

31. The parabolas y2=4x and x2=4y divide the square region bounded by the lines x=4,

y = 4 and the coordinate axes. If S1 , S 2 , S 3 are respectively the areas of these parts
numbered from top to bottom, then S1 : S 2 : S 3 is
(a) 2 : 1 : 2 (b) 1 : 1 :1 (c) 1 : 2 : 1 (d) 1 : 2 : 3
32. The part of circle x2+y2=9 in between y=0 and y=2 is revolved about y-axis. The
volume of generating solid will be
46 
(a) (b) 12 (c) 16 (d) 28 
3

33. The area bounded by the curves y = x , 2y + 3 = x and x − axis in the 1st quadrant is

27
(a) 9 (b) (c) 36 (d) 18
4

34. The area of the smaller segment cut off from the circle x2+y2=9 by x = 1 is
1
(a) (9 sec −1 3 − 8 ) (b) 9 sec −1 (3) − 8 (c) 8 − 9 sec−1(3) (d) None of these
2

35. The area of region {(x, y) : x 2 + y 2  1  x + y} is

2 2 2
    1
(a) (b) (c) (d) −
5 2 3 4 2

36. Area enclosed by the parabola ay = 3(a 2 − x 2 ) and x-axis is

(a) 4 a 2 sq. unit (b) 12 a 2 sq. unit (c) 4 a 3 sq. unit (d) None of these

37. Area inside the parabola y 2 = 4 ax, between the lines x = a and x = 4 a is equal to
2 2
a a
(a) 4 a2 (b) 8a2 (c) 28 (d) 35
3 3

38. The area of the curve xy 2 = a 2 (a − x) bounded by y-axis is

(a) a2 (b) 2a2 (c) 3a2 (d) 4a 2

39. The area formed by triangular shaped region bounded by the curves y = sin x, y = cos x and
x = 0 is

(a) 2 −1 (b) 1 (c) 2 (d) 1 + 2

40. The area bounded by the x-axis, the curve y = f(x) and the lines x = 1, x = b is equal to

b2+1 − 2 for all b > 1, then f(x) is

 x
(a) x−1 (b) x+1 (c) x2+1 (d)
1+x2

42. Area under the curve y = 3x+4 between x=0 and x = 4, is

56 64
(a) sq. unit (b) sq. unit (c) 8 sq. unit (d) None of these
9 9

43. The area bounded by curve y 2 = x, line y=4 and y-axis is

16 64
(a) (b) (c) 7 2 (d) None of these
3 3

44. For 0  x   , the area bounded by y = x and y = x + sin x, is

(a) 2 (b) 4 (c) 2 (d) 4

45. The area bounded by the straight lines x = 0, x = 2 and the curves y=2x,y=2x−x2

(a) 4 − 1 (b) 3
+
4 (c)
4
−1 (d) 3 4
−
3 log 2 log 2 3 log 2 log 2 3

46. The area between the curve y=4+3x−x2 and x-axis is

(a) 125/6 (b) 125/3 (c) 125/2 (d) None of these


47. The area between the curve y = sin 2 x, x − axis and the ordinates x=0 and x= is
2

(a) (b)  (c) 
(d) 
2 4 8

48. Area bounded by the curve xy − 3 x − 2y − 10 = 0, x-axis and the lines x = 3, x = 4 is

(a) 16 log 2 − 13 (b) 16 log 2 − 3 (c) 16 log 2 + 3 (d) None of these

49. The area of the triangle formed by the tangent to the hyperbola xy = a 2 and co-ordinate
axes is
(a) a2 (b) 2a 2 (c) 3a 2 (d) 4a2
 AREAS
HINTS AND SOLUTIONS

1. (d) Required area is A +A =  y dx +  y dx = 4 sq.unit

2
1 2

0 
Y

A1 2
O  A2 X

2. (c) Required area = 

1
4
x dy =  4

1 2
y dy

2
=1. | y 3/2| 4= 7
sq. unit.
1
2 3 3

a a 2
3. (b) Required area is y dx = xe x dx
0 0

dt
We put x 2 = t  dx = as x = 0  t = 0 and x = a  t = a2 , then it reduces to
2x

e −1
1 2 a2
a2 t [et ]a
= = sq. unit.

1 e dt
0
2 0 2 2

4  x 4 4 255
x dx =   =
(d) Required area
+ sq. unit.
3
4. =
1  4 1 4

+
2
5. (b) Area of smaller part =2 4 − x 2 dx
1
Y
x=1

(2, 0) X
(0,0)

8
= − 3 .
3
2
2
2 x3 4 x 16
6. (a)  (x − 4 x)dx = 

−  = sq. unit.
0  3 2 0 3

/3 /4 /3

(d) A1 = 0 cos x dx,
7. A2 =

0
cos 2x dx −
/ 4
cos 2x dx .


9
8. (d) Shaded area A=2 4 ax dx
4
Y x=4 x=9

y2=4ax
 2 152 a
A = 4 a  [x 3 / 2 ]9 = .
4
3 3

9. (b) Let the ordinate at x = a divide the area into two equal parts
Y

(2, 3) A  3
C B  4, 
 2

X
O M D N
(4,0)

Area of 4  8   8 4
AMNB =


1 + 2  dx = x − x  = 4
x 
2
 2
 

Area of ACDM = a 1 + 8
=2
 2 dx
2  x 

On solving, we get a = 2 2 ;Since a0 a=2 2 .

10. (c) Given curve y = log x and x=1, x=2.
2
Hence required area =  log x dx = (x log x − x)2 1
1

= 2 log 2 − 1 = (log 4 − 1) sq. unit.

11. (a) Required area 2 y dx =  y0 . dx +  y 2. dx =
5
sq. unit.
−1 −1 0 2
Y

x =2
(–1, 0)
X
x =1 (2,0)

2 3
12. (b)  2 kx dx =  22k − 1 = 3k . Now check from options, only (b) satisfies the above condition.
0 log 2

13. (a) y 2 = x and 2y = x  y 2 = 2y  y = 0, 2

2
area = ( y2 − 2 y)dy sq. unit.
 Required
 0

/3
14. (c) Required area = 2 tan x dx = 2[log sec x]0 / 3 = 2 log(2)
0 .


x

3 2
15. (c) Required area = dx + 4 − x 2 dx
0 3 3

Y
x=3y

(2,0) X
16. (d)
Y x = /4

X
(– / 2,0) (/4,0) ( /2, 0)

 /2
Shaded area =  tan x dx = [log sec x ] 0
/4

= log sec ( / 4) − log sec 0 = log 2 − log 1 = log 2 .

17. (c) Given equations of curves y = e x ; y = e −x and straight line x=1 We know that area of the
figure
1
bounded by the curves and straight line
1
= (e x − e −x )dx = [e x + e −x ] 1 = e + − 2.
0
0
e

18. (b) y = x − 1, if x  1 and y = −(x − 1), if x  1

Y x =1

y=1

x=2

x+y =1
y=x–1

Area 1
2  x 2 1  x 2 2
=  (1 − x)dx + (x − 1)dx = x − 2
 + 
−
x 
0 1  0  2 1
 1  1  1 1
=
− + −  − 1 = + =1.
 2   2
1 
   2 2

19. (d) Given parabolas are x2=1+y, x2=1−y

Y

(0, 1)
x2 = 1 + y

(0, –1) x2 = 1 – y
 1
1  x3 8
 = .
Required area =4
 0
(1 − x 2 ) dx = 4 x −
 3  0 3

b   dy  2 
20. (b) Curved surface =  a
2 y 1 +  dx  dx
   

Given that a = 2, b = 3 and y=x+1.

On differentiating with respect to x ,

dy dy
=1+0 or =1
dx dx

Therefore, curved surface

3 3
= + 2 (x +1)
2
[1 + (1)2 ] dx = 2
2 (x + 1) 2
dx

3
3 (x + 1)2 
2 

= 2 2  (x + 1) dx = 2
2  2  2


22
=  [(3 + 1)2 − (2 + 1)2 ] = 2 (16 − 9) = 7 2 = 7 2 .
2

21. (b) Given, y = −x 2 + 2 x + 3 and y = 0

Therefore, x = −1 and x=3

3
= (−x 2 + 2 x + 3)dx
 Required area  −1

3
 3 + x 2 + 3 x 32
= −x = .
 3   −1 3

22. (b) y 2 = 8 x and y = x  x 2 = 8 x  x = 0 ,8

 Required area =  (2
8
2x − x)dx
0

4 2 x 2 
8

=
128 64 32
x 3/2− = − = sq. unit.
 3 2 0 3 2 3


23. (c) Given equations of curves y = cos x and y = sin x and ordinates x = 0 to x= . We know that
4
 /4  /4
area bounded by the curves =   
x2
ydx = cos xdx − sin x dx
x 1 0 0
 (c) Area = 20 
a a
24. y dx = 2 4 ax dx
0

(a, 2a)

(a, 0)
X
O
(a –2a)

2 a 8
2  2 a  x 3 / 2 = a2 sq. unit.
3 0 3

25. (c) We have y=4x−x2 and y = 0 ; x = 0 , 4

=  (4 x − x
4x 2 x 3 4
4
Required area 2
)dx =  − 
0  2 3  0

26. (b)
Y
2 y=x
y=–x y=x

(–1,1) (1,1)
X

Required area = 2 (shaded area in first quadrant)

= 2
1 1 1
(x − x 2 ) dx = 2  = .
0 6 3

27. (d) Equations of curves y2=4x and x 2 = 4 y. The given equations may be written as y=2 x and
x2
y= .
4
4 4x 2 32 16 16
= x dx −
We know that area enclosed by the parabolas

0
2  0 4
dx =
3
−
3
=
3
sq. unit.

28. (a) Given curves are y = x 2 and y = x

On solving, we get x=0, x=1

O
X

1
A = (x 2 − x)dx
Therefore, required area  0

1
2
 x 3 − x  = 1 − 1 =1 sq. unit.
=
3 2  0 3 2 6
29. (d) Solving the two equations, we have x 4 = 64 a 3 x
 x = 0, 4a

Y
x2=4ay

y2=4ax

A P (4a,4a)
B
X
O M(4a,0)

4a
Required area =  
4a
2a
1 /2
x 1 / 2 dx − x2
dx
0 0 4a

30. (a) y = x2 . ... (i)

y=2−x2 . ... (ii)
 By equation (i) and (ii) , we get, x = 1

 y = 1 Y

y = x2
(0, 2)

(–1, 1) (1, 1)

X
y = 2 – x2

area = 2 
1
 Required
− x 2)dx − x 2 dx 
1

0 (2  0  

 2 x 3 1  x 3 1 2 8
= 2  2 x− 3  = 4  x −  = 4  3 = 3 .
 3
 0  0  

31. (b) y 2 = 4 x and x2=4y are symmetric about line y=x

 Area bounded between is 

4 8
y2=4x and y=x (2 x − x)dx =
0 3

(4, 4)
S1
S2
S3
X
  As 2
=
16
and As 1 = AS 3 =
16
3 3

 AS : AS : AS :: 1 : 1 : 1 .
1 2 2

32. (a) The part of circle x2+y2=9 in between y = 0 and y = 2 is revolved about y-axis. Then a
frustum of sphere will be formed.
The volume of this frustum
2 2


=  x 2 dy =  (9 − y 2 )dy
0  0

 
=  9y − 1 3  =  9  2− 1 (2)3 − (9.0 − 1 .0)
2
y 

 3  0 
 3 3 
 8  46
= − =  cubic unit.
 18
 3  3
33. (a) Solving y 2 = x and x = 2y + 3

4 y 2 = (x − 3)2 , 4 x = x 2 − 6 x + 9

 x 2 − 10 x + 9 = 0  (x − 1)(x − 9) = 0  x = 1, 9

B x=9
A
X
O (3,0)

x =1

= − 4 [x log x − x ]10 = −4(−1) = 4 sq. unit,

(□ lim x log x = 0) .
x→0

Required area = A+B = 3 xdx +  9
−  x − 3 
 x  dx
0 3  2 


1
34. (d) Required area =2 ( y − 1) dy ,(From the symmetry)
1/ 4

Y
y =(x+1)2 y=(x–1)2

1
(0,1)
y= 4
X
 On solving, we get required area = 1 sq. unit.
3

+
3
35. (b) Area of smaller part I=2 9 − x 2 dx
1

1 x 3    1 
= 2. x 9−x 2 + 9 sin−1  = 9 − − 9 sin−1  
8
2 3 1 2  3 
Y

(1, 0)

(3, 0)
X
(0, 0)

x=1

36. (d) x 2 + y 2 = 1, x + y = 1 meetwhen

x 2 + (1 − x)2 = 1  x 2 + 1 + x 2 − 2 x = 1

B (0, 1)
x2+y2 = 1

A (1, 0)
O y=1–x

 2 x 2 − 2 x = 0  2 x(x − 1) = 0

 x = 0, x = 1  y = 1, y = 0 , i.e., A (1,0); B (0,1)

1
Required area = +0[ 1 − x 2 − (1 − x)] dx

3
37. (a) The parabola meets x-axis at the points, where (a 2 − x 2 ) = 0  x = a. So the required area
a

 
a 3 6 a
= (a 2 − x 2 )dx = (a 2 − x 2 )dx = 4 a 2 sq. unit.
−a a a 0

38. (c) We have y 2 = 4 ax  y=2 ax

We know the equations of lines x=a and x = 4a

 The area inside the parabola between the lines

  x 2 
1 3 4a
A= 4ya dx = 4 2a
dx = 2 4a
x 2 dx = 2
  ax a  a 
3
a a a  
 2 a

a a−x
39. (a) Since the curve is symmetYrical about x-axis, therefore Required area A = 
2 a
0 x
dx

Put x = a sin2 
x=a
X
 dx = 2a sin . cos  d

/ 2
a cos 2 
A =2  0
a
a sin 2 
a sin 2 d

 /.2 cos 
= 2a 2  2 sin cos  d
0 sin

/2
A = 4a2  cos 2  d  1 
A = 4 a 2 . . = a 2
0 22

40. (a) Given required area has been shown in the figure.

x= is the point of intersection of both curve
4
Y
y = cos x

y = sin x
X
x =  /4

/ 4

Required area =  0(cos x − sin x) dx

 1 1 − 
= [sin x + cos x ] / 4
0 = + 1
 2 2 

2
= − 1=2−1.
2

(d) +1
b
f(x)dx = b 2 + 1 − 2 = b 2 + 1 − 1 +1 = [ x + 1]
2 b
41. 1

d
 f(x) = x2+1 =
2x
=
x
.
dx 2x +1 2
x +1
2

4 4
(3x + 4)3 / 2
42. (d) Area = 0 3x + 4 dx = 3.(3 / 2)
0

y=3x+4
B

O X

A
 2 112
=  56 = sq. unit.
9 9

43. (b) Required area = area of OABC – area of OBC

Y
y= 4
A (16,4)B

X
O C

y2 = x

16
16 x dx = 64 −  x 3 / 2 64
= 16  4 −
 0
3/2
 0
= 3.

44. (a) The curves y = x and y = x + sin x intersect at (0, 0) and ( ,  ) . Hence area bounded by the two
curves
  


= (x + sin x)dx −
0
 x dx =  sin x dx
0 0

= [− cos x ]0 = − cos  + cos 0 = −(−1) + (1) = 2 .

2
45. (d) Required area =  [2 x − (2 x − x 2 )]dx
0

2
2x x 3
=  − x 2+ 
log 2 3 0

46. (a) Solving y = 0 and y = 4 + 3 x − x 2 , we get x = −1, 4 . Curve does not intersect x-axis between x = −1 and

x=4.

 Area =  4(4 + 3 x − x 2 )dx = 125 .

−1 6
 /2  /21 − cos 2 x 
(b) Required area A=
 sin x . dx =

2
47. dxa
 
0 0  2 
3 x + 10
48. (c) Given curve is y(x − 2) = 3 x + 10  y =
x−2
3 x + 10
Required area is  
4 4
y dx = dx
3 3x −2

=[3 x + 16 log( x − 2)]43= 3 + 16 log 2 sq. unit.

 2
a
49. (b) Given xy = a 2 or y =.................................... (i)
x

There are two points on the curve (a, a),(– a,– a)

2a
X
2a
(–a,–a)

The equation of the line at (a, a) is,

 2
y−a=
 dy 
 (x − a) =  − a  (x − a)
 


dx 2
 (a, a)
 (a, a)

y − a = −(x − a) therefore, equation of the tangent at (a, a) is x + y = 2 a .The interception of line

x + y = 2a with x-axis is 2a and with y-axis is 2a.
1  2a  2a = 2a2 .
 Required area =
2
 AREAS
PRACTICE EXERCISE

1. The area bounded by y = 5x - x2 - 4 and the x -axis

9 3 9
1) sq.units 2) 9 sq.units 3) sq.units 4) sq.units
4 8 2 2

2. The area bounded by the curve y = (x - 1)2 - 25 and the x-axis is

200 300 400 500
1) sq.units 2) sq.units 3) sq.units 4) sq.units
3 4 3 3
3. The area bounded by x2 = 4y, x = 4y - 2
1) 9/8 sq.units 2) 9/4 sq.units 3) 9/16 sq.units 4) 3/2 sq.units
4. The area bounded by y2 = 4x and the line y = 2x-4
1) 18 sq.units 2) 9/2 sq.units 3) 9 sq.units 4) 3/2 sq.units
5. The area between the curves y = 8 - x2 and y = x2 in sq.units is
1) 32/3 2) 64/3 3) 16/3 4) 8/3
6. The area enclosed within the curve |x| + |y| = 1 is
1) 4 sq.units 2) 1 sq.unit 3) 2 sq.unit 4) 8 sq.unit
7. Area of the region bounded by y = 1 - |x| and the x-axis
1) ½ 2) 1 3) ¼ 4) 2
8. Area of the region bounded by y = [x], the x-axis and the coordinates x = 1,
x = 2 is
1) 2 2) 1 3) ½ 4) 1/3
9. The area bounded by y = x3 - 6x2 + 8x and the x - axis
1) 8 sq.units 2) 4 sq.units 3) 16 sq.units 4) 4 sq.units
2 2
10. The whole area bounded by x 3 + y 3 = 4 in sq. units

1) 24 2) 48 3) 12 4) 36

11. The area of the region bounded by the curve y=sinx and the x-axis between - and 
is

1) 8 sq.units 2) 4 sq.units 3) 2 sq.units 4) 1 sq.unit

12. The area bounded by one of the ac of y = cosax and the x-axis is
1 1 2
1) 2) 3) 4) 2
|a| a a |a|

13. The area between the curves y2 = x/2 and 3y2 = x + 1 in sq.units is
1) 4/3 2) 2/3 3) 8/3 4) 16/3
x2
x 2
14. The area between the curves y = and y = 3 - in sq.units is
4 2
1) 8 2) 16/3 3) 8/3 4) 12
1 1 1
15. The area of the region between the x-axis and the curve f(x) = x2 + x − in [0, 2] is
3 3 3 4 4 2
1) sq.units 2) sq.units 3) sq.units 3
4) sq.units
4 2 8 5
16. The area bounded by the x - axis, part of the curve y = 1 + 8/x2 and the ordinates at x=2
and x=4 is divided by the ordinate x = a into two equal parts. Then a =
1) 2 2) 2 3) 4 4)
17. The area bounded by the curve ay2 = x3, the x-axis and the ordinate x=a
2a 2
4a2 3a2
8a sq.units
2
2) sq.units 3) sq.units 4) sq.units
1)
3 5 5 5
18. The whole area bounded by a2y2 = a2x2 - x4 is
2 8 4 5a2
1) a2 sq.units 2) a2 sq.units 3) a2 sq.units 4) sq.units
3 3 3 3
19. The area bounded by a2y2 = x3(2a-x)
a2 a2
1) a2 sq.units 2) sq.units 3) 2a2 sq.units 4) sq.units
2 4
20. The area bounded by the line y = x curve and y = x3 is
1) 1 sq. units 2) ½ sq. units 3) 1/3 sq. units 4) ¼ sq. units
21. Area bounded by y = (x - 1)(x - 2)(x - 3) between x = 0, x = 3 in sq. units is
9
1) 2) 11 3) 7 4) 3
4 4 4 4

22. Area of the region bounded by y = ex and y = e-x and the line x = 1 in sq. units is
1 1 1 1
1) e + 2) e - 3) e + +2 4) e + - 2
e e e e
 23. The area bounded by y = x2, y = [x + 1], x  1, and the y-axis in sq. units is
1) 1 3 2) 2 3 3) 1 4) 7 3

24. The area bounded by the curve xy=4 and x-axis the ordinates x=2, x=4 in sq. units is
1) 4 log 2 2) 2 log 2 3) 8 log 2 4) log 2
25. The area of the curve x = acos3t, y = bsin3t in sq. units is
3 3  
1) ab 2) ab
3) ab 4) ab
4 8 4 8
26. The area bounded by one arc of y = sin2x and x-axis in sq. units is
1) 1 2) 2 3) 3 4) 4
27. The area bounded by the curve y = sinx-cosx. X-axis and x = 0, x = /2 in sq.units is
1) -1 2) 2( - 1) 3) 2( - 1) 4) 2( + 1)

28. Area of the region bounded by y = tanx, and tangent at x = and the x-axis in sq. units
4

2) log 2 + 
1 2
1) log - 3) log 2 − 4) log
4 16 4
29. The area bounded by y = cosx, y = x +1 and y = 0 in the second quadrant in sq. units is
1) 1/2 2) 3/2 3) 1/4 4) 5/4
x
2
y
2
x y
30. The area between + = 1 and the straight line + = 1 in sq. units is
a2 b2 a b
1 ab ab ab ab
1) ab 2) 3) 4) −
2 2 4 4 2
31. The area of the triangle formed by the positive x-axis and the normal and tangent to the
circle x2 + y2 = 4 at (1, 3 ) in sq. units is
1
1) 2) 3) 2 4) 2/ 3

32. The area of the region bounded by the curves y = |x-2|, x = 1, x = 3 and the x-axis is
1) 1 sq. units 2) 4 sq. units 3) 3 sq. units 4) 2 sq. units
 AREAS

Key for Practice Exercise

1 2 3 4 5 6 7 8 9 10

4 4 1 3 2 3 2 2 2 1

11 12 13 14 15 16 17 18 19 20

1 2 1 1 1 1 3 3 1 2

21 22 23 24 25 26 27 28 29 30

4 4 2 1 2 1 1 1 1 4

31 32

3 1