CHAPTER 72 AREAS UNDER AND BETWEEN CURVES
EXERCISE 283 Page 771
1. Show by integration that the area of the triangle formed by the line y = 2x, the ordinates x = 0 and
x = 4 and the x-axis is 16 square units.
A sketch of y = 2x is shown below.
Shaded area =
∫
4
0
yd=
x
∫
4
0
2x d =
x
[ x 2 ]=0
4
16 − 0 = 16 square units
2. Sketch the curve y = 3x2 + 1 between x = –2 and x = 4. Determine by integration the area enclosed
by the curve, the x-axis and ordinates x = –1 and x = 3. Use an approximate method to find the
area and compare your result with that obtained by integration.
A sketch of y = 3x2 + 1 is shown below
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© 2014, John Bird
Shaded area =
∫
3
−1
y d x=
Width of interval =
X
y = 3x2 + 1
∫ ( 3x
3
2
−1
+ 1) d x=
[ x3 + x ] −1=
3
(27 + 3) − (−1 − 1) = 32 square units
3 − −1
= 0.5
8
–1
4.0
– 0.5
1.75
0
1.0
0.5
1.75
1.0
4.0
1.5
7.75
2.0
13.0
2.5
19.75
3.0
28
Hence, using Simpson’s rule,
1
1

Area ≈ (0.5)  ( 4.0 + 28 ) + 4(1.75 + 1.75 + 7.75 + 19.75) + 2(1.0 + 4.0 + 13.0) 
3
2

=
1
(0.5) [16.0 + 124 + 36.0] = 29.33
3
If a greater number of intervals is chosen the area would be close to 32 square units
3. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = 5x;
x = 1, x = 4
A graph of y = 5x is shown below.
Shaded area =
∫
4
1
y=
dx
∫
4
1
4
 5x2 
5 x=
dx  =
(40) − (2.5) = 37.5 square units
 2  1
4. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = 2x2 – x + 1;
x = –1, x = 2
A sketch of y = 2 x 2 − x + 1 is shown below
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© 2014, John Bird
Shaded area =
∫
2
−1
yd=
x
2
 2 x3 x 2

 16
  2 1 
∫ −1 ( 2 x 2 − x + 1) d =x  3 − 2 + x  −=1  3 − 2 + 2  −  − 3 − 2 − 1
2
= 7.5 square units
5. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = 2 sin 2θ;
θ = 0, θ =
π
4
A sketch of y = 2 sin 2θ is shown below.
Shaded area =
∫
π /4
0
yd x = ∫
π /4
0

π /4
 π 
2sin 2θ d x = [ − cos 2θ ] 0 =  − cos 2    − ( − cos 0 )
 4 

= – 0 – –1 = 1 square unit
6. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
θ = t + et;
t = 0, t = 2
2
t2

4

Shaded area = ∫ (t + e ) d t =  + et  =  + e 2  − ( 0 + e0 ) = 8.389 square units
0
2
0  2

2
t
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© 2014, John Bird
7. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = 5 cos 3t;
t = 0, t =
π
6
A sketch of y = 5 cos 3t is shown below.
Shaded area =
∫
π /6
0
y d=
x
∫
π /6
0
5cos 3t d=
t
5
5
3π
5
π
π /6
[sin 3t ] 0= sin − sin 0= sin
3
3
6
2
 3
=
5
= 1.67 square unit
3
8. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = (x – 1)(x – 3);
x = 0, x = 3
A sketch of y = (x – 1)(x – 3) is shown below
Shaded area
=
∫
1
0
3
( x − 1( x − 3) d x − ∫ ( x − 1)( x − 3) d =
x
1
∫ (x
1
0
2
− 4 x + 3) d x − ∫
3
1
( x 2 − 4 x + 3) d x
 1
 
 x3
  x3


1

=  − 2 x 2 + 3 x  −  − 2 x 2 + 3 x  =  − 2 + 3  − ( 0 )  − ( 9 − 18 + 9 ) −  − 2 + 3  
3
0  3
 1  3

3

 
1
3
2
 1  1
= 1  −  −1  =2 = 2.67 square units
3
 3  3
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© 2014, John Bird
EXERCISE 284 Page 773
1. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
y = 2x3; x = –2, x = 2
A sketch of y = 2x3 is shown below.
Shaded area =
∫
2
0
0
2
−2
0
yd x−∫ yd x =∫
2
0
 2x4   2x4 
−
2x d x − ∫ 2x d x = 
−2
 4  0  4  − 2
3
0
3
= [(8 – 0) – 0 – 8)]
= 16 square units
2. Find the area enclosed between the curve, the horizontal axis and the given ordinates:
xy = 4; x = 1, x = 4
A sketch of xy = 4, i.e. y =
4
is shown below.
x
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© 2014, John Bird
Shaded area =
∫
π /6
0
yd x =
∫
4
1
4
4
d x = [ 4 ln x ]1 = 4 ln 4 − 4 ln1 = 4 ln 4 = 5.545 square units
x
3. The force F newtons acting on a body at a distance x metres from a fixed point, is given by:
x2
F = 3x + 2x2. If work done = ∫ F d x , determine the work done when the body moves from the
x1
position where x = 1 m to that when x = 3 m.
Work done =
∫
x2
x1
3
 3 x 2 2 x3   27
 3 2
F d x = ∫ ( 3x + 2 x ) d x = 
+
=  + 18  −  +  = 29.33 N m

1
3 1  2
 2
 2 3
3
2
4. Find the area between the curve y = 4x – x2 and the x-axis.
y = 4x – x 2 = x(4 – x). When y = 0, x = 0 and x = 4. A sketch of y = 4x – x 2 is shown below
4
x3 
64 


Shaded area = ∫ ( 4 x − x ) d x =  2 x 2 −  =  32 −  − ( 0 ) = 10.67 square units
0
3 0 
3 

4
2
5. Determine the area enclosed by the curve y = 5x2 + 2, the x-axis and the ordinates x = 0 and x = 3.
Find also the area enclosed by the curve and the y-axis between the same limits.
A sketch of y = 5x2 + 2 is shown below
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© 2014, John Bird
Shaded area =
∫
3
0
3
 5 x3

y d x = ∫ (5x + 2) d x = 
+ 2 x  = (45 + 6) − (0) = 51 square units
0
 3
0
3
2
The area enclosed by the curve y = 5 x2 + 2 (i.e. x =
y−2
), the y-axis and the ordinates y = 2 and
5
y = 47 (i.e. area ABC in the sketch above) is given by:
Area =
∫
y = 47
y =2
x d y=
∫
47
2
y−2
d y=
5

3
1  ( y − 2) 2

3
5 

2
1
1 47
2 d y=
−
y
(
2)
∫
5 2
47




  2
 3

1  45 2
=
− 0   = 90 square units



1.5
5


6. Calculate the area enclosed between y = x3 – 4x2 – 5x and the x-axis.
y = x3 − 4 x 2 − 5 x = x ( x 2 − 4 x − 5 ) = x( x − 5( x + 1)
Hence, when y = 0, x = 0 or 5 or –1
When x = 2, y = 2(–3)(1) = –6
A sketch of the graph y = x3 − 4 x 2 − 5 x is shown below
Shaded area =
0
5
 x 4 4 x3 5 x 2 
 x 4 4 x3 5 x 2 
−
−
−
−
−
=
−
−
−
−
−
x
4
x
5
x
d
x
x
4
x
5
x
d
x
(
)
(
)
∫ −1
∫0
 4
3
2  −1  4
3
2  0
0
3
2
5
3
2


 1 4 5    625 500 125 
= ( 0 ) −  + −   − 
−
−
 − ( 0 )
3
2 
 4 3 2    4


= (0.91666) – (–72.91666) = 73.83 square units
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© 2014, John Bird
7. The velocity v of a vehicle t seconds after a certain instant is given by: v = (3t2 + 4) m/s.
Determine how far it moves in the interval from t = 1 s to t = 5 s.
Distance moved = area under v/t graph =
∫
5
1
vdt =
∫ ( 3t
5
1
2
+ 4 ) d t = [t 3 + 4t ]1 = (125 + 20) − (1 + 4)
5
= 140 m
8. A gas expands according to the law pv = constant. When the volume is 2 m3 the pressure is
250 kPa. Find the work done as the gas expands from 1 m3 to a volume of 4 m3 given that work
done =
∫
v2
v1
pdv .
pv = k When v = 2 m3 and p = 250 kPa then k = pv = 500 kPa m3 = 500 k
v2
Work done = ∫ =
pdv
v1
4
k
dv ∫
∫=
v
1
4
1
500
dv
=
v
kN 3
m = 500 kNm
m2
= 500 kJ
4
ln v ]1 ( 500 ln 4 − 500 ln1)
[500=
= 500 ln 4
= 693.1 kJ
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© 2014, John Bird
EXERCISE 285 Page 775
1. Determine the coordinates of the points of intersection and the area enclosed between the parabolas
y2 = 3x and x2 = 3y.
x2
x4
or y 2 =
3
9
y 2 = 3 x and x 2 = 3 y i.e. y =
Equating y 2 values gives:
i.e.
3x =
x4
9
i.e. 27x = x 4
x 4 – 27x = 0 i.e. x ( x3 − 27 ) = 0
Hence, x = 0 or x3 − 27 =
0 from which, x3 = 27 and x =
Thus,
3
27 = 3
x = 0 and x = 3
When x = 0, y = 0 and when x = 3, y = 3
Hence, (0, 0) and (3, 3) are the points of intersection of the two curves
A sketch of the curves is shown below.
3


2
3
3/2 
3  x2
3

x
x
x



1/2
Shaded area = ∫  − 3 x  d x =
3

 − 3 x  d x =+
∫
0
0
3 
9
 3

 3



2 0


33 
= 9 − 3
 − ( 0 ) = 9 – 6 = 3 square units

1.5 

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© 2014, John Bird
2. Sketch the curves y = x2 + 3 and y = 7 – 3x and determine the area enclosed by them.
The two curves intersect when x2 + 3 = 7 – 3x
i.e.
x2 + 3x – 4 = 0
i.e.
(x + 4)(x – 1) = 0
i.e. when x = –4 and x = 1
The two curves are shown below.
Area enclosed by curves =
∫
1
−4
(7 − 3 x) d x − ∫
( x 2 + 3) d x = ∫ −4 (7 − 3x) − ( x 2 + 3) d x = ∫ −4 ( 4 − 3x − x 2 ) d x
−4
1
1
1
1
3x 2 x3 

= 4 x −
− 
2
3  −4

3 1 
64 

=  4 − −  −  −16 − 24 + 
2 3 
3 

2
1
2
 1 
=  2  −  −18 = 2 + 18
3
6
3
 6 
= 20
5
square units or 20.83 square units
6
3. Determine the area enclosed by the curves y = sin x and y = cos x and the y-axis.
A sketch of y = sin x and y = cos x is shown below
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© 2014, John Bird
When sin x = cos x then
Shaded area =
sin x
= 1 , i.e. tan x = 1
cos x
and x = tan −1 1 = 45° or
π /4
π
∫ ( cos x − sin x ) d x = [sin x + cos x ]
0
0
/4
π
4
rad
π
 π
=  sin + cos  − ( sin 0 + cos 0 )
4
4

= (0.7071 + 0.7071) – (0 + 1)
= 0.4142 square units
4. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5
y = –2x + 5 and y = 3x intersect when –2x + 5 = 3x
i.e. when
5 = 5x
x
x
intersect when –2x + 5 =
2
2
i.e. when
5 = 2.5x i.e. when x = 2
y = –2x + 5 and y =
i.e. when x = 1
The three straight lines are shown below
Shaded area =
∫
1
0
2
x
x

 3 x −  d x + ∫ 1 (−2 x + 5) − d x
2
2

 3 1 
 
1 
x2 
 3x 2 x 2   2

= 
−  +  − x + 5 x − =  −  − (0)  + ( −4 + 10 − 1) −  −1 + 5 −  
4 0 
4  1  2 4 
4 
 2

 
1
2
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© 2014, John Bird
1 1
 1 
 3 
= 1  + ( 5 ) −  3   =1 + 1 = 2.5 sq units
4 4
 4 
 4 
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© 2014, John Bird