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© 2014, John Bird

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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 4 4200 0

d 2 d 16 0y x x x x= = = −∫ ∫ = 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

© 2014, John Bird

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Shaded area = ( ) [ ]3 3 32 3

11 1d 3 1 d (27 3) ( 1 1)y x x x x x

−− −= + = + = + − − −∫ ∫ = 32 square units

Width of interval = 3 1 0.58− −

=

X –1 – 0.5 0 0.5 1.0 1.5 2.0 2.5 3.0 y = 3x2 + 1 4.0 1.75 1.0 1.75 4.0 7.75 13.0 19.75 28

Hence, using Simpson’s rule,

Area ( )1 1(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 36.03

+ + = 29.33

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 = 424 4

1 11

5d 5 d (40) (2.5)2xy x x x = = = − ∫ ∫ = 37.5 square units

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 = 22 1x x− + is shown below

© 2014, John Bird

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Shaded area = ( )23 22 2

21 1

1

2 16 2 1d 2 1 d 2 2 13 2 3 3 2x xy x x x x x

− −−

= − + = − + = − + − − − − ∫ ∫

= 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 = 4π

A sketch of y = 2 sin 2θ is shown below.

Shaded area = [ ] ( )/4 /4 /4

00 0d 2sin 2 d cos 2 cos 2 cos 0

4y x x

π π π πθ θ = = − = − − −

∫ ∫

= – 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

Shaded area = ( )222

2 00

0

4( e )d e e 0 e2 2

t ttt t + = + = + − + ∫ = 8.389 square units

© 2014, John Bird

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7. Find the area enclosed between the curve, the horizontal axis and the given ordinates:

y = 5 cos 3t; t = 0, t = 6π

A sketch of y = 5 cos 3t is shown below.

Shaded area = [ ]/6 /6 /6

00 0

5 5 3 5d 5cos3 d sin 3 sin sin 0 sin3 3 6 3 2

y x t t tπ π π π π = = = − = ∫ ∫

= 53

= 1.67 square unit

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 3 1 3

2 20 1 0 1( 1( 3)d ( 1)( 3)d 4 3 d 4 3 dx x x x x x x x x x x x− − − − − = − + − − +∫ ∫ ∫ ∫

= ( ) ( )1 33 3

2 2

0 1

1 12 3 2 3 2 3 0 9 18 9 2 33 3 3 3x xx x x x − + − − + = − + − − − + − − +

= 1 1 21 1 23 3 3

− − = = 2.67 square units

© 2014, John Bird

1137


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 04 42 0 2 0

3 30 2 0 2

0 2

2 2d d 2 d 2 d4 4x xy x y x x x x x

− −−

− = − = − ∫ ∫ ∫ ∫

= [(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 = 4x

is shown below.

© 2014, John Bird

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Shaded area = [ ]/6 4 4

10 1

4d d 4ln 4ln 4 4ln1 4ln 4y x x xx

π= = = − =∫ ∫ = 5.545 square units

3. The force F newtons acting on a body at a distance x metres from a fixed point, is given by:

F = 3x + 2x2. If work done =2

1d

x

xF x∫ , determine the work done when the body moves from the

position where x = 1 m to that when x = 3 m.

Work done = ( )2

1

32 332

11

3 2 27 3 2d 3 2 d 182 3 2 2 3

x

x

x xF x x x x = + = + = + − + ∫ ∫ = 29.33 N m

4. Find the area between the curve y = 4x – x2 and the x-axis. y = 4x – 2x = x(4 – x). When y = 0, x = 0 and x = 4. A sketch of y = 4x – 2x is shown below

Shaded area = ( ) ( )434

2 20

0

644 d 2 32 03 3xx x x x − = − = − − ∫ = 10.67 square units

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

© 2014, John Bird

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Shaded area = ( )333 3

20 0

0

5d 5 2 d 2 (45 6) (0)3xy x x x x = + = + = + − ∫ ∫ = 51 square units

The area enclosed by the curve y = 5 x2 + 2 (i.e. x = 25

y − ), the y-axis and the ordinates y = 2 and

y = 47 (i.e. area ABC in the sketch above) is given by:

Area =

473

147 47 47 22

2 2 2

2

2 1 1 ( 2)d d ( 2) d 35 5 52

y

y

y yx y y y y=

=

− −

= = − =

∫ ∫ ∫

= 321 45 0

1.55

−

= 90 square units

6. Calculate the area enclosed between y = x3 – 4x2 – 5x and the x-axis. y = 3 24 5x x x− − = ( )2 4 5 ( 5( 1)x x x x x x− − = − +

Hence, when y = 0, x = 0 or 5 or –1

When x = 2, y = 2(–3)(1) = –6

A sketch of the graph y = 3 24 5x x x− − is shown below

Shaded area =

( ) ( )0 54 3 2 4 3 20 5

3 2 3 21 0

1 0

4 5 4 54 5 d 4 5 d4 3 2 4 3 2x x x x x xx x x x x x x x

−−

− − − − − = − − − − − ∫ ∫

= ( ) ( )1 4 5 625 500 1250 04 3 2 4 3 2

− + − − − − −

= (0.91666) – (–72.91666) = 73.83 square units

© 2014, John Bird

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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 5 52 3

11 1d 3 4 d 4 (125 20) (1 4)v t t t t t= + = + = + − +∫ ∫

= 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 = 2

1d

v

vp v∫ .

pv = k When v = 2 m3 and p = 250 kPa then k = pv = 500 kPa m3 = 500 k2

kNm

m3 = 500 kNm

= 500 kJ

Work done = [ ] ( )2

1

4 4 4

11 1

500d d d 500ln 500ln 4 500ln1v

v

kp v v v vv v

= = = = −∫ ∫ ∫ = 500 ln 4

= 693.1 kJ

© 2014, John Bird

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1. Determine the coordinates of the points of intersection and the area enclosed between the parabolas y2 = 3x and x2 = 3y.

2 3y x= and 2 3x y= i.e. 2

3xy = or

42

9xy =

Equating 2y values gives: 3x = 4

9x i.e. 27x = 4x

i.e. 4x – 27x = 0 i.e. x ( )3 27x − = 0

Hence, x = 0 or 3 27 0x − = from which, 3 27x = and x = 3 27 = 3

Thus, 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.

Shaded area =

3

2 2 3 3/23 31/2

0 0

0

3 d 3 d 3 33 3 92

x x x xx x x x

− = − = +

∫ ∫

= ( )339 3 0

1.5

− −

= 9 – 6 = 3 square units

© 2014, John Bird

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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 1 1 1

2 2 24 4 4 4(7 3 )d 3 d (7 3 ) 3 d 4 3 dx x x x x x x x x x

− − − −− − + = − − + = − −∫ ∫ ∫ ∫

= 12 3

4

342 3x xx

−

− −

= 3 1 644 16 242 3 3

− − − − − +

= 1 2 1 22 18 2 186 3 6 3

− − = +

= 5206

square units or 20.83 square units

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

© 2014, John Bird

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When sin x = cos x then sin 1cos

xx= , i.e. tan x = 1 and x = 1tan 1− = 45° or

4π rad

Shaded area = ( ) [ ] ( )/4 /4

00cos sin d sin cos sin cos sin 0 cos 0

4 4x x x x x

π π π π − = + = + − + ∫

= (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 i.e. when x = 1

y = –2x + 5 and y = 2x intersect when –2x + 5 =

2x i.e. when 5 = 2.5x i.e. when x = 2

The three straight lines are shown below

Shaded area = 1 2

0 13 d ( 2 5) d

2 2x xx x x x − + − + −

∫ ∫

= ( )1 22 2 2

2

0 1

3 3 1 15 (0) 4 10 1 1 52 4 4 2 4 4x x xx x − + − + − = − − + − + − − − + −

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