CHAPTER 72 AREAS UNDER AND BETWEEN CURVES
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 1136
© 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
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 1136
© 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
