So according to this calculation, the area enclosed by the curve and the x-axis is zero. But we know that this is not the case, because we have a sketch to prove it. Clearly what has happened is
that the ‘signed’ values of the two areas have been added together in the process of integration, and they have cancelled each other out.
Thus the value of the integral evaluated between two ordinates is not necessarily the value of the area between the curve, the x-axis and the two ordinates. So we must be very careful when
calculating areas to avoid this particular trap. The best way is always to draw a sketch of the curve over the required range of values of x.
Key Point
When calculating the area between a curve and the x-axis, you should carry out separate calcu- lations for the parts of the curve above the axis, and the parts of the curve below the axis. The
integral for a part of the curve below the axis gives minus the area for that part. You may find it helpful to draw a sketch of the curve for the required range of x-values, in order to see how
many separate calculations will be needed.
3. Some examples
Example Find the area between the curve y = xx − 3 and the ordinates x = 0 and x = 5.
Solution If we set y = 0 we see that xx − 3 = 0, and so x = 0 or x = 3. Thus the curve cuts the x-axis
at x = 0 and at x = 3. The x
2
term is positive, and so we know that the curve forms a U-shape as shown below.
y = xx − 3
5 3
A B
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From the graph, we can see that we need to calculate the area A between the curve, the x-axis and the ordinates x = 0 and x = 3 first, and that we should expect this integral to give a negative
answer because the area is wholly below the x-axis:
A = Z
3
y dx =
Z
3
x
2
− 3xdx =
x
3
3 −
3x
2
2
3
= [
27 3
−
3×9 2
] − [
3
−
3×0 2
] = [9 −
27 2
] − [0] = −4
1 2
. Next, we need to calculate the area B between the curve, the x-axis, and the ordinates x = 3
and x = 5: B =
Z
5 3
y dx =
Z
5 3
x
2
− 3xdx =
x
3
3 −
3x
2
2
5 3
= [
125 3
−
3×25 2
] − [
27 3
−
3×9 2
] = 41
2 3
− 37
1 2
− 9 + 13
1 2
= 8
2 3
. So the total actual area is 4
1 2
+ 8
2 3
= 13
1 6
units of area. Example
Find the area bounded by the curve y = x
2
+ x + 4, the x-axis and the ordinates x = 1 and x = 3. Solution
If we set y = 0 we obtain the quadratic equation x
2
+ x + 4 = 0, and for this quadratic b
2
− 4ac = 1 − 16 = −15 so that there are no real roots. This means that the curve does not cross the x-axis. Furthermore, the coefficient of x
2
is positive and so the curve is U-shaped. When x = 0, y = 4 and so the curve looks like this.
3 1
A y = x
2
+ x + 4
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The required area A is entirely above the x-axis and so we can simply evaluate the integral between the required limits:
A = Z
3 1
y dx =
Z
3 1
x
2
+ x + 4dx =
x
3
3 +
x
2
2 + 4x
3 1
= [
27 3
+
9 2
+ 12] − [
1 3
+
1 2
+ 4] = 25
1 2
− 4 5
6
which equals 20
2 3
units of area. Exercises
1. Find the area enclosed by the given curve, the x-axis, and the given ordinates. a
The curve y = x, from x = 1 to x = 3. b
The curve y = x
2
+ 3x, from x = 1 to x = 3 c
The curve y = x
2
− 4 from x = −2 to x = 2 d
The curve y = x − x
2
from x = 0 to x = 2 2. Find the area contained by the curve y = xx − 1x + 1 and the x-axis.
3. Calculate the value of Z
1 −
1
xx − 1x + 1dx . Compare your answer with that obtained in question 3, and explain what has happened.
4. Calculate the value of Z
6
4x − x
2
dx . Explain your answer.
4. The area between two curves
