Understanding Engineering Mathematics

or
A(x+h)−A(x)
h

=

1

2

[f(x+h)+f(x)]

and this approximation improves ashgets smaller. Indeed, taking the limith→0weget

lim

h→ 0

[

A(x+h)−A(x)

h

]

=

1

2

[f(x+h)+f(x)]

=f(x)

From Section 8.2.2 (

➤

) we recognise the left-hand side asdA/dx:

lim

h→ 0

[

A(x+h)−A(x)

h

]

=

dA

dx

=f(x)

So, using the first viewpoint of the integral, as the inverse of differentiation, we have

A(x)=

∫x

a

f(x)dx

for the area under the curve.
This argument for the equivalence of finding an area and reversing an integration can be
extended to all other cases in Figure 10.12. The review questions illustrate what happens
in practice.

Solution to review question 10.1.5

A. (i) The area under the curvey= 4 x^2 +1 betweenx=0andx= 2

is given by the definite integral

∫ 2

0

( 4 x^2 + 1 )dx=

[

4

3

x^3 +x

] 2

0

=

38

3

(ii) Fory=xexbetweenx=0andx=1 we have the area

∫ 1

0

xexdx=

[

xex−ex

] 1

0 =^1

B. A sketch is always useful in this kind of question. The curvesy=

x^2 −xandy= 2 x−x^2 intersect where

x^2 −x= 2 x−x^2

orx= 0 , 3 /2. They are sketched in Figure 10.15.

The area required is shaded. You can now see the reason for the sketch.

Betweenx=0 and 1, the area for the curvey=x^2 −xis negative.