IntegrationP1^
6
EXAMPLE 6.7 Evaluate the definite integral 49(^32)
∫ xxd^
SOLUTION
4
9
4
9
4
9
(^32)
(^52)
(^52)
52 5
5
2
2
5
2
594
∫ =
=
=−
xxxxd222
5243 32(^8425)
()
=−()
=.
This gives the shaded area in figure 6.12.Definite integralsExpressions like ∫
2
− 1 (4x(^3) + 4) dx and
4
(^932)
∫ xxd^ in Examples 6.6 and 6.7 are called
definite integrals. A definite integral has an upper limit and a lower limit and can
be evaluated as a number. In the case of Example 6.6 the definite integral is 27.
Note
In Example 6.6 you found that the value of ∫
2
–1 (4x
(^3) + 4) dx was 27. If you evaluate
∫
–1
2 (4x
(^3) + 4) dx you will find its value is –27.
Consider ∫
b
af(x) dx^ = F(b) − F(a),
So ∫
a
bf(x) dx^ = F(a) − F(b)
= −(F(b) − F(a))
= −∫
b
af(x) dx
In general, interchanging the limits of a definite integral has the effect of reversing
the sign of the answer.
O 4 9 x
y
y = x^3 –^2
Figure 6.12
To divide by a fraction,
invert it and multiply.
3
2 1
5
+= 2