### d x.

```→
Definite
Integral
in terms of x
from (lower bound) x=a
to (upper bound) x=b
Use the substitution
u = f (x)
d u = f ' (x) d x
c = f (a) (lower bound)
d = f (b) (upper bound)
to get
→
=D
Definite
Integral
in terms of u
From u=c to u=d
Then, INTEGRATE and
use F.T.C to get
=D
Note: D is a real number
Slide 7-1
EXAMPLE
e²
Evaluate
I = ∫1 (-6) [x ( 7 - 3 ln x )]
-1
d x.
Let u = 7 - 3 ln x , so that
d u = - 3 x -1 d x , [and thus x -1 d x = d u / (-3)
x = 1, then u = 7 - 3 ln (1) = 7 – 3(0) = 7
x = e², then u = 7 - 3 ln (e²) = 7 – 3(2) = 1
Then with these substitutions, we get:
1
I=∫
(-6) u-1 d u / (-3)
7
1
lower bd in x < upper bd in x
i.e. 1 < e²
lower bd in u > upper bd in u
i.e. 7 > 1
1
= ∫7 (-6)/(-3) u d u = 2 ∫7 u -1 d u
1
= 2 ln |u| |
7
= 2 [ ln (1) – ln (7) ] = - 2 ln (7) ≈ - 3. 892
-1
Note: The definite integral in terms of u is equal to a real number.
And thus the definite integral in terms of x is also equal to that
real number. The value of a definite integral that is not
representing
anEducation,
area, Inc.
does not need to be positive.
Slide 7-2