Calculus- Early Transcendentals, 2021a

294 Techniques of Integration
Now using the definition of improper integral for
∫ 1
0
lnxdx = lim
∫ 1
R→0 + R
∫
= xlnx −
1dx
= xlnx − x +C
∫ 1
0
lnxdx:
lnxdx= lim
R→0 +(xlnx − x) ∣ ∣∣∣
1
R
= −1 − lim lim
R→0 +(RlnR)+ R
R→0 +
Note that lim R = 0. We next compute lim First, we rewrite the expression as follows:
R→0 + R→0 +(RlnR).
lnR
lim lim
x→0 +(RlnR)= R→0 + 1/R .
Now the limit is of the indeterminate type (−∞)/(∞) and l’Hôpital’s Rule can be applied.
lnR
lim lim
R→0 +(RlnR)= R→0 + 1/R = lim 1/R
R→0 + −1/R 2 = lim
R→0 −R2
+ R = lim
R→0 +(−R)=0
Thus, lim
R→0 +(RlnR)=0. Thus ∫ 1
lnxdx= −1,
0
and the integral is convergent to −1.
Graphically, one might interpret this to mean that the net area under lnx on [0,1] is −1(theareainthis
case lies below the x-axis).
Example 7.51: Integral of a Square Root
∫ 4 dx
Determine if √ is convergent or divergent. Evaluate it if it is convergent.
4 − x
0
♣
1
Solution. Note that √
4−x
is discontinuous at the endpoint x = 4. We use a u-substitution to compute
∫
√dx
4−x
.Weletu = 4 − x, thendu = −dx, giving:
∫
∫
dx
√ = 4 − x
− du
u 1/2