Page 26 - Integration Problems
P. 26
2 1
= ∫ (1 − )
1
Performing the integration,
2 1
2
∫ (1 − ) = [ − ln(| |)]
1
1
2
[ − ln(| |)]
1
Substitute back,
= 1 +
2
[1 + − ln(|1 + |)]
1
This becomes,
2
1
1
2
[1 + − ln(|1 + |) − 1 − + ln(|1 + |)]
or
1 + 1
2
1
− + ln ( )
1 + 2
The keystrokes for your fx calculator to perform this
calculation are,
qH2$pqH1$+ha1
+qH1R1+qH2$$)
=
22
