1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
Z3
1.
0
MATH 151-Fall 02
December 10
x2 + 2x
dx =
(3x2 + x3 + 1)2
let u = 3x2 + x3 + 1 then du = (6x + 3x2 )dx or 13 du = (2x + x2 )dx.
Z3
0
x2
+ 2x
dx =
(3x2 + x3 + 1)2
x=3
Z
x=0
1 1
du =
3 u2
x=3
Z
x=0
1 −2
u
3
3
x=3
−1 −1 −1
du =
=
u =
2 + x3 + 1) 3
3(3x
x=0
0
−1
−1
18
−
=
165
3
55
Notice that on this problem we did not change the limits of integration. We could have done
that. If x = 0, then u = 1 and if x = 3 then u = 55.
=
Z3
0
x2
(3x2
+ 2x
dx =
+ x3 + 1)2
Z
2.
Z55
1
1 1
du =
3 u2
Z55
1
1 −2
u
3
55
−1 −1
−1
18
du =
=
−
=
3u 165
3
55
1
x
dx =
(x + 2)5
let u = x + 2 then du = dx
Z
Z
x
x
dx
=
du
5
(x + 2)
u5
we can not integrate yet, because there is still an x in the problem. From the substitution, we
notice that u − 2 = x.
Z
Z
Z
Z
x
x
u−2
dx =
du =
du = u−4 − 2u−5 du =
(x + 2)5
u5
u5
=
Z1
3.
u−3 2u−4
1
−1
+
+C
−
+C =
−3
−4
3(x + 2)3 2(x + 2)4
2
3
u cos(1 − u ) du =
0
Z1
x2 cos(1 − x3 ) dx =
0
If the variable of u causes a problem, then rewrite everything in terms of x. Otherwise to a
v-substitution.
Let u = 1 − x3 then du = −3x2 dx or
Z1
x2 cos(1 − x3 ) dx =
0
Z1
x=0
x2 cos(1 − x3 ) dx =
0
Z
4.
x=1
Z
−1
3 du
= x2 dx
1
x=1
−1
−1
−1
−1
−1
3 =
cos(u) du =
sin(u)
sin(1 − x ) =
sin(0) −
sin(1)
3
3
3
3
3
1
sin(1)
3
1
dx =
x ln(x)
let u = ln(x) then du = x1 dx.
Z
1
1
dx =
du = ln(u) + C = ln(ln(x)) + C
x ln(x)
u
Z
x=0
0