Math 121: Midterm solutions
1. (a) Since limx→∞ e− x/2 x α−1 = 0, there exits T > 0 such that e− x/2 x α−1 ≤ 1 if x ≥ T.
Thus,
Z
Z
∞
0≤
and
R∞
T
e− x x α−1 dx
T
e− x x α−1 dx ≤
∞
T
e− x/2 dx = 2e−T/2
converges by the comparison theorem. If α > 0, then
0≤
Z T
0
Z T
e− x x α−1 dx <
x α−1 dx
0
converges by Theorem 2(b). Thus the integral defining Iα converges.
(b)
Iα =
Z ∞
0
e− x x α−1 dx =
lim
Z R
c→0+ ,R→∞ c
e− x x α−1 dx,
Let U = x α−1 , dV = e− x dx, dU = (α − 1) x α−2 dx, and V = −e− x ,
Iα =
lim
c→0+ ,R→∞
(−e−x x α−1 |cR + (α − 1)
= 0 + ( α − 1)
2. Let x = tan 2θ , dθ =
Z
2
dx,
1+ x 2
cos θ =
dθ
=
1 + sin θ + cos θ
=
Z ∞
Z R
e− x x α−2 dx = (α − 1) Iα−1 .
0
1− x 2
,
1+ x 2
sin θ =
2x
.
1+ x 2
So
( 1+2x2 )dx
Z
2
x
) + ( 1+2xx2 )
1 + ( 11−
+ x2
Z
θ
dx
= ln |1 + x | + C = ln |1 + tan | + C.
1+x
2
3. The arc length element for the parabola y = x2 is ds =
surface area is
Z 1 p
S = 2π
x 1 + 4x2 dx,
√
0
let u =
1 + 4x2
c
e− x x α−2 dx )
and du = 8xdx, so
π 5 1/2
S =
u du
4 1
π 3/2 5
u |1
=
6
π √
=
(5 5 − 1)squareunits.
6
Z
1
1 + 4x2 dx, so the required
12
4. Pick points (− 59 , 0), ( 16
5 , 0) and (0, 5 ) to construct the right triangle, the hypotenuse
lies on the x-axis. A horizontal strip has area
25 12
( − y)dy,
12 5
dA =
therefore, the mass of the plate is
m=
12
5
Z
25 12
( − y)ydy.
12 5
0
And we have
M x =0
1
=
2
My = 0 =
12
5
Z
[
0
12
5
Z
0
25 12
( − y)]2 ydy,
12 5
25 12
( − y)y2 dy.
12 5
Thus, the center of mass of plate can be expressed as: ( Mmx=0 ,
5.
g( x ) =
Z ex
1
cos( x ) sec((ln t) )dt = cos( x )
g0 ( x ) = − sin( x )
= − sin( x )
so we have
2
Z ex
1
Z ex
1
Z ex
1
My =0
m ).
sec((ln t)2 )dt.
sec((ln t)2 )dt + cos( x ) sec((ln e x )2 )e x
sec((ln t)2 )dt + cos( x ) sec( x2 )e x ,
g0 (0) = 0 + 1 = 1.
2