Summer 2006 Final Solutions
1. (a)
4
Z
dx
1
=
16 + x2
16
0
Let u =
x
4
dx
4 .
so that du =
1
4
Z
1
0
Z
4
dx
1 + ( x4 )2
0
This gives
du
1
= (arctan 1 − arctan 0)
1 + u2
4
π
16
=
(b) Pick u = x2 so that du = 2x dx. Now
Z
(x2 )
x3
1
dx =
2
Z
3u du =
3u
+C
2 ln 3
2
=
3(x )
+C
2 ln 3
(c) This was done in class.
(d) Pick u = ln x so that du =
Z
1
e
dx
x .
Now
(ln x)2
dx =
x
=
Z
1
u2 du
0
1
3
(e) Integrate by parts twice. Pick u = ex so that du = ex dx. Pick
dv = sin x dx so that v = − cos x. Now
Z
ex sin x dx = −ex cos x +
Z
ex cos x dx
Pick now u = ex and dv = cos x dx so that v = sin x. Now we have
Z
ex sin x dx = −ex cos x + ex sin x −
Z
ex sin x dx
Add the integral on the right to both sides and divide by 2 to get
Z
ex sin x dx =
1
1 x
(e sin x − ex cos x)
2
R
2. (a) v(t) = a(t) dt = −6t + C1 . To solve for C1 notice that 3 = v(0) =
−6(0) + C1 = C1 so that C1 = 3. Plug this in to get
v(t) = −6t + 3
(b) x(t) = v(t) dt = −3t2 + 3t + C2 . To solve for C2 notice that 0 =
x(0) = C2 . Plug this in to get
R
x(t) = −3t2 + 3t
(c) x(1) = 0
3 ≥ 0 iff t ≤
1
2
R1
|v(t)| dt =
R1
| − 6t + 3| dt. Notice that −6t +
R1
R1
so that this integral turns into 02 3 − 6t dt + 1 6t − 3 dt.
(d) Distance traveled is
0
Evaluate these integrals to get
3. (a) F (0) = 0 and F (4) =
3
2
0
2
units.
4
3
(b) F 0 (x) = x2 − 4x + 3 = (x − 3)(x − 1). This is negative for 1 < x < 3 (F
is decreasing here), and positive for x < 1 and for x > 3 (F is increasing
here). 1 and 3 are critical points, the first being a local max and the
second being a local min.
(c) F 00 (x) = 2x − 4 This is negative for x < 2 (F is concave down here),
and positive for x > 2 (F is concave up here). Thus x = 2 is an inflection
point for F .
(d) F (1) = 23 , F (2) = 23 , and F (3) = 0.
4. (b) Uf (P) = 31 and Lf (P) = 7.
5. (a) To find the intersection points, set y1 = y2 :
x2 + 1 = 2x + 1
x2 − 2x = 0
x(x − 2) = 0
so that the intersection points are 0 and 2. To find which function is the
upper bounding function, plug x = 1 into both and get y1 (1) = 2 and
y2 (1) = 3. Thus y2 is the upper bounding function. The area between the
curves is
Z
2
Z
y2 − y1 dx =
0
2
2x − x2 dx
0
=4−
2
8
4
=
3
3