Name:
Math 151, A Review for the First Exam.
1. Compute the following:
Z 2x
Z x3 √
i
dh e √ 2
d
(a)
4t + 2 dt and (b)
t3 − t dt.
dx 0
dx x2
Z
1 2+h √
2. Find lim
1 + t3 dt.
h→0 h 2
3. Evaluate
each of the following integrals.
Z
Z 1
1 5
3
√
+
dx,
(a) (6x + 20x + 20x + 49) dx, (b)
2
4 + x2
4
−
x
Z Z 2
Z eπ
x x
x5 + 1
sin(ln x) dx
√
+
(c)
dx, (d)
dx, (e)
,
2
6
2
4
+
x
x
+
6x
+
1
x
4
−
x
0
1
Z
Z
Z
Z 1
−x4
−2x
2
xe
dx, (h) xe
dx, (i) (3x2 + 1) cos(3x) dx,
(f) tan x dx, (g)
−1 Z
Z
Z
(10x + 15) dx
(x3 − x + 1) dx
3 dx
(j)
, (k)
, (l)
2
4
3
4
2x + 7x + 3
x −x
x + 5x2 + 4
Z 7√
49 − x2 dx by interpreting it as an area.
4. Find the value of
0
5. Find the total (area) between y = x3 − x and the x-axis for x ∈ [0, 4].
Rb
6. Find the interval [a, b] for which the value of a (2 + x − x2 ) dx is a maximum.
Hint: Graph the integrand!
7. Given the region between the two curves y = cos x and y = 1 for x ∈ [0, π2 ],
set up, but do not integrate, the following integrals for:
(a) The volume obtained from rotating this region about the x-axis.
(b) The volume obtained from rotating this region about the line y = −1.
(c) The volume obtained from rotating this region about the y-axis.
(d) The volume obtained from rotating this region about the line x = 2.
8. Compute the lengths of the curves:
x3
1
+
for x ∈ [1, 2].
6
2x
Z e2x √
(b) y =
4t2 + 2 dt for x ∈ [0, 2].
(a) y =
0
9. If f is a continuous function, show that
Ra
0
f (x) dx =
Ra
0
f (a − x) dx.
Solutions.
√
√
√
1. (a) 2e2x 4e4x + 2, (b) 3x2 x9 − x3 − 2x x6 − x2
2. 3
3. (a) 6x6 + 5x4 + 10x2 + 49x + C
(b) arcsin( x2 ) + 12 arctan( x2 ) + C
√
(c) − 4 − x2 + 12 ln(4 + x2 ) + C
(d) 16 ln 77
(e) 2
(f) tan x − x + C
(g) 0
(h) − x2 e−2x − 14 e−2x + C
(i) 31 (3x2 + 1) sin(3x) + 23 x cos(3x) − 29 sin(3x) + C
(j) 3 ln |x + 3| + 2 ln |2x + 1| + C
(k) 2x12 + ln |x − 1| + C
(l) arctan x − 12 arctan( x2 ) + C
4. Quarter circle area: 49
π
4
R1
R4
5. 0 −(x3 − x) dx + 1 (x3 − x) dx =
113
2
6. [−1, 2], the biggest interval giving positive area.
Rπ
7. (a) Washers: 02 π(1 − cos2 x) dx
Rπ
(b) Washers: 02 π[(1 − (−1))2 − (cos x − (−1))2 ] dx
Rπ
(c) Cylindrical Shells: 02 2πx(1 − cos x) dx
Rπ
(d) Cylindrical Shells: 02 2π(2 − x)(1 − cos x) dx
8. (a)
17
12
8
(b) e + 1
9. RLet w = a − x andR dw = −dx. So, R
Ra
a
a
0
f
(a
−
x)
dx
=
f
(w)
·
(−dw)
=
f
(w)
dw
=
f (x) dx.
0
a
0
0