Math 143 – Practice Midterm Exam 1
1. Find the following integrals:
Z
(lnx)2 dx. Solution: This was the trickiest one. Set x = ey , then dx = ey dy. Then simplify
a)
and use integration by parts.
Z
b)
x2 sin(πx)dx. Solution: Use integration by parts twice. u = x2 , dv = sin(πx)dx first,
then
Z ....
c)
sin4 (x)cos3 (x)dx. Solution: Use cos2 x = 1 − sin2 x, then u substitution where u = sinx.
Z
2
1
dx. Solution: there was a small typo. dx should be dt. Now use trig
t2 − 1
2
√
substitution. Set t = sec(x). Since x goes from 2 to 2, t goes from π/4 to π/3. Also dt =
Z π/3
Z π/3
Z π/3
1
sec(x)tan(x)
dx
=
dx
=
sec(x)tan(x)dx, so the integral is
cos2 (x)dx =
3
2
π/4 sec (x)
π/4
π/4 sec (x)tan(x)
√
Z π/3
1
1
3−2
(1 + cos(2x))dx = (π/12 +
)
2
4
π/4 2
Z
2x + 3
e)
. Solution: Standard partial fractions question.
(x − 1)(x2 + 1)2
Z 1
x
dx. Solution: Need to complete the square, then do partial fractions. In
f)
2 + 4x + 13
x
0
Z 1
x
particular, x2 + 4x + 13 = x2 + 4x + 4 − 4 + 13 = (x + 2)2 + 9. So
dx =
x2 + 4x + 13
0
Z 1
Z 3
Z 3
Z 3
x
u−2
u
2
dx. Set u = x+2, so get
du, which equals
du−
du.
2+9
2+9
2+9
2+9
(x
+
2)
u
u
u
0
2
2
2
1
2
−1
3
The first integral is a substituion v = u + 9, and the second integral is 2 3 tan (u/3)]2 .
d)
√
t3
√
2. Calculate the following limits:
x3x
xln(3)3x + 3x
.
Solution:
L’Hopital
gives
lim
. Divide top and bottom by 3x
x→0 3x − 1
x→0
ln(3)3x
1 + xln(3)
= 1/ln(3).
and get lim
x→0
ln(3)
a) lim
lim (x − lnx)
b) lim (x − lnx). Solution: Raise e to the limit. So consider ex→∞
= lim e(x−lnx) =
x→∞
x→∞
x
e
x −ln(x)
lim e e
= lim
. L’Hopital says this last limit is ∞. Therefore, the original limit is
x→∞
x→∞ x
also ∞ (since if e to something is ∞, then that original something is ∞).
c) lim (tan2x)x . Solution: Standard, did something like this in class. Write lim (tan2x)x =
x→0+
lim eln((tan2x)
x→0+
x
)
lim+ ln((tan2x)x )
= ex→0
x→0+
= ...
cos(2x)sin(6x)
,
x→0
sin(2x)
d) lim cot2x sin6x. Solution: Turn into sines and cosines: lim cot2x sin6x = lim
x→0
x→0
then use L’Hopital.
x
e) lim
. Solution: L’Hopital.
x→0 tan−1 (4x)
1
Rπ
3. (a) Find the approximations T6 , M6 , S6 for 0 sinxdx and the corresponding errors ET , EM , ES .
Solution: Let’s do S6 : The formula is S6 = ∆x
3 (f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 ) +
4f (x5 ) + f (x6 )). ∆x = π/6, and we have x0 = 0, x1 = π/6, x2 = 2π/6, x3 = 3π/6, x4 =
π
4π/6, x5 = 5π/6, x6 = 6π/6 = π. Then S6 = 18
(sin(0)+4sin(π/6)+2sin(2π/6)+4sin(3π/6)+
√
π
2sin(4π/6) + 4sin(5π/6) + sin(π)) = 18 (2 3 + 8).
(b) How large do we have to choose n so that the approximations Tn , Mn , Sn to the integral
in part (a) are accurate to within 0.00001? Solution: Let’s do the problem for Mn : f 00 (x) =
−sin(x). Then |f 00 (x)| = |−sin(x)| ≤ 1 for every 0 ≤ x ≤ π. So we can take K = 1. Recall the
3
K(b−a)3
≤ .00001.
formula |EM | ≤ K(b−a)
24n2 . We want |EM | ≤ .00001. In other words, we want
24n2
3
K(b−a)3
π3
π3
= 1(π−0)
= 24n
Thus, we want 24n
2.
2 ≤ .00001.
24n2
24n2
q
3
3
3
100,000π
100,000π
π
n2 ≥ 24(.00001) =
, so that we want n ≥
.
24
24
But
In other words, we want
4. Determine whether each improper integral is convergent or divergent. Evaluate those that are
convergent.
Z 0
(a)
xe2x dx. Solution: Type 1 improper integral, then use integration by parts.
−∞
8
Z
4
dx. Solution: Type 2 improper integral, then use u substitution.
3
6 (x − 6)
Z 1
Z 9
1
1
√
√
.
Solution:
Type
2
improper
integral,
noting
that
we
must
write
=
(c)
3
3
x
−
1
x
−
1
0
0
Z 9
Z 9
1
1
√
√
+
, and then use limits. Also, it is a simple u substitution later.
3
3
x−1
x−1
1
0
Z ∞
1
(d)
dx. Solution: Type 1 improper integral. To solve the integral from 1 to t, use
2
x +x
1
partial fractions.
(b)
5. Use the Comparison Theorem to determine whether the integral is convergent or divergent.
Z ∞
−x
2 + e−x
(a)
dx. Solution: 2+ex ≥ x2 ≥ 0 for x ≥ 1 since 2 + e−x ≥ 2 for x ≥ 1. So we can
x
1
Z ∞
Z ∞
2
2 + e−x
use comparison theorem. Show that
dx diverges, so
dx diverges as well.
x
x
1
1
Z 1
sec2 x
√ dx. Solution: since 0 < cos(x) ≤ 1 for 0 ≤ x ≤ 1, we have that 0 < cos2 (x) ≤ 1,
(b)
x
x
0
so that 0 ≤ 1 ≤ sec2 (x) for 0 ≤ x ≤ 1. So we can use comparison theorem. Show that
Z 1
Z 1
1
sec2 x
√ dx diverges, so
√ dx diverges as well.
0 x x
0 x x
2