MATH 107-153 Recitation 4-5
JD Nir
Avery 230 • Office Hours: W 4-5 R 11-2
[email protected]
www.math.unl.edu/∼jnir2/107-153.html
September 8, 2015
2
1 + y2
dy exactly [as in ln(3π)], using the Fundamental They
1
orem, and numerically [ln(3π) ≈ 2.243].
Z
p. 130 #59: Evaluate
If we’re going to use the Fundamental Theorem, we’ll need an antiderivative.
At first, this integral doesn’t appear to be in the form we’re used to working with, but note that:
Z
Z
1 + y2
1 y2
dy =
+
dy when y 6= 0.
y
y
y
That note about y 6= 0 is important; this function is not defined at y = 0. Fortunately we will only
be focusing on the interval [1, 2] so we can use this equality.
Now we can split up our integral into two parts and write each part in a way that we know how to
find the antiderivative:
Z
Z
Z
1 y2
1
+
dy =
dy + y dy
y
y
y
Then we use our antiderivative rules:
Z
Z
1
1
dy + y dy = ln(y) + y 2 + C
y
2
We can choose any antiderivative for the Fundamental Theorem, so let’s choose C = 0. Then we
can say
Z 2
1 + y2
1
1
dy = F (2) − F (1) = (ln(2) + 22 ) − (ln(1) + (1)2 )
y
2
2
1
Noting that ln(1) = 0 we get
Z
1
2
1 + y2
3
dy = ln(2) + ≈ 2.193
y
2
1
MATH 107-153 Recitation 4-5
Z
p. 338 #30: Find
et + 1
dt. Check your answer by differentiating.
et + t
Note that the derivative of the denominator is the numerator! Looks like a good candidate for
u-substitution.
Let u = et + t. Then
du
dt
= et + 1 and dt = etdu
+1 . So
Z t
e +1
et +
1 du
dt
=∈
et + 1
u et +
1
Z
1
=
du
u
= ln |u| + C
= ln |et + t| + C
Finally,
1
d t
1
et + 1
d
[ln |et + t|] = t
·
[e + t] = t
· et + 1 = t
dx
e + t dx
e +t
e +t
Z
p. 338 #40: Find
x cos(x2 )
p
dt. Check your answer by differentiating.
sin(x2 )
The denominator is certainly the complicated part of p
the expression, but if we’re going to use
2
2
u-substitution should we set u = x , u = sin(x ) or u = sin(x2 )? If we use u = x2 we’ll have a x
du
2
stuck in the numerator so let’s try u = sin(x2 ). Then du
dx = cos(x ) · 2x so dx = 2x cos(x2 ) . Then
Z
2
x cos(x
)
du
x cos(x2 )
p
√
dx =
2
2
2
·
x
cos(x
u
sin(x )
)
Z
1
√ du
=
2 u
Z
1
1
=
u− /2 du
2
1 u1/2
=
+C
2 12
√
= u+C
p
= sin(x2 ) + C
Finally,
d p
1
1
d
1
1
d
1
1
x cos(x2 )
[ sin(x2 )] = · p
· [sin(x2 )] = · p
·cos(x2 )· [x2 ] = · p
·cos(x2 )·2x = p
dx
2 sin(x2 ) dx
2 sin(x2 )
dx
2 sin(x2 )
sin(x2 )
Note that because of a typo in the problem, we used dx instead of dt. However, if the problem
really asked for the integral with respect to dt, then the answer would just be
x cos(x2 )
p
·t+C
sin(x2 )
because we could consider anything with x to be a constant!
2
MATH 107-153 Recitation 4-5
Z
p. 338 #74: Find
√
(z + 2) 1 − z dz.
Let’s try u = 1 − z to simplify the square root a little bit. Then
back in, we get
Z
√
(z + 2) u(−du).
du
dz
= −1 so dz = −du. Plugging
Notice that the z didn’t cancel out. This might mean we need a different choice for u, but nothing
stands out. Instead, we can solve for z in terms of u. Since u = 1 − z, we know u − 1 = −z or
z = 1 − u. Then we can instead solve
Z
√
− ((1 − u) + 2) u du.
Now this is in a form we can solve:
Z
Z
√
1
− ((1 − u) + 2) u du = − (3 − u)u /2 du
Z
Z
1/2
3/2
=−
3u du + −u du
!
u3/2 u5/2
+C
=− 3 3 − 5
2
3/2
= −2u
2
2 5
+ u /2 + C
5
2
5
3
= (1 − z) /2 − 2(1 − z) /2 + C
5
3