Homework 2 - Math 105, Section 204
Due at the beginning of lecture on February 17, 2012
Name:
SID:
1. A curve in the (x, y) plane is defined by the equation
Z g(x)
2
2
ex −t dt where g(0) = 1.
y=
0
If the curve has slope −3 at the point x = 0, find the value of g 0 (0).
(10 points)
First, we rewrite the equation as follows:
Z g(x)
2
x2
e−t dt.
y=e
0
This is valid by either checking that both versions give the same value for
fixed x, or by noting that x is independent with respect to t, so it can be
treated as a constant. In either case, we are now in a position to take the
derivative. If we write
Z x
2
e−t dt,
F (x) =
0
we see that
2
2
y 0 = 2xex F (g(x)) + ex F 0 (g(x))g 0 (x).
Hence as given in the problem, when x = 0 we have
−3 = F 0 (g(0))g 0 (0) = F 0 (1)g 0 (0).
Therefore we need to determine F 0 (1) to determine g 0 (0). To evaluate
F 0 (1), we apply the Fundamental Theorem of Calculus, which says
Z x
d
2
2
0
F (x) =
e−t dt = e−x ,
dx 0
so we get
−3 = e−1 g 0 (0),
that is
g 0 (0) = −3e.
1
2
2. Evaluate the indefinite integral
Z
x 5 + x3
dx.
x4 + 1
(Hint: Write x5 + x3 = x(x4 + 1) + x3 − x. )
(10 points)
Solution: Following the hint, we can rewrite the integral as follows:
Z
Z
Z
Z 5
x(x4 + 1)
x3
x
x + x3
dx =
dx +
dx −
dx
4
4
4
4
x +1
x +1
x +1
x +1
Z
Z
Z
x
x3
dx −
dx.
=
x dx +
4
4
x +1
x +1
The first integral can be evaluated directly. We get
Z
1
x dx = x2 + C.
2
To evaluate the second integral, we use the substitution u = x4 + 1. In
that case, we get du = 4x3 dx, so the integral becomes
Z
1
1
1
1
du = log u + C = log |x4 + 1| + C.
4
u
4
4
Note: we can replace the absolute values by brackets since x4 + 1 is
non-negative. To evaluate the last integral, we use the substitution
u = x2 . We then get du = 2x dx, so the integral becomes
Z
1
1
1
1
du = arctan u + C = arctan(x2 ) + C.
2
2
u +1
2
2
Combining our three answers gives
1
1 2 1
x + log |x4 + 1| − arctan(x2 ) + C.
2
4
2
3
3. Use integration by parts to find an antiderivative of f (x) = sec4 x.
(10 points)
2
2
To integrate by parts, we pick u = sec x and dv = sec x, so that
du = 2 sec2 xZtan x and v = tan x. Hence Z
f (x) dx = sec2 x tan x − 2
tan2 x sec2 x dx.
Using the identity tan2 x = sec2 x − 1, we get
Z
Z
2
f (x) dx = sec x tan x − 2 (sec2 x − 1) sec2 x dx
Z
Z
2
4
= sec x tan x − 2 sec x dx + 2 sec2 x dx
Z
2
= sec x tan x + 2 tan x − 2 f (x) dx.
Therefore
Z
3
that is
f (x) dx = sec2 x tan x + 2 tan x,
Z
2
1
sec2 x tan x + tan x + C.
3
3
We also have theZ alternate solutions
2
f (x) dx = sec2 x tan x − tan3 x + C,
3
Z
1
f (x) dx = tan3 x + tan x + C.
3
f (x) dx =