Download Math 2153

Math 2153
Quiz 2, February 2
1. (5 points each) Describe the first step you would take in computing each
of the following integrals. Right now our options are “algebraic simplification” (describe the simplification), “guess-and-fix” (describe the initial
guess), “u-substitution” (state u and du), “integration by parts” (state
u and dv), “use a reduction formula” (write the result of the formula),
“trigonometric substitution” (state the rewritten integral), and “use a table” (state the formula you (expect to) find in the table, and the values of
any parameters such as u or a). If instead we don’t know a way to replace
the integral with a less difficult problem, say that instead.
Z
√
(a)
( x − 1)2 dx
Z
√
√
Solution: Do algebra to it. The new integral is
( x)2 − 2( x)(1) + (1)2 dx =
Z
√
x − 2 x + 1 dx, which is just a polynomial.
Z
x2 dx
√
(b)
6x2 − 49
Solution: Use a trigonometric substitution. 6x2 − 49 looks kind of
like sec2 θ − 1 = tan2 θ, so make the substitution 6x2 = 49 sec2 θ,
i.e. x = √76 sec θ, which makes dx = √76 sec θ tan θ dθ. The integral becomes
Z 49 sec2 θ √7 sec θ tan θ dθ
Z
2
6
x dx
6
√
=
49(sec2 θ − 1)
6x2 − 49
Z
49 · 7 · sec2 θ · sec θ tan θ dθ
√
=
6 · 6 · 49 tan2 θ
Z
7 sec3 θ dθ
√
,
=
6 6 tan θ
which we can finish by using various trigonometric techniques
(ask in office hours).
Z
(c)
sec x tan4 x dx
2
2
Solution: Do algebra to it. tan4 x = tan2 x = sec2 x − 1 =
Z
Z
4
4
2
sec x−2 sec x+1, so sec x tan x dx =
sec5 x − 2 sec3 x + sec x dx .
We can finish this off using formulas 48 and 13 from the end of
the textbook.
It isZ disappointing that the textbook doesn’t include a formula
for secm u tann u du, but the fact is that it doesn’t, so we have
to be able to rewrite integrals like this using other techniques.
Z
(d)
e3x cos 4x dx
Z
Solution: Use formula 20,
The solution is
eau cos bu du, with a = 3, b = 4, and u = x.
e3x
(3 cos 4x + 4 sin 4x) + C.
+ 42
32
Alternatively, use integration by parts with u = e3x and dv = cos 4x dx
or use integration by parts with u = cos 4x and dv = e3x dx . This
Z
appears to simply transform the problem into
e3x sin 4x dx,
which is no easier, but then doing it again will yield the formula
Z
Z
3 3x
9
1 3x
3x
e3x cos 4x dx.
e cos 4x dx = e sin 4x+ e cos 4x−
4
16
16
Now we can finish the problem using algebra: move the integral
9
over to the left and divide both sides by (1 + 16
).
Extra Credit (3 points): Let C be the part of the curve x2 + y 2 = 4 in the first
quadrant. Express the length of C as a definite integral, in a form that
a Math 2144 student would understand at the end of the semester. Then
evaluate the integral, using techniques from this class.
Z end p
Solution: The arc-length formula is
dx2 + dy 2 .
start
In our case, the curve starts at (2, 0) and ends at (0, 2) (or, perhaps,
goes in the other direction).
Differentiating the formula for C, we get 2xdx + 2ydy = d(4), i.e.,
x
2xdx + 2ydy = 0. So dy = − dx.
y
Z x=2 s
x2
dx2 + 2 dx2 , which still doesn’t make
The integral becomes
y
x=0
sense to a Calculus I student.
Z x=2 r
x2
2
2
Substituting y = 4 − x , we get
dx2 +
dx2 , which
4 − x2
x=0
is at least
closer. Extracting the common factor of dx2 turns it into
Z x=2 s
Z x=2 r
x2
x2
1+
dx2 =
dx , which has the
1+
2
4−x
4 − x2
x=0
x=0
virtue of looking like integrals are “supposed” to look, and should be
recognizable to a student at the end of calculus I.