Download Document

Math 129-1
Exam 1 Solutions, Problems 1-6 (from one version of the exam)
None of these problems requires a calculator for doing the integration. In all cases, in order to receive full
credit, you should show enough work that it is clear that you can do the problem correctly
and completely without using a calculator.
1. For the following integral, choose an appropriate trigonometric substitution, and make the
substitution completely in the integral. After completing the substitution (so that there is no
variable x in the integral), IT IS NOT NECESSARY TO SIMPLIFY OR TO FIND THE INTEGRAL.
∫
x2
10 +
x2
dx .
SOLUTION. Let x = √10 tan θ. [NOTE that you are asked for a TRIGONOMETRIC substitution.]
Then dx = √10 sec2 θ dθ. Substituting,
∫
x2
10 + x 2
dx =
∫
10tan2θ
10 + 10tan2θ
10 sec 2θ dθ .
2. Show how to expand in partial fractions (i.e., how to split into partial fractions). You don’t have
to do any calculations. Just show what a partial fraction expansion would look like.
x
.
2
(x + 9)( x − 3)2
Ax + B
C
D
=
+
+
x−3
(x 2 + 9)( x − 3)2
x2 + 9
(x − 3)2
[NOTE: The original rational function has degree 4 in the denominator. Hence, the partial fraction
decomposition has to have a total degree of 4 in the denominator.
Also, as pointed out repeatedly in class, since the denominator has degree 4, there will be 4
“unknowns”, A, B, C, D.]
SOLUTION.
x
3. Split (decompose) into partial fractions, using the quick short method discussed in
1
class for finding the constant numerators in the partial fractions:
.
2
x − 25
1
1
A
B
SOLUTION. 2
=
=
+
.
(x − 5)(x + 5)
x−5
x+5
x − 25
Multiplying through by x - 5 and then setting x = 5 gives A = 1/10.
Multiplying through by x + 5 and then setting x = -5 gives B = -1/10.
1
1 1
1 1
So, 2
=
.
10 x − 5 10 x + 5
x − 25
4. Evaluate the following integral, without using tables.
x2
∫ x 2 − 25 dx
COMMENT. Note that this is almost the same function as in Problem 3. DON’T DO THE WORK
ALL OVER AGAIN. USE THE RESULT of Problem 3 to do this integral. (See next page.)
4. Evaluate the following integral, without using tables.
1
x2
∫ x 2 − 25 dx
1 1
1 1
10 x − 5 10 x + 5
x 2 − 25
SOLUTION. The integrand is a rational function, so we try partial fractions. The degree of the
numerator is not less than the degree of the denominator, so we have to correct for that.
One can use long division, but much, much easier is to use common sense:
⎡1 1
x2
x 2 − 25 + 25
25
1 1 ⎤
=
=
1
+
=
1
+
25
⎢
⎥ from Problem 3
⎣10 x − 5 10 x + 5 ⎦
x 2 − 25
x 2 − 25
x 2 − 25
5⎡ 1
1 ⎤
= 1+ ⎢
⎥
2 ⎣ x−5
x+5⎦
Integrating, we easily get
As a reminder, here is the answer to Problem 3:
x2
∫ x 2 − 25 dx
= x+
5
[ ln|x-5| - ln |x+5|] + C
2
Check: This is easy to check. The derivative is 1 +
=
x 2 − 25
x 2 − 25
=
+
25
=
x 2 − 25
x2
x 2 − 25
5⎡ 1
1 ⎤
5 10
.
⎢
⎥ = 1+
2 ⎣ x−5
x+5⎦
2 x 2 − 25
.
5. Without using any integral formulas or techniques, verify (check) the formula
∫ sec( x) dx = ln [ sec(x) + tan(x) ] + C
(For simplicity when dealing with the ln, you may assume sec(x) + tan(x) > 0, although this doesn’t really matter.)
COMMENT. The way to check an antiderivative is to differentiate it.
This is the Fundamental Idea of Calculus, mentioned MANY times in class.
COMMENT. A basic algebraic fact, which everyone in the course should be familiar with:
ln (a + b) ≠ ln(a) + ln(b). The derivative of ln (a + b) is not 1/a + 1/b..
COMMENT. A basic calculus fact, which everyone in the course should be familiar with:
1
The derivative of ln [ f (x)} is not
. You have to use the chain rule.
f (x )
d
1
SOLUTION.
ln [ sec(x) + tan(x) ] =
[ sec(x) tan(x) + sec 2 (x) ]
dx
sec( x) + tan( x )
sec( x) + tan( x )
=
sec(x) = sec(x) .
sec( x) + tan( x )
∫ x sec( x
6. Evaluate
2
) dx .
SOLUTION. Let u = x 2, so du = 2x dx. Substituting, we get
[USE Problem 5. That’s why it’s there.]
∫ x sec( x
2
) dx =
1
2
∫ sec(u ) d u
=
1
1
ln [ sec(u) + tan(u) ] + C = ln [ sec(x 2) + tan(x 2) ] + C
2
2