# MATH 1310 Integral Calculus with Applications Winter 2009 May

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```					                                 MATH 1310
Integral Calculus with Applications
Winter 2009

May 2
Worksheet 3 Partial Solutions
√              √
1. Using the hint, u4 + 1 = (u2 − 2u + 1)(u2 + 2u + 1). We therefore seek a partial
fraction decomposition of the form
1       Au + B        Cu + D
=      √       +     √        .
u4   +1   u 2−   2u + 1 u 2+   2u + 1
Adding the terms on the right-hand side with a common denominator of u4 +1, and
then equating the numerators of the left- and right-hand sides yields the equation
√                          √
1 = (A + C)u3 + (B + D + 2(A − C))u2 + (A + C + 2(B − D))u + (B + D)

Equating coeﬃcients (i.e., viewing the left-hand side as 0u3 + 0u2 + 0u + 1), yields
four equations for the four unknowns A, B, C, D. These are easily solved to yield
1             1
B = D = 1/2 and A = − 2√2 , C = 2√2 . Thus we have the partial fraction
decomposition
1
1
− 2√ 2 u + 1
2
1
√ u+ 1
2 2     2
4+1
=        √         +      √        ,
u         u 2−     2u + 1 u 2+     2u + 1
which it’s useful to rewrite as
√              √                  √           √
1        1       −2u + 2                 2           2u + 2             2
= √             √        +         √       +       √       +      √           .
u4 + 1    4 2 u2 − 2u + 1 u2 − 2u + 1 u2 + 2u + 1 u2 + 2u + 1

Integrating using our usual techniques yields
1                      √                     √
du = − 4√2 ln |u2 − 2u + 1| + 2√2 tan−1 ( 2u − 1)
1                      1
u4 + 1
√                     √
+ 4√2 ln |u2 + 2u + 1| + 2√2 tan−1 ( 2u + 1) + C.
1                      1

2. The integral is already in the form given by partial fraction decomposition. To
carry out the integration, it’s useful to rewrite Ax + B in terms of the derivative of
x2 + bx + c, which is 2x + b, as Ax + B = A (2x + b) + (B − A b). Thus the original
2                 2
integral splits up as
Ax + B          A           2x + b                              1
dx =                        dx + (B − A b)
2
dx.
(x2   + bx + c)3      2     (x2   + bx + c)3                  (x2   + bx + c)3
Of the two integrals on the right, the ﬁrst one is easily solved by the substitution
1
u = x2 + bx + c. Thus if we can ﬁnd (x2 +bx+c)3 dx, then we evaluate the entire

1
2
original integral. Now, the quadratic x2 +bx+c is irreducible if and only if c− b4 > 0.
b2
Thus we may set γ =      c−     4
to be the positive square root and write x2 +bx+c =
b
(x + 2 )2 + γ 2 , completing the square. Applying the substitution
b
x+     = γ tan θ or, in other words, θ = tan−1 ((x + b/2)/γ),
2
yields, after simpliﬁcation,
1           1
dx = 5           cos4 θ dθ,
x2    + bx + c     γ
which can be handled by known techniques (either a reduction formula, or repeated
application of the formula cos2 α = (1 + cos(2α))/2).
3. (a) One can use integration (as done in lecture to analyze radioactive decay) to
obtain the general solution to (2), y(x) = Ce−x . So a particular solution, for
example, is y(x) = e−x (corresponding to C = 1).
(b) If y(x) = u(x)e−x , then y (x) = (u (x) − u(x))e−1 . So y + y = u e−x , and,
in terms of u, equation (1) becomes u e−x = x. This can be rewritten as
u = xex , which can be integrated to yield u = (x − 1)ex + C. And therefore
y = ue−x = x − 1 + Ce−x . This is the general solution to equation (1).
4. This is very similar to the proof given in lecture, whereby the theorem that ev-
ery bounded increasing sequence converges was used to prove that every bounded
decreasing sequence converges.
5. (a) The ﬁrst four terms of the sequence are: 1, 1 + 1 , 1 + 1 + 9 , 1 + 1 + 1 +
4       4
1
4   9
1
16
.
(b) This is a special case of the integral test (see the lecture notes from May 1).
One can verify by drawing the graph of 1/t2 that
n                     ∞
1                     1
an < 1 +                   dt < 1 +              dt = 2.
1       t2            1       t2
This shows that the given sequence is bounded (since it’s positive). Since the
sequence is also increasing, it necessarily converges.
6. Here there was an error in the question! It should read “Prove that if a sequence
converges, then it is necessarily bounded.” The converse is false; for example the
sequence 1, −1, 1, −1, 1, −1, · · · is bounded but does not converge.

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