1. Let Jn(a) = Z a
   0
   1
   (x
   2 + a
   2
   )
   n
   dx where a is a real number and n is a non-negative integer. Evaluate
   J0(a) and J1(a). As well, for a 6= 0 and n ≥ 1, find a formula for Jn+1(a) in terms of Jn(a). As
   part of your solution, state the formula for integration by parts.
2. Suppose f(x) has a continuous second derivative for all x ∈ R. For any x ∈ R, prove
   f(x) = f(0) + f
   0
   (0)x +
   Z x
   0
   (x − s)f
   00(s)ds.
3. Provide an example of an indefinite integral that when integrating requires sin(x) as a substitution. As well, evaluate your indefinite integral.
4. Let g be a continuous function. Prove Z b
   a
   g(x)dx =
   Z a+b
   2
   a
   (g(x) + g(a + b − x))dx.
5. Find and explain the errors (if any) in the following solution.
   We evaluate Z
   1
   1 + t
   4
   dt by first factorizing the denominator as follows. Notice 1 + t
   4 = (1 +
   t
   2
   )
   2 − 2t
   2 and factoring we have
   1 + t
   4 = (1 + t
   2
   )
   2 − 2t
   2 = (1 + t
   2 −
   √
   2t)(1 + t
   2 +
   √
   2t).
   Considering the concept of partial fractions we have
   1
   1 + t

4

At + B
t
2 +
√
2t + 1
+
Ct + D
t
2 −
√
2t + 1
(which implies 1 = (At + B)(t
2 −
√
2t + 1) + (Ct + D)(t
2 +
√
2t + 1)) where A, B, C and D are
determined by solving the following system of equations
1 = B + D,
0 = A −
√
2B + C +
√
2D,
0 = −
√
2A + B +
√
2C + D,
0 = A + C.
The solution to this system of equations is A = −C =
1
2
√
2
, B = D =
1
2
. We can now write
1
1 + t
4
as
1
1 + t

4

1
2
√
2
t +
1
2
t
2 +
√
2t + 1
+
−
1
2
√
2
t +
1
2
t
2 −
√

2t + 1

1
2
√
2
t +
√
2
(t
2 +
√
2t + 1)
−
1
2
√
2
t −
√
2
(t
2 −
√

2t + 1)

1
2
√
2
t +
√
2
((t +
√
2
2
)
2 +
1
2
)
−
1
2
√
2
t −
√