MATH 105 101
Assignment 3 Solutions
Due date: October 16, 2014
MATH 105 101 Assignment 3 Solutions
All work must be shown for full marks.
1. Let f be a differentiable function on an interval [1, 5] and f (1) = 0. Show that:
Z x
Z x
d
d
f (t)dt =
(f (t)) dt
(3 marks)
dx
1 dt
1
Solution: By The Fundamental Theorem of Calculus Part I, we have that:
Z x
d
f (t)dt = f (x).
dx
1
On the other hand, by the definition of anti-derivatives, f (t) is an anti-derivative
of dtd (f (t)), so by the Fundamental Theorem of Calculus Part II:
Z x
d
(f (t)) dt = f (x) − f (1) = f (x) − 0 = f (x).
1 dt
Therefore,
d
dx
Z
x
Z
f (t)dt =
1
1
x
d
(f (t)) dt = f (x),
dt
as desired.
2. Evaluate the following integrals:
Z
1
√ dx
a) (3 marks)
3+ x
√
√ , or equivalently,
Solution: Using substitution, let u = 3 + x. So, du = 2dx
x
√
dx = 2 xdu = (2u − 6)du. Therefore,
Z
Z
1
2u − 6
√ dx =
du
u
3+ x
Z
Z
Z
Z
2u
6
6
=
du −
du = 2 du −
du
u
u
u
√
√
= 2u − 6 ln(|u|) + C = 2(3 + x) − 6 ln (|3 + x|) + C
Page 1 of 4
MATH 105 101
Assignment 3 Solutions
Z
b) (3 marks)
Due date: October 16, 2014
3x3 (x2 + 4)5 dx
Solution: Using substitution, let u = x2 + 4, so du = 2xdx, . Therefore,
Z
Z
3 2 2
3 2
5
3x (x + 4) dx =
x (x + 4)5 2xdx
2
Z
Z
3
3 6
5
=
(u − 4)u du =
(u − 4u5 ) du
2
2
3 u7 4u6
=
−
+C
2 7
6
3(x2 + 4)7
=
− (x2 + 4)6 + C
14
Z
7π
sin (x2011 ) − x dx
c) (2 marks)
−7π
Solution: Recall that f is an odd function if f (−x) = −f (x). Consider the
function f (x) = sin (x2011 ) − x, we first show that it is an odd function:
f (−x) = sin ((−x)2011 ) − (−x) = sin (−x2011 ) + x
= − sin (x2 011) + x = −(sin (x2 011) − x) = −f (x)
Z a
If f is an odd function, then
f (x) dx = 0 for any a.
−a
Z 7π
So,
sin (x2011 ) − x dx = 0.
−7π
Z
d) (5 marks)
√
e
x
dx
Solution: Using substitution, let t =
√
dx
x, so dt = √ , or equivalently, dx =
2 x
2t dt. Therefore,
Z
√
e
x
Z
dx =
et 2t dt
Using integration by parts, u = t and dv = et dt. So, du = dt, and v = et .
Page 2 of 4
MATH 105 101
Assignment 3 Solutions
Due date: October 16, 2014
Therefore,
Z
t
Z
Z
u dv = 2uv − 2 v du
Z
t
= 2te − 2 et dt = 2tet − 2et + C
√
√ √
= 2 xe x − 2e x + C
e 2t dt = 2
Z
e) (3 marks)
√
q
√
x x x + 1 dx
√
3 x
Solution: Using substitution, let u = x x+1. Then, du =
dx. Therefore,
2
√
Z q
Z
q
√
√
√
3 x
2
x x x + 1 dx =
x x+1
dx
3
2
Z √
2 u
=
du
3
3
4 √
4 3
= u 2 + C = (x x + 1) 2 + C
9
9
√
Z
f) (3 marks)
√
r5 r3 + 1 dr
Solution: Using substitution, let u = r3 + 1, so du = 3r2 dr. Then,
Z √
Z
1 3√ 3
5
3
r r + 1 dr =
r r + 1(3r2 dr)
3
Z
Z
√
1 3 √
1
(u − 1) u du =
(u 2 − u) du
=
3
3
5
3
2 5 2 3
2
2
= u 2 − u 2 + C = (r3 + 1) 2 − (r3 + 1) 2 + C.
15
9
15
9
Z
4
g) (3 marks)
p
9 − (x − 1)2 dx (using geometry)
−2
p
Solution: Let y = 9 − (x − 1)2 . Then, y 2 = 9 − (x − 1)2 , or (x − 1)2 + y 2 = 9,
which is an equation for a circle of radius 3 centered at (1, 0). So, the integral
that we want to evaluate corresponds to the area under the curve of the upper
Page 3 of 4
MATH 105 101
Assignment 3 Solutions
Due date: October 16, 2014
Z
4
semi-circle. The area of the whole circle is 9π, therefore,
−2
9π
.
2
Total: 25 marks.
Page 4 of 4
p
9 − (x − 1)2 dx =