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Dawson College - Winter 2007 - Mathematics Department
Final Examination - Calculus 2 (201-203-DW)
1. (5 points ) Find f ( ) x given (^) ( )
5 4 3
2
2 x 6 x 5 x f x x
′ (^) = and f ( − 3 ) = 4.
2. (5 points ) The population of a small town is changing at the rate of (^) ( )
2
3
t P t t
where P
is the number of residents, and t is the number of years from now. Today the population is 2000
people. What will the population be 7 years from now?
3. (5 points ) Find the indefinite integral. dx
x
x ∫ − 1
4. (4 points ) Find the indefinite integral. (^) ( )
4 4 3 sin
x x ⋅ e ⋅ e dx ∫
5. (5 points ) Use the definition of definite integral (Reimann Sum) to evaluate the definite integral.
( )
4 2
0
3 x + 2 x − 1 dx ∫
6. (5 points ) Find the average value of the function (^) ( ) 2
f x 1 x
= + over [ 2 , 8 ].
7. (5 points ) Consider the functions f ( x ) = 5 − 2 x and
( ) 3 5
2 g x = x + x +. Find the area of the region completely
enclosed by the graphs of f and g.
8. (5 points ) The supply function for a particular brand of computer is given by
( )
2 p = S x = 6 x + 30 x + 500 where p is the unit price in dollars and x is the quantity
demanded. Determine the producers’ surplus if the unit price is set at $800.
00 .
9. (5 points ) Find the indefinite integral.
9 3
x x e dx ∫
Winter 2007 Final Examination
10. (5 points ) Find the indefinite integral.
( )( )( )
2 4 19
x x dx x x x
∫
11. (6 points ) Use the Trapezoidal Rule ( n = 4 )to approximate the definite integral.
7
0
x dx x + ∫
12. (6 points ) Find each limit. (a)
( )
ln 3 1 lim 2 −
→∞ (^) x
x
x
(b) e x
x x
x x cos 2
lim 5
3
→
13. (4 points ) Determine whether the improper integral converges or diverges. If it converges find
its value.
dx ∫ x
∞
8
4 / 3
14. (5 points ) Determine whether
x y = xe is a solution of y ′′ + y = 2 y ′.
15. (5 points ) Solve for y given the equation
4
2
x y y
′ = and the condition y ( − 1 ) = − 2.
16. (5 points ) Find the 4
th Taylor Polynomial at x = 0 for (^) ( )
2 x f x = e.
17. (5 points ) Does the series
2
2 1
n^3
n n
n
∞
=
∑ converge or diverge?
18. (5 points ) Evaluate the series (^) ∑
∞
=
1
1 3
n
n
n
.
19. (5 points ) Use the Integral Test to determine whether (^) ∑
∞
3
2
n n^1
n converges or diverges.
You may assume that the function (^) ( )
2
3 1
x f x x
is decreasing.
20. (5 points ) Determine whether the series 3 1
n n^4
∞
∑ converges or diverges.
Winter 2007 Final Examination
SOLUTIONS
1. ( ) ( )
5 4 3 3 2 2
x x x f x f x x x x x
( ) (^) ( ) (^) ( ) ( ) ( ) ( )
3 2 1 4 1 3 1 2 1 4 3 5 2 4 3 2 2 2
f x = 2 x + 6 x + 5 x dx = 2 x + 6 x + 5 x + C → f x = x + 2 x + x + C ∫
( ) [ ] [ ] [ ] ( )
1 4 3 5 2 1 4 3 5 2 2 2 2 2
f − 3 = 4 → − 3 + 2 − 3 + − 3 + C = 4 → C = − 5 → f x = x + 2 x + x − 5
2. Let
3 2 1 2 3
u = t + 1 → du = 3 t dt → du = t dt
( ) ( ) ( ) (^) ( ) ( )
3 1/ 2^2 1/ 2 1/ 2 1/ 2 1 3
P t 15 t 1 t dt 15 u du 5 u du 5 2 u C
− (^) − (^) − = + = = = + ∫ ∫ ∫
( )
3 → P t = 10 t + 1 + C
( ) [ ]
3 P 0 = 2000 → 10 0 + 1 + C = 2000 → C = 1990
( ) ( ) [ ] ( )
3 3 P t = 10 t + 1 + 1990 → P 7 = 10 7 + 1 + 1990 → P 7 =2175.
7 years from now the population of this small town will be approximately 2175.
3. Let u = x − 1 → du = dx and x = u + 1
( ) ( )
1/ 2 1/ 2 2 3 / 2 1/ 2 3
x u^ du u dx du u u du u u C x u u u
−
+ ⎛^ ⎞
∫ ∫ ∫ ∫
( )
2 3 / 2 3
x dx x x C x
∫
4. Let
4 4 1 4 4
x x x u = e → du = e dx → du = e dx
( ) (^ ) ( ) (^ )^ (^ )^ ( )
4 4 1 3 3 3 4 4 4 4 4
3 sin 3 sin sin cos cos
x x x ⋅ e ⋅ e dx = u ⋅ du = u du = ⎡− u ⎤+ C = − e + C ∫ ∫ ∫ ⎣^ ⎦
5. We subdivide the interval (^) [ 0, 4] into n equal subintervals of width
x n
∆ = with
i^0
i x n
( xi is the x-value of the right endpoint of each subinterval ) and
2 yi = 3 xi + 2 xi − 1 ( y (^) i is the
y -value of the corresponding rectangle ). (^2 )
2
i i
i i i^ i y y n n (^) n n
2 2
, (^3 2 3 ) 1
n
i n i n i
i i i i P y x S n n n^ n n n =
∑
continued...
Winter 2007 Final Examination
( ) (^ )^ ( )
2 3 2 1 1 1
n n n
n i i i
S i i n (^) = n = n =
3 2
n
n n n n n S n n n n
n
n n^ n^ n n S n n n n n
⎛ +^ ⎞ ⎛ +^ ⎞ ⎛ + ⎞
1 1 1 S (^) n 32 1 2 16 1 4 n n n
→ = ⋅ + ⋅ + + ⋅ + −
( ) ( ) ( ) ( ) (^ )^ ( )
4 2
0
3 2 1 lim (^) n lim 32 1 2 16 1 4 32 2 16 1 4 n n
x x dx S →∞ →∞ n n n
Therefore ( )
4 2
0
3 x + 2 x − 1 dx = 76
6. The average value of f over [^ a , b ]is given by the formula f ( ) x^ dx
b a
b
a
8 2 1 8 2 2
1 1 1 10 10 1 39 13
8 2 6 6 8 2 6 4 8
yave 1 10 x dx x 10 x 8 2
− −
−
= ⋅ + = ⎡^ − ⎤ = ⎡^ − ⎤^ − ⎡^ − ⎤ = =
7. Set ( ) ( ) ( )
2 2 f x = g x → 5 − 2 x = x + 3 x + 5 → x + 5 x = 0 → x x + 5 = 0
The graphs of f and g intersect at x = − 5 and at x = 0. Furthermore, f > g over [ −5, 0 ].
0 0 0 2 2
5 5 5
area f x g x dx 5 2 x x 3 x 5 dx x 5 x dx
− − −
= ⎡ − ⎤ = ⎡^ − − + + ⎤^ = ⎡^ − − ⎤
[ ] [ ] [ ] [ ]
(^0 3 2 3 ) 3 2 5
1 5 1 5 1 5 125 3 2 3 2 3 2 6
x x 0 0 5 5 −
= ⎡^ − − ⎤ = ⎡−^ − ⎤^ − ⎡−^ − − − ⎤ =
Therefore the required area is 125 6
2 2 2 p = 6 x + 30 x + 500 = 800 → 6 x + 30 x − 300 = 0 → 6 x + 5 x − 50 = 0
→ 6 ( x + 10 )( x − 5 ) = 0 → x = 5 ( x > 0 )
Therefore p^ = 800 → x = 5 (the quantity demanded)
5 2 3 2 5 0 0
CS = 800 5 − 6 x + 30 x + 500 dx = 4000 − ⎡ 2 x + 15 x + 500 x ⎤ = 875
The Producers’ surplus is $875.
00 .
Winter 2007 Final Examination
12.b This is a limit of the form 0
. Apply L’Hopital’s Rule to find the value of this limit.
( )
( ) 5
4
5
5 2 sin 2
lim cos 2
lim 5
2
(^50)
3
0
→ → e x
x
e x
x x
x x x x
13. ( ) [ ]
( ) 2
15 (^81) / 3 1 / 3
1 / 3
8
4 / 3
8
4 / 3 8
lim 5 5 lim 3 15 lim
→−∞
− →∞
− →∞
∞
∫ ∫ k
dx x dx x x a
k
k
b
b
This improper integral converges. Its value is 2
15 .
x x x x x y = xe → y ′= xe + e → y ′′= xe + 2 e
LHS ( )
x x x x x = y ′′+ y = xe + 2 e + xe = 2 xe + 2 e
RHS ( )
x x x x = 2 y ′ = 2 xe + e = 2 xe + 2 e
LHS = RHS. Therefore
x y = xe is a solution of y ′′^ + y = 2 y ′.
15. y dy x dx y dy x dx y x C y x C
y
x
dx
dy = → = → = → = + → = + ∫ ∫
3 5 1
2 4 2 4 3 5 2
4 5 3 5
1 3
1
( 1 ) 2 5 [ 2 ] 3 [ 1 ] 37 5 3 37
(^3 ) y − =− → − = − + C → C =− → y = x − or (^) ( )
1/ 3 3 5 37 5 5
y = x −
16. ( ) ( ) 0 1 0 ( ) 0 1
2 f x = e → f = → T = f =
x
f ( ) x e f ( ) T f ( ) ( x ) x
x 2 0 2 1 0 0 2
2 ′ = → ′ = → = ′ ⋅ − =
( ) ( )
( ) ( )
(^22) 2
2 0 2 2!
4 0 4 x x
f f x e f T
x ⋅ − =
( ) ( )
( ) ( )
3 3 3
2 3
4 0 3!
8 0 8 x x
f f x e f T
x ⋅ − =
( ) ( )
( ) ( )
4 4
( 4 )
4
( 4 ) 2 ( 4 ) 3
2 0 4!
16 0 16 x x
f f x e f T
x = → = → = ⋅ − =
The 4
th Taylor polynomial is P 4 (^) ( x (^) ) = T 0 (^) + T 1 (^) + T 2 (^) + T 3 (^) + T 4.
( )
2 3 4 (^4 )
2 3
4 P x = 1 + 2 x + 2 x + x + x
Winter 2007 Final Examination
17. Apply the Test for Divergence ( (^) find lim with thehelpofL'Hopital'sRuletwice n
an → ∞
lim 0 3
lim 6
lim 6
lim 3 4
lim lim 3
4 2
2
⎟= →^ ≠ ⎠
→∞ →∞ →∞ →∞ →∞ →∞ n n n n n n n n
a n
n
n
n n a
According to the Test for Divergence this series is divergent.
18. The first few terms of this series are ... 243
16 81
8 27
4 9
2
This is a geometric series with 9
2 a = and 3
2 r =. Since r < 1 this sequence converges.
2
3
2
9
2
1 r 1
a The sum of this (convergent) infinite series is 3
2 .
19. Apply the Integral Test. I.e. find the improper integral
2
3 1 1
x dx
x
∞
∫
2 3 2 3 3
1 1 1 1 3 3 3 3
4 ln ln 4 1
x dx du u x du x dx u x x u
⎢ =^ +^ →^ =^ →^ =^ =^ =^ + ⎥
∫ ∫
2 2 3 3 (^3 3 ) 1 1
1 1 1 3 3 3
lim lim ln 4 lim ln 4 ln 5 1 1
k k
k k k
x dx x dx x k x x
∞
→∞ →∞ →∞
= = ⎡^ + ⎤^ = ⎡^ + − ⎤ = ∞
∫ ∫
This improper integral diverges.
Therefore, by the Integral Test, the series (^) ∑
∞
3
2
n n^1
n diverges.
20. Use the Comparison Test. Let 3
a n n
. Then 3 3
a n bn n n
The series (^) ∑
∞
n = 1
bn is a p -series with p = 3 → p > 1 → The series (^) ∑
∞
n = 1
bn converges.
Furthermore, an < bn for all n (shown above).
Therefore, according to the Comparison Test, the series 3 1
n n^4
∞
∑ converges.
