Calculus II
Final Exam
May 23, 2002
Name____________________
R.  Hammack
Score______


(1) Evaluate the following integrals.
(a)   ∫ (1/x^2 + 1/x + x^2 + 5) dx =-1/x + ln(x) + x^3/3 + 5 x + C


(b) ∫ (ln x)^(1/2)/x dx =∫ (ln(x))^(1/2) 1/x dx =∫ u^(1/2) du =2/3 u^(3/2) + C = 2/3 ln(x)^(1/2)^3 + C
u = ln(x)
du = 1/x dx


(c)   ∫ x/(1 + x^2) dx =1/2 ∫ (2 x)/(1 + x^2) dx =1/2 ∫ 1/u du = 1/2 ln(u) + C = 1/2 ln(1 + x^2) + C

u = 1 + x^2
du = 2x dx


(d)   ∫ 1/(1 + x^2) dx =tan^(-1) (x ) + C


(e)   ∫ x^4 e^x^5 dx = 1/5 ∫ e^x^5 5 x^4 dx = 1/5 ∫ e^u du = 1/5 e^u + C = 1/5 e^x^5 + C

u = x^5
du = 5 x^4 dx


(f)    ∫ e^(5 x) dx =1/5 e^(5 x) + C


(g)   ∫ x e^(5 x) dx =x 1/5 e^(5 x) -∫ 1/5 e^(5 x) dx = 1/5 x e^(5 x) - 1/25 e^(5 x) + C

u = x              dv = e^(5 x) dx   
du = dx          v = 1/5 e^(5 x)

(2) Evaluate the following definite integrals

(a)  ∫ _ (-1)^2 x^3 dx =[x^4/4] _ (-1)^2 = 2^4/4 - (-1)^4/4 = 4 - 1/4 = 15/4


(b) ∫ _ (-1)^2 | x^3 | dx = ∫ _ (-1)^0 | x^3 | dx + ∫ _ 0^2 | x^3 | dx =∫ _ (-1)^0 -x^3 dx + ∫ _ 0^2 x^3 dx = [-x^4/4] _ (-1)^0 + [x^4/4] _ 0^2= 1/4 + 16/4 = 17/4


(c) ∫ _ 0^π^(1/2) 5 x cos(x^2) dx =5/2 ∫ _ 0^π^(1/2) cos(x^2) 2 x dx =5/2 ∫ _ 0^2^π^(1/2)^2 cos(u) 2 du =[sin(u)] _ 0^π= 0 - 0 = 0

u = x^2
du = 2x dx

(3) Evaluate the following integrals.
(a) ∫ sin^3(x) cos^2(x) dx =∫ sin^2(x) cos^2(x) sin(x) dx =-∫ (1 - cos^2(x)) cos^2(x) (-1) sin(x) dx =
-∫ (1 - u^2) u^2 du =-∫ (u^2 - u^4)    du =-u^3/3 + u^5/5 + C =-cos^3(u)/3 + cos^5(u)/5 + C

(b) ∫ dx/(x^2 + 3 x - 4) =∫ dx/((x + 4) (x - 1)) =∫ ( (- 1 / 5)/(x + 4) + (1/5)/(x - 1)) =-1/5 ln(x + 4) + 1/5 ln(x - 1) + C = ln((x - 1)/(x + 4)^(1/5)) + C



1/((x + 4) (x - 1)) = A/(x + 4) + B/(x - 1)

1 = A(x - 1) + B ( x + 4 )

Set x = -4.  Get A = -1/5
Set x = 1.  Get B = 1/5



(4) Evaluate the following definite integrals. (Notice that they are both improper.)
(a)   ∫ _ 1^∞ e^(-x) dx =Underscript[lim , l -> ∞] ∫ _ 1^l e^(-x) dx =Underscript[lim , l -> ∞][-e^(-x)] _ 1^l =Underscript[lim , l -> ∞][-1/e^x] _ 1^l =Underscript[lim , l -> ∞][-1/e^l + 1/e^1] = 1/e


(b) ∫ _ 0^(π/6) (cos x)/(1 - 2 sin x)^(1/2) dx =Underscript[lim , l -> π/6 ] ∫ _ 0^l (cos x)/(1 - 2 sin x)^(1/2) dx= -1/2 Underscript[lim , l -> π/6 ] ∫ _ 0^l (-2 cos x)/(1 - 2 sin x)^(1/2) dx = -1/2 Underscript[lim , l -> π/6 ] ∫ _ (1 - 2 sin 0)^(1 - 2 sin l) 1/u^(1/2) du= -1/2 Underscript[lim , l -> π/6 ] ∫ _ (1 - 2 sin 0)^(1 - 2 sin l) u^(-1/2) du= -1/2 Underscript[lim , l -> π/6 ][2 u^(1/2)] _ 1^(1 - 2 sin l)= -1/2 Underscript[lim , l -> π/6 ][2 (1 - 2 sin l)^(1/2) - 2 1 ^(1/2)]= -1/2[2 (1 - 2 sin(π/6))^(1/2) - 2]= -1/2[2 0^(1/2) - 2] = 1

u = 1 - 2 sin x
du = -2 cos x

(5) Consider the region between the graph of y = x - x^2 and the x-axis.



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(a) This region is rotated around the y-axis. What is the volume of the resulting solid?

Volume by shells:
∫ _ 0^1 2 πx(x - x^2) dx =2 π ∫ _ 0^1 (x^2 - x^3) dx =2 π[x^3/3 - x^4/4] _ 0^1 =2 π[1^3/3 - 1^4/4] = 2 π[1/12] = π/6cubic units



(b) This region is rotated around the x-axis. What is the volume of the resulting solid?

Volume by slicing:
∫ _ 0^1 π(x - x^2)^2 dx =π ∫ _ 0^1 (x^2 - 2 x^3 + x^4) dx =π[x^3/3 - 2 x^4/4 + x^5/5] _ 0^1 =π[1/3 - 2/4 + 1/5] = π 2/60 = π/30cubic units

(6) Find the area enclosed between the curves y = x^2,   y = x^(1/2),   x = 1/4,  and  x = 1.

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∫ _ (1/4)^1 (x^(1/2) - x^2) dx = [(2 x^(1/2)^3)/3 - x^3/3] _ (1/4)^1 = ... = 49/192

(7) Find the derivative of the function   F(x) = ∫ _ 2^x^3 (1 + sin t)^(1/2) dt

F ' (x) = (1 + sin x^3)^(1/2) 3 x^2

(8) Decide if the following sequences converge or diverge. In the case of convergence, find the limit.

(a) {(8 e^n)/(e^n + 1)} _ (n = 1)^∞

Underscript[lim , n -> ∞] (8 e^n)/(e^n + 1) = 8  The sequence converges to 8.


(b)  {1/n ln(1/n)} _ (n = 1)^∞

Underscript[lim , n -> ∞] 1/n ln(1/n) = Underscript[lim , n -> ∞] -ln(n)/n =  Underscript[lim , n -> ∞] (-1/n)/1 = 0 (Using L'Hopital's Rule)
The sequence converges to 0.


(9) Decide if the following series converge or diverge.
(a) Underoverscript[∑ , k = 1, arg3] (3/5)^k (k - 1)

Let's try the ratio test on this one (it works well with powers).
Underscript[lim , n -> ∞] ((3/5)^(k + 1) (k + 1 - 1))/((3/5)^k (k - 1)) = Underscript ... ∞] (3/5 (k + 1 - 1))/(k - 1) = Underscript[lim , n -> ∞] 3/5 k/(k - 1) = 3/5 < 1
Thus, the series converges by the ratio test.

(b)  Underoverscript[∑ , k = 1, arg3] 1/k^(1/2) P-series with p = 1/2 < 1, so it diverges.


(c)  Underoverscript[∑ , k = 1, arg3] (-1)^k/k^(1/2) Alternating series that meets conditions of the Alternating Series Test, so it converges.



(d)  Underoverscript[∑ , k = 1, arg3] 3/(2^k + k^(1/2))
The given series converges by the comparison by comparison with the convergent series whose kth term is 3/2^k.


(10)   Both of the following series converge. Say what number they converge to.

(a)   3 - 3/2 + 3/4 - 3/8 + 3/16 - 3/32 + 3/64 - ...   = 3/(1 - (-1/2)) = 2
(Because it's a geometric series with a = 3 and r = -1/2)


(b)   1 + 1/1 ! + 1/2 ! + 1/3 ! + 1/4 ! + 1/5 ! + 1/6 ! + 1/7 ! + ...   = e
(because it's the Maclaurin series for e^xwith x = 1.)


(11) The remaining problems on this exam are based on the function f(x) = ln(1 + x).

(a) Derive the Maclaurin series for the function f(x) = ln(1 + x).   Show all of  your work.
f(x) =      ln(1 + x)         &nbs ... nbsp;               f(0) = 0
f ' (x) =              1/(1 + x) & ...                f ' (0) = 0 !
f '' (x) =          -1/(1 + x)^2    &nb ... bsp;              f '' (0) = -1 !
f ''' (x) =         ((2) (1))/(1 + x)^3    & ... bsp;              f ''' (0) = 2 !
f^(4)(x) = (-(3) (2) (1))/(1 + x)^4          &nbs ... bsp;              f^(4)(0) = -3 !
f^(5)(x) = ((4) (3) (2) (1))/(1 + x)^5          & ... nbsp;              f^(5)(0) = 4 !
:

f^(k)(x) = ((-1)^(k + 1) (k - 1) !)/(1 + x)^k          ... bsp;           f^(k)(0) = (-1)^(k + 1) (k - 1) !

Maclaurin Series: Underoverscript[∑ , k = 0, arg3] f^(k)(0)/k ! x^k = Underoverscript[∑ , k = 1, arg ... arg3] ((-1)^(k + 1) (k - 1) !)/k ! x^k = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k x^k



(b) Find the interval of convergence of the Maclaurin series from part (a).  Show all of  your work.

Let's use the ratio test to check for absolute convergence.

Underscript[lim , n -> ∞] (| (-1)^(k + 1 + 1)/(k + 1) x^(k + 1) |)/(| (-1)^(k + 1)/k ... k + 1) (k + 1)) | x | = Underscript[lim , n -> ∞] (k^2 + k)/(k^2 + 2 k + 1) | x | = | x |

Thus, we get convergence provided that | x | = ρ < 1and divergence when | x | = ρ > 1
Thus the series converges absolutely on the interval (-1, 1) and diverges on (-∞ - 1) and (1, ∞)

What about at x = 1?  Then the series is Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k 1^k = 1 - 1/2 + 1/3 - 1/4 ... which is the convergent alternating harmonic series.

What about at x = -1?  Then the series is Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k (-1)^k = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 2)/k = 1 + 1/2 + 1/3 + 1/4 ... which is the divergent  harmonic series.

Thus the interval of convergence is (-1, 1].

(c) Use your answer to part (a) above to express ln(2) as an infinite series.

Part (a) and (b) say ln(x + 1) = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k x^k on the interval (-1, 1].  Since 1 is on this interval, we have:

ln(2) =ln( 1+1) = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k 1^k = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...



(d) Use your answer to part (a) above to find a power series representation of the function ln(1 + x^2).

ln(1 + x^2)= Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k (x^2)^k = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/k x^(2 k)



(e) Write ∫ _ 0^1 ln(1 + x^2) dx  as an infinite series. (You may just write out the first 5 or 6 terms.)

∫ _ 0^1 ln(1 + x^2) dx = ∫ _ 0^1 (Underoverscript[∑ , k = 1, arg3] (-1)^(k ... (2 k) dx) = Underoverscript[∑ , k = 1, arg3] ((-1)^(k + 1)/k[x^(2 k + 1)/(2 k + 1)] _ 0^1) =
Underoverscript[∑ , k = 1, arg3] ((-1)^(k + 1)/k 1/(2 k + 1)) = Underoverscript[∑ , k = 1, arg3] (-1)^(k + 1)/(2 k^2 + k) = 1/3 - 1/10 + 1/21 - 1/36 + 1/55 - 1/78 + ...