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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 11

Lesson 4: AP Calculus BC 2008

# AP Calculus BC exams: 2008 1 c&d

parts c and d of problem 1 in the 2008 AP Calculus BC free response. Created by Sal Khan.

## Want to join the conversation?

• At , I've put the exact same definite integral into my calculator (Casio fx-9860 GII) and it just doesn't seem to give me the same answer. I get approx. 10.227. I'm positive that I have the integral entered correctly and I'm positive that I understand the concept. Is there anything I can check to determine whether or not I have it right?
• I encountered the same problem while using my TI-84 plus. Turns out the calculator was in degree mode. That means that the input for the trigonometric functions (sin, cos, tan, etc.) would be in degrees. But when we do problems like these, we want the input to be in radians, the more fundamental unit. That way, the zeroes and critical points of the function 'sin(pi*x)' will align with rational values of 'x'. Just change your calculator to radian mode, and you will get the right answer. In the TI-84, you can do this by pressing (MODE) and selecting (RADIAN). Hope I helped!
• If you wanted to evaluate the ∫ [0,2] (sin(pi*x)-x^3+4x)^2 dx normally, it's not "impossible" or "stupidly long" to do, it does take some time, so in a test i'm not sure if you could do it, it depends i guess, anyway here goes nothing:

first expand out (sin(pi*x)-x^3+4x)^2
For the sake of knowing what's going on in there :
a=sin(pi*x), b=x^3, c= 4x
(a-b+c)^2 = (a-b+c)(a-b+c)= a^2 -2ab+2ac+b^2-2bc+c^2
if we subtitute back a b and c, we get:
V=∫ [0,2] sin^2(pi*x) -2x^3 sin(pi*x) +8x sin(pi*x) + x^6 -8x^4 +16x^2 dx

we'll split the integral to treat each parts separately :
A= ∫ [0,2] sin^2(pi*x) dx
-2B= -2 ∫ [0,2] x^3 sin(pi*x) dx (this one will be annoying)
8C=8 ∫ [0,2] x sin(pi*x) dx
D = ∫ [0,2] x^6-8x^4+16x^2 dx

So V = A-2B+8C+D

A= ∫ [0,2] sin^2(pi*x) dx
∫ [0,2] 1/2(1-cos(2pi*x)) dx (trig identities)
1/2 ∫ [0,2] 1-cos(2pi*x) dx
1/2 [x - sin(2pi*x)/2pi] [0,2]
1/2 [(2-0)-(0-0)] = 2/2 = 1
A=1

-----------------------------------------
B) Now the hard/long one, lots of integration by parts, the goal is to get "x^3" down to 1 using integration by parts:

B=∫ [0,2] x^3 sin(pi*x) dx

Integration by parts, u=x^3, v'=sin(pi*x)
=x^3 (-cos(pi*x)/pi) - ∫ 3x^2 (-cos(pi*x)/pi) dx [0.2]

Clean up, pull the constant out of the integral
= -x^3/pi cos(pi*x) + 3/pi ∫ x^2 cos(pi*x) dx [0.2]

Integration by parts, u=x^2, v'=cos(pi*x)
= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - ∫ 2x sin(pi*x)/pi dx ) [0.2]

Clean up
= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi ∫ x sin(pi*x) dx ) [0.2]

Integration by parts, u=x, v'=cos(pi*x)
= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (x * -cos(pi*x)/pi - ∫ -cos(pi*x)/pi dx )) [0.2]
Clean up
= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (-x/pi * cos(pi*x) +1/pi ∫ cos(pi*x) dx )) [0.2]

Evaluate ∫ cos(pi*x) dx
= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (-x/pi * cos(pi*x) +1/pi sin(pi*x)/pi)) [0.2]

Distribute
= -x^3/pi cos(pi*x) +3x^2/pi^2 sin(pi*x)+6x/pi^3 cos(pi*x) -6/pi^4 sin(pi*x) [0,2]

Making it this far without making any careless mistakes is a miracle, so check if you didn't miss anything :) Now for the fun part, it may seem daunting to evaluate but if we look carefully, we can see that when x = 0 wherever there's a x, the whole part is = to 0, and the only part without 'x's running around by themselves, we have sin(pi*x), which when x=0 is also equal to 0, so that's one less thing to worry about. Also wherever there's a sin(pi*x) if x=2, we have sin(2pi) and that's also equal to 0. Which means that we only have 2 parts to evaluate at 2, which is great !

= -2^3/pi * cos(2pi) + 6*2/pi^3 cos(2pi)
= -8/pi + 12/pi^3

Rewritten that gives us:
B= 12-8pi^2 / pi^3

I guess we deserve a drum roll here, we're not done but still !

-------------------------------------------------
C= ∫ [0,2] x * sin(pi*x) dx
Integration by parts, u = x, v' = sin(pi*x)
= x * (-cos(pi*x)/pi) - ∫ (-cos(pi*x)/pi) dx [0,2]
Clean & evaluate
= -x/pi cos(pi*x) + 1/pi ∫ cos(pi*x) dx [0,2]
= -x/pi cos(pi*x) + 1/pi (sin(pi*x)/pi) [0,2]
= -x/pi cos(pi*x) + 1/pi^2 sin(pi*x) [0,2]

When x = 0, everything is equal to 0, and when x = 2, the second part with sin(pi*x) is also equal to 0, so that leaves us with just :
-2/pi * cos(2pi)

C= -2/pi

------------------------------------------------------------
D = ∫ [0,2] x^6-8x^4+16x^2 dx
= (x^7)/7 - 8(x^5)/5 + 16(x^3)/3 [0,2]
= (2^7)/7 - 8(2^5)/5 + 16(2^3)/3
A cool thing to notice here is that we only have a bunch of power of 2s:
= (2^7)/7 - 2^3(2^5)/5 + 2^4(2^3)/3
= (2^7)/7 - (2^8)/5 + (2^7)/3
we can factor out 2^7
=(2^7) (1/7 - 2/5 + 1/3)
=(2^7)/(7*5*3) (3*5-2(3*7)+7*5)
=(2^7)/(105) (15-42+35)
=(2^7)/(105) (8)=(2^7)(2^3)/(105)=(2^10)/105

D=1024/105
Done!
--------------------------------------------------
All that's left now is to evaluate V,
Quick recap :
V= A-2B+8C+D
A=1,
B= 12-8pi^2 / pi^3
C=-2/pi
D=1024/105

V= 1-2(12-8pi^2 / pi^3)+8(-2/pi)+1024/105
= (105+1024)/105 + (-24+16pi^2)/pi^3 - 16/pi
= 1129/105 + (-24+16pi^2-16pi^2 )/pi^3

V=1129/105 - 24/pi^3

It doesn't take "forever" but in a test idk how long you've got, but that's just one part of the question. lol, good exercize nonetheless. There might be faster ways to do it, i don't know, and it may contain some errors, i checked but i'm only human! The final answer is correct tho.

Random bragging, V = 1 + 2^10/(3*5*7) - 4!/pi^3, it's cute!