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Simple antiderivative

  • Thread starter Rasine
  • Start date
  • #1
205
0
Find f if f '(t)=8cost+sec^2t and f(pi/3)=4 .

so f(t)=8sint+tant+c

4=8sin(pi/3)+tan(pi/3)+c

c=-3^1/2

so f(t)=8sint+tant-3^1/2 is that right
 

Answers and Replies

  • #2
2,063
2
You need to re-evaluate the values of sin and tan at pi/3.
 
  • #3
205
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thank you very much
 
  • #4
205
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sin(pi/3) is sqroot3/2 and tan(pi/3) is sqroot3 and 8(sqroot3/2)+sqroot3 is
4sqroot3+sqroot3=5sqroot3

is that right?
 
  • #5
2,063
2
Yes.

But for the sake of clarity, use something like sqroot(3)/2 instead of sqroot3/2 when referring to [tex]\frac{\sqrt{3}}{2}[/tex]. :smile:
 
  • #6
205
0
ok. so now i have (5sqroot(3)/2)+c=4

then c=4-5sqroot(3)/2 so that would be -sqroot3

is that right
 
  • #7
2,063
2
then c=4-5sqroot(3)/2
Why the 2 in the denominator?

so that would be -sqroot3
I don't understand. You've already solved for c, so...

f(t) = 8sin(t) + tan(t) + 4-5sqroot(3)
 

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