Asked by Lele
Find F(x) such that F'(x) = sec^4 (x) and F(pi/4) = 0
F'(x) = sec^4
F(x) = (1/3)tan^3 (x) + tan(x) + C
Now I don't understand the last part of the question?
Just replace x by pi/4?
F'(x) = sec^4
F(x) = (1/3)tan^3 (x) + tan(x) + C
Now I don't understand the last part of the question?
Just replace x by pi/4?
Answers
Answered by
Reiny
your integral is correct.
so replace x with π/4
F(x) = (1/3)tan^3 (x) + tan(x) + C
F(π/4) = (1/3)tan^3 (π/4) + tan(π/4) + C
What part of this don't you understand?
You should of course
know that tan(π/4) = 1 , (tan45° = 1)
so replace x with π/4
F(x) = (1/3)tan^3 (x) + tan(x) + C
F(π/4) = (1/3)tan^3 (π/4) + tan(π/4) + C
What part of this don't you understand?
You should of course
know that tan(π/4) = 1 , (tan45° = 1)
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