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Calculus antiderivative question (has square roots)?
I'm given f'(x) = 8/sqrt(1-x^2), that is, eight over the square root of (one minus [x squared])
Also given is f(1/2) = 2
I'm usually able to do these, but the square root thing is really messing me up, help would be appreciated!
That's f PRIME equals that blob of an equation, it's hard to see the apostrophe.
2 Answers
- 1 decade agoFavorite Answer
Thats actually a definition XD
d/dx [ arcsin(x) ] = 1/sqrt(1 - x^2)
f(x) = 8arcsin(x) + C
2 = 8arcsin(1/2) + C
1/4 = arcsin(1/2) + C
1/4 = pi/6 + C
C = 1/4 - pi/6
f(x) = 8arcsin(x) + 1/4 - pi/6
- Anonymous1 decade ago
im guessing you have to find f(x)
So just integrate that, but first rewrite it.
f ' (x) = int [8(1-x^2)^(-1/2) dx]
Let u = (1-x^2)^(-1/2)
du = (-1/2) ah screw this
Its annoying to write it out entirely so ill just do it my way.
f(x) = 8 [2(1-x^2)(1/2) (1/-2x) ] + c
f(x) = (16/-2x)sqrt(1-x^2) + c
f(1/2) = (16/-2(.5)) sqrt(1 - .5^2) + c = 2
= -13.85640646 + c = 2
c = -15.85640646
Plug c in for c in the f(x) equation and thats it. Sorry too lazy to do square roots and stuff on computer
Nevermind mine is wrong. Lol. I'm so prepared for the AP test. Dang arc formulas
