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What is #f(x) = int -cos2x dx# if #f(pi/3) = 0 #?

Answer 1

Paisley Johnson

#f(x)=-1/2sin2x+sqrt3/4.#

#f(x)=int-cos2xdx=-(sin2x)/2+C#

To determine #C#, we make use of the cond. that, #f(pi/3)=0#

#f(pi/2)=0 rArr -1/2sin(2(pi/3))+C=0#

#:. C=1/2sin(pi-pi/3)=1/2sin(pi/3)=1/2*sqrt3/2=sqrt3/4#

# :. f(x)=-1/2sin2x+sqrt3/4.#

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Answer 2

Ezra Adams

To find ( f(x) = \int -\cos(2x) , dx ) if ( f(\frac{\pi}{3}) = 0 ), integrate the function ( -\cos(2x) ) with respect to ( x ).

[ \int -\cos(2x) , dx = -\frac{1}{2} \sin(2x) + C ]

Where ( C ) is the constant of integration.

Now, substitute ( \frac{\pi}{3} ) for ( x ) and ( f(\frac{\pi}{3}) = 0 ) into the equation.

[ -\frac{1}{2} \sin\left(2\left(\frac{\pi}{3}\right)\right) + C = 0 ]

[ -\frac{1}{2} \sin\left(\frac{2\pi}{3}\right) + C = 0 ]

[ -\frac{1}{2} \sin\left(\frac{\pi}{3}\right) + C = 0 ]

[ -\frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) + C = 0 ]

[ -\frac{\sqrt{3}}{4} + C = 0 ]

[ C = \frac{\sqrt{3}}{4} ]

So, the function ( f(x) = \int -\cos(2x) , dx ) with ( f(\frac{\pi}{3}) = 0 ) is:

[ f(x) = -\frac{1}{2} \sin(2x) + \frac{\sqrt{3}}{4} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.