What is #f(x) = int -sinx dx# if #f(pi/3) = -6 #?

Answer 1

Christopher Allers

#cosx-13/2#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(intsinx dx=-cosx)color(white)(2/2)|)))#

#rArrint-sinxdx=cosx+c#

where c is the constant of integration.

We find c, using #f(pi/3)=-6#

#rArrcos(pi/3)+c=-6#

#rArr1/2+c=-6rArrc=-13/2#

#rArrF(x)=cosx-13/2#

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Answer 2

Elijah Abram

To find ( \int -\sin(x) , dx ), integrate the function ( -\sin(x) ) with respect to ( x ).

[ \int -\sin(x) , dx = \cos(x) + C ]

Given that ( f(\frac{\pi}{3}) = -6 ), we can find ( C ) by substituting ( \frac{\pi}{3} ) into ( \cos(x) + C ) and equating it to ( -6 ).

[ \cos\left(\frac{\pi}{3}\right) + C = -6 ]

[ \frac{1}{2} + C = -6 ]

[ C = -6 - \frac{1}{2} = -\frac{13}{2} ]

So, ( f(x) = \cos(x) - \frac{13}{2} ).

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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.