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Antiderivative of a function problem

  • Thread starter Slimsta
  • Start date
  • #1
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Homework Statement


[tex]$g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt$[/tex]
whats g'(x)...


Homework Equations





The Attempt at a Solution


how to find the antiderivative of sqrt(1-t^2)?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Try t=sin(u) or t=cos(u)
 
  • #3
33,084
4,787
I have edited your formula slightly so that it now shows g(x), not just (x). You had extra $ characters that shouldn't have been there.

Homework Statement


[tex]g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt[/tex]
whats g'(x)...


Homework Equations





The Attempt at a Solution


how to find the antiderivative of sqrt(1-t^2)?
Use the Fundamental Theorem of Calculus to find g'(x). You will also need the chain rule since your integral isn't strictly a function of just x, but is a function of sin(x).
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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249
Hi Slimsta! :smile:
how to find the antiderivative of sqrt(1-t^2)?
Use the Fundamental Theorem of Calculus to find g'(x) …
Just to add to what Mark44 :smile: says …

the beauty of using the Fundamental Theorem of Calculus is that you don't need to know the antiderivative. :wink:
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,770
911

Homework Statement


[tex]g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt[/tex]
whats g'(x)...


Homework Equations





The Attempt at a Solution


how to find the antiderivative of sqrt(1-t^2)?
Try t=sin(u) or t=cos(u)
The problem does not ask you to find the anti-derivative nor is it necessary.
Letting y= sin(x), this is
[tex]g(y)=\int_2^y \sqrt{1- t^2} dt[/itex]
You can find dg/dy directly from the "Fundamental Theorem of Calculus" and then use the chain rule to find dg/dx.
 
  • #6
190
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The problem does not ask you to find the anti-derivative nor is it necessary.
Letting y= sin(x), this is
[tex]g(y)=\int_2^y \sqrt{1- t^2} dt[/itex]
You can find dg/dy directly from the "Fundamental Theorem of Calculus" and then use the chain rule to find dg/dx.
okay, that make sense but what if i have a function like this:
[tex]$\int _{3\pi /4}^{\pi }(3 \sec ^2x -\frac{6 }{\pi })dx$ [/tex]

this is confusing me :S
 
  • #7
867
0
That's a definite integral, so that gives you the signed area between the limits of integration, or just a number. What's the derivative of a constant?

A variable (other than x since x would be the dummy variable) in either of the limits of integration would make it a function.
 
  • #8
190
0
hh i already figured that out.. i just took out the 3 and then it becomes tanx - 2/pi x and then its easy..
 

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