Fundamental Theorem of Calculus?
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. Thank You.
- сhееsеr1Lv 71 decade agoFavorite Answer
Well, you have:
∫ u³ / (1 - u² ) du
You can't use the theorem until the x part is in the top limit of integration, so first you have to:
- ∫ u³ / (1 - u² ) du
Remember, if you switch the two limits of integration, you need to add a negative sign. So then you remember the rule, which is:
(d/dx) ∫ f(u) du = f(g(x)) g'(x)
for any functions f and g, and any constant u. The g'(x) that comes out there is essentially the chain rule in effect here.
Applying it to our case, you have:
g(x) = 1-3x
g'(x) = -3
f(g(x)) = (1-3x)³ / ( 1 - (1-3x)² )
So your answer is:
- f(g(x)) g'(x)
= 3 (1-3x)³ / (1 - (1-3x)² )
= 3 (1-3x)³ / ( 1 - (1 - 6x + 9x² ) )
= 3 (1-3x)³ / ( 6x - 9x² )
= (1-3x)³ / ( 2x - 3x² )