## Trending News

# Fundamental Theorem of Calculus?

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. Thank You.

### 1 Answer

- сhееsеr1Lv 71 decade agoFavorite Answer
Well, you have:

3

∫ u³ / (1 - u² ) du

1-3x

You can't use the theorem until the x part is in the top limit of integration, so first you have to:

1-3x

- ∫ u³ / (1 - u² ) du

3

Remember, if you switch the two limits of integration, you need to add a negative sign. So then you remember the rule, which is:

g(x)

(d/dx) ∫ f(u) du = f(g(x)) g'(x)

a

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 )