Integration Problem with a Variable for an Upper Limit

[carlodelmundo]

Homework Statement

If f(x) = integration from 0 to x of ( 1 / sqrt [ (t^3) + 2] ) dt , which of the following is FALSE?

$$int({1}/{sqrt(t^3 + 2)},x=a..b)$$

a.) f(0) = 0
b) f is continuous at x for all x >= 0
c) f(1) > 0
d) f'(1) = 1 / sqrt [3]
e) f(-1) > 0.

I know how to do a,b, and d. But how can I test for c and d if I don't know how to integrate such a weird function? I tried thinking of an arctan integration... but it looks "weird".

Thanks

 

[Dick]

You don't necessarily have to integrate a function to know something about it's values. For t in [-1,1], what is the sign of 1/sqrt(t^3+2)?

 

[carlodelmundo]

I can see that the graphof 1 / sqrt(t^3 + 2) from [-1,1] has a positive area. However, is this enough to prove that the integration is going to be positive?

I graphed the function... and when we take the integration from 0 to -1... will yield a negative error. Therefore, e) is wrong, correct?

 

[Dick]

Right. And, yes, if you have a positive area, the integral is positive. The integral measures area.