Anyone willing to point me into the right direction?

>>214700
I would derivate the function and then evaluate by 6

>>214712
So take the derivative of sqrt(1+t^3) , And evaluate that for 6? How does the sum from x --> x^2 factor in here?

>>214713
Since you're evaluating a cubic function, you're calculating the area under the curve limited by x and x^2, so you should replace t with x and x^2 and just solve the equation, after derivating, that is

>>214717
I get the derivative as

1/2(1+t^3)^(-1/2) * 3t^2

If I solve it for 6, and 36, what do I do with those numbers?

>>214727
You should replace t with x and x^2

>>214731

Derivative if t = x
1/2(1+x^3)^(-1/2) * 3x^2

Derivative if t = x^2
1/2(1+x^6)^(-1/2) * 6x^5

So now I solve by subbing in 6 for x, but I get two values. What do i do with the two values?

File: 123.png (23KB, 811x420px) Image search: [Google]

23KB, 811x420px

not op
but please help

>>214738
You add them, since what you're doing is basically calculating area

>>214700
A simple example:
F(x) = integral from x to x^2 of t dt

the inverse derivative of t is t^2/2 + constant c, so F(x) = ((x^2)^2/2 + c) - (x^2/2 + c) = (x^4 - x^2)/2

F'(x) = (4x^3 - 2x)/2 = 2x^3 - x

Note that the first term corresponding to x^2 changed, but the second term corresponding to x remained the same (obviously).

The first term can be computed with the chain rule .

Let g(x) be the inverse derivative of the given function f(x), here g(x) = x^2/2 + c and f(x) = x

For the first term you need to compute
(g(x^2))' = g'(x^2)(x^2)' = f(x^2)(2x) = 2x^3
by the chain rule.

The second term is obviously (g(x))' = f(x) = x

In your exercise you have f(x) = sqrt(1+t^3)

So instead of taking the antiderivative and derivating it again, with the shortcut above you obtain:

F'(x) = f(x^2)(2x) - f(x) = sqrt(1+x^6)2x - sqrt(1+x^3)

Not sure if everything is correct, it has been ages since I did these problems.

>>214763
This would give me a numerical answer, when I use 6. But the question says I should get a symbolic answer, i.e pi should be in there somewhere, shouldn't it?

>>214763

Nevermind, it worked. Thank you!

>>214763
*antiderivative

>>214769
how about 12sqrt(46657) - sqrt(217)

but as I said, it has been ages.