Thread: How to do this Anti derivative problem.

Well I'm back again after failing Calculus again....

And It just makes me and wanna shoot myself for not being able to solve this problem even though since I'm taking the same class three times in a row but I still can't solve this argh



I ended up with ((x^2)/2)/((x^3/2)/(3/2)) - (6x)/((x^3/2)/(3/2)) but I know that my process is incorrect.

How do I solve this?

Originally Posted by masterofcheese

Well I'm back again after failing Calculus again....

And It just makes me and wanna shoot myself for not being able to solve this problem even though since I'm taking the same class three times in a row but I still can't solve this argh



I ended up with ((x^2)/2)/((x^3/2)/(3/2)) - (6x)/((x^3/2)/(3/2)) but I know that my process is incorrect.

How do I solve this?

x/Sqrt(x) = Sqrt(x) = x^(1/2) and its antiderivative is (2/3)x^(3/2)

6/Sqrt(x) = 6*x^(-1/2) and its antiderivative is 12*x^(1/2)

Tonio

Originally Posted by tonio

x/Sqrt(x) = Sqrt(x) = x^(1/2) and its antiderivative is (2/3)x^(3/2)

6/Sqrt(x) = 6*x^(-1/2) and its antiderivative is 12*x^(1/2)

Tonio

So when I am dividing by square root, I should bring the square root up and turn the exponent for x^(-1/2) then work on antiderivative from there?

*feeling very stupid as usual

Originally Posted by masterofcheese

So when I am dividing by square root, I should bring the square root up and turn the exponent for x^(-1/2) then work on antiderivative from there?

*feeling very stupid as usual

As you must know in order to deal with these things, Sqrt(x) = x^(1/2), and the sign minus in the eponent just means the square root is in the denominator.
It's not that you "should", but it makes things way easier.

Tonio

Franly, it appears that the reason you are failing Calculus again is that you have never learned algebra well.