Recent content by rhothi

Yes that is what I meant. How would I solve the equation in terms of y? I have no idea how to compute the integral in terms of x.

I meant that it is (4x^4−4x^2+1)/(x^2). After adding 1 underneath the radical I received ((4x^4 -3x^2 +1)/(x^2))^((1/2)).

I used the formula for calculating arc length. The formula is ((1 + (f'(x))^(2))^(1/2) from a to b. As a result, I got f'(x) to be 2x - 1/x. When I squared the function i received ((4x^4 -4x^2 +1). From there I plugged f'(x)^2 into the equation to get the integral that I mentioned.

Find the length of the curve y = x^2 − ln(x), 1 ≤ x ≤ 2. I have have to find the arc-length of the given equation. I have checked my work several times and i ultimately come up with the integral of ((4x^4 -3x^2 +1)/(x^2))^(1/2), which I just cannot seem to solve. Any help would be appreciated.