Remember that arc length is given by int_a^bsqrt(1+(dy/dx)^2)dx∫ba√1+(dydx)2dx.
We can see that we need the derivative of the function in question, so let's get to that first:
f(x)=(x-1)(x+1)=x^2-1f(x)=(x−1)(x+1)=x2−1
f'(x)=2x
Next step is to plug this into the formula:
int_0^1sqrt(1+(2x)^2)dx
Before we do anything, we should note that this integral merits a trig substitution. It resembles a right triangle (see below).
![enter image source here]()
From the picture, we know that tan(theta)=(2x)/1=2x. That means tan(theta)/2=x, and furthermore, (dx)/(d theta)=sec^2(theta)/2. Multiplying both sides by d theta tells us dx=sec^2(theta)/2d theta.
Making these substitutions into the integral,
int_0^1sqrt(1+(2(tan(theta)/2))^2)sec^2(theta)/2d theta
Note that while the stuff inside the integral is in terms of theta now, the limits of integration (0 and 1) are still in terms of x. To fix this issue, recall that tan(theta)=2x. Taking the inverse tangent of both sides yields theta=tan^(-1)(2x). Now we can get x=0 and x=1 in terms of theta:
theta=tan^(-1)(2(0))->theta=tan^(-1)(0)=0
theta=tan^(-1)(2(1))->theta=tan^(-1)(2)~~1.107
Unfortunately, there is no exact answer for tan^(-1)(2), so we'll have to live with an approximation. Our integral finally becomes
int_0^1.107sqrt(1+(2(tan(theta)/2))^2)sec^2(theta)/2d theta
Simplifying:
1/2int_0^1.107sqrt(1+(tan(theta))^2)sec^2(theta)d theta
1/2int_0^1.107sqrt(1+tan^2(theta))sec^2(theta)d theta
If you remember your trig well, you'll know that 1+tan^2(theta)=sec^2(theta).
1/2int_0^1.107sqrt(sec^2(theta))sec^2(theta)d theta
1/2int_0^1.107sec^3(theta)d theta
Evaluating this integral is unnecessary because we can consult a table of integrals to get a result.
After doing this, we see that intsec^3(theta)d theta=1/2sec(theta)tan(theta)+1/2lnabs(sec(theta)+tan(theta)). The final step is finding what this is on the interval [0, 1.107]:
1/2[1/2sec(theta)tan(theta)+1/2lnabs(sec(theta)+tan(theta))]_0^1.107
=1/2(1/2sec(1.107)tan(1.107)+1/2lnabs(sec(1.107)+tan(1.107)))-1/2(1/2sec(0)tan(0)+1/2lnabs(sec(0)+tan(0)))
=1.478-0=1.478
