# IB Math question: Fresnel integral?

P moves along x-axis such that its velocity, v, at time t is given by $v=\cos(t^2)$. Find the time at which the total distance travelled by P is 1. (all in meters, meters/sec). So the total distance travelled should be integral of $\left| \cos{t^2} \right|$, and so I'm supposed to solve the equation: $$\int_0^x \left| \cos{t^2} \right| \, dt=1$$

No idea how to do this. Apparently this is a Fresnel function, can't easily be done by hand; is there an easy way to do this using a TI-84? Everything I have found about Fresnel integrals seems to be in terms of definite intervals from 0 to infinity; here I need one from 0 to $x$.

• I tried solving it in Mathematica which gave a value of $x=1.3858077886$ but I'm not sure how you would translate the code to the TI-84. But this is the code I needed to use to get it to evaluate it: FindRoot[NIntegrate[Abs[Cos[t^2]], {t, 0, x}] - 1, {x, 1}]. So maybe that could be translated into the TI-84's language? – Jay Mar 5 '14 at 22:44
• Yeah that's the answer that the markscheme gave. I'm not exactly familiar with coding, and I'm not sure how that would get translated onto the TI-84 – Matei Dan Mar 5 '14 at 22:49
• It didn't let me just evaluate it directly in Mathematica that's why I needed to use the FindRoot tool but just looking at the documentation for the TI-84 my guess would be: solve( fnInt(abs(t^2)),t,0,x)=1 ,x,1,{0,E99}). But I'm not sure, I've never used one before. – Jay Mar 5 '14 at 22:59

## 1 Answer

In my opinion, the major problem is with the absolute value. However, the integrand cancels at $\sqrt{\frac{\pi }{2}}$; integrating from $0$ to this value gives as a result $$\sqrt{\frac{\pi }{2}} C(1)$$ which is slightly smaller than $1$ ($0.977451$) which is the target value. So, still forgetting the absolute value and computing the integral from $\sqrt{\frac{\pi }{2}}$ to $x$ assuming that $x<\sqrt{\frac{3\pi }{2}}$, one can obtain the second area (below the axis) as $$\sqrt{\frac{\pi }{2}} \left(C\left(\sqrt{\frac{2}{\pi }} x\right)-C(1)\right)$$ So, the value of the integral is given by the difference of both results, that is to say that the equation to solve is $$f(x)=\sqrt{\frac{\pi }{2}} \left(2 C(1)-C\left(\sqrt{\frac{2}{\pi }} x\right)\right)-1=0$$ This function cancels for $x=1.11605$ which is smaller than $\sqrt{\frac{\pi }{2}}$ ($1.25331$) and this solution must be discarded and for $x=1.38581$ as already given by Jay.

Now, we can try to approximate the solution using a Taylor expansion of $f(x)$ built at $x=\sqrt{\frac{\pi }{2}}$; this gives as an approximation $$\left(\sqrt{\frac{\pi }{2}} C(1)-1\right)+\sqrt{\frac{\pi }{2}} \left(x-\sqrt{\frac{\pi }{2}}\right)^2+O\left(\left(x-\sqrt{\frac{\pi }{2}}\right)^3\right)$$ which cancels for $$x=\sqrt{\sqrt{\frac{2}{\pi }}-C(1)}+\sqrt{\frac{\pi }{2}}=1.38745$$