Suppose that k(t) is a continuous function with positive values. Show that for any t (or at least for any t not too large), there is a unique τ so that τ =∫ (k(η)dη,0,t); conversely
any such τ corresponds to a unique t. Provide a brief explanation on why there is such a 1-1 correspondence.
The Attempt at a Solution
Stuck on it but here are some of my thoughts and reasoning:
I first view τ as function dependent upon t. since k(t) is positive and continuous, that will mean that the antiderivative of k(t) will only give us increasing values for increasing t. The new k(η) function is essentially same as k(t) except with η as the independent var. Hence since the k(t) is positive then k(η) is also positive. Then the integral of k(η) must be increasing for each increasing t. Hence for t2 and t1 the integral of k(η) from 0 to t2 is greater than the integral of k(η) from 0 to t1. This makes sure the for every different t substitute into the integral have a different output. And as we said τ is
The problem is how do I show the unique τ for each t part.
Thank you very much in advance for any help :)