Question

Find a positive continuous function \(f\) such that the area under the graph of \(f\) from \(0\) to \(t\) is \(A\,(t)\, = \,{t^3}\) for all \(t > 0\)

Short answer

The positive continuous function \(f\) whose area is \(A(t) = {t^3}\) is \(f(t) = 3{t^2}\).

Step-by-step solution

The area under a curve

The area under a curve between two points is found out by doing a definite integral between the two points. To find the area under the curve\(y = f\left( x \right)\)between\(x = a\)&\(x = b\), integrate\(y = f\left( x \right)\)between the limits of\(a\) and\(b\).

The area under the given graph 

The area under the curve for the graph \(f\) for 0 to \(t\) is given by

\(\begin{aligned}{}A = \int_0^t f (x)dx\\{t^3} = \int_0^t f (x)dx\end{aligned}\)

Now differentiating both sides with respect \(t\), we have

\(\begin{aligned}{}\frac{d}{{dx}}\left( {{t^3}} \right) = \frac{d}{{dx}}\left( {\int_0^t f (x)dx} \right)\\3{t^2} = f(x)\\f(x) = 3{t^2}\end{aligned}\)

The function \(f(x) = 3{t^2}\) is positive and continuous.