Let be a function such that the points and are on the graph of . Assume that for every point on the graph of , the graph is above the line segment joining to . Further, assume that the area of the region bounded by the graph of and the line segment is . Find a formula for the function .

**Incomplete.**

From our given information, we wish to find a function f(t) such that the area enclosed by the function and the line segment from (0,1) to (x, f(x)) is always equal to x^3. Since the line segment goes through (0, 1) and (x, f(x)), we can write the equation of the line in point-slope form:

Solving for slope m gives us:

And the equation of the line y = mt + b becomes:

And since we know that f(t) is always greater than y = mt + b on (0, x), we can write it out the area of the enclosed region through the following integrals:

Evaluating the polynomial-termed integrals gives us the following

And from our given information, we know that the area of the enclosed region is always x^3. So:

Differentiating both sides with respect to x gives us:

Now, since the right-hand side is a second degree polynomial, it may be the case that f(x) is one as well. If f(x) is a second-degree polynomial

Then, its derivative is

And

Solving for A and C, we get:

And using our given that f(1) = 0, we can solve for B

And our polynomial f(x) is

To confirm that this is indeed the function that satisfies all the initial conditions, it should suffice to evaluate the original integrals using f(t) = -6t^2 + 5t + 1. Or, to check the back of the book and see that it matches :)