1. The problem statement, all variables and given/known data Suppose C:y=f(x) with f a twice-differentiable function such that f''(x)> 0 for each x on the closed interval [0,a] where a is a positive constant. Suppose T is the tangent line to C at a point P= (r,f(r)) on C where r is in the open interval (0,a). Let A be the area of the plane region bounded by the y-axis, the vertical line K: x=a, the curve C, and the tangent line T. Express A as a function of r. 3. The attempt at a solution The curve is concave up on the closed interval [0,a]. After sketching the curve, drawing a tangent line at a point P on the interval between 0 and a that lies below the curve, I realized that the area A lies between the curve C, and the tangent line T. The area A could be the integral of C:y=f(x) subtracting the area under the tangent line. A friend and I then thought of interpreting the area below the tangent line T to be a trapezoid (f(0)+f(a))/2. Although the area of the trapezoid doesn't use x as a variable and seems to work, I can't seem to find an expression to substitute r for x in the integral for the area of the curve C. This problem has been bugging me for the past few days, I would sleep a lot better if I can solve it! I would really appreciate it if someone could help me out.