Sketch the region enclosed by the curves and compute its area

Question

The curves are: x-2=4y and x-14=(y-3)^2

Answer 1

To sketch the area, it is best to graph both functions in terms of y. The first equation is linear, with a slope of 4 (run over rise) and an x-intercept of 2. The second equation is a quadratic given in vertex form, with a vertex at (14,3). To find the region to integrate, we must find the points of intersection. We solve the equation 4y+2=(y-3)^2 + 14. This gives us the values y=3,7. To set up the integral in terms of y, we take the rightmost function and subtract the leftmost function, integrating between the bounds. The resulting integral is of the expression (4y-2)-[(y-3)^2 + 14] from y=3 to y=7. Resulting from this integral, the area between the two curves is 296/3.