Question

Then find the area bounded by the two graphs of y=2x^2−24x+42 and y=7x−2x^2

Answer 1

To find the area under a curve, we integrate the function. To find the area bound between two curves, we integrate the difference of the functions. That is, we find:

$\int\limits^a_b {(f(x)-g(x))} \, dx$

If you think about it, we are really doing this with all integration, only the second function is just y=0.

First, we need to figure out which function is on top. In this case we know that $2x^2-24x+42$ is a positive parabola while $7x-2x^2$ is negative, so the negative parabola will be on top. It is always a good idea to draw a rough sketch of the graphs because the curves could intercept multiple times, flipping which graph is on top at different intervals.

Next, we need to determine the bounds. These will be where the two graphs intercept, so we can just set them equal to each other and solve for x:

$2x^2-24x+42=7x-2x^2$

Combine like terms:

$4x^2-31x+42$

Now factor and find the zeros. We can use the quadratic formula:

$\frac{31+ \sqrt{31^2-4(4)(42)} }{8}$

and

$\frac{31- \sqrt{31^2-4(4)(42)} }{8}$

x = 1.75 and 6

$\int\limits^6_{1.75} {((7x-2x^2)-(2x^2-24x+42))} \, dx$

$\int\limits^6_{1.75} {(-4x^2+31x-42)} \, dx$

Solve:

$\frac{-4x^3}{3} + \frac{31x^2}{2} - 42x$

Plug in bounds:

$\frac{-4(6)^3}{3} + \frac{31(6)^2}{2} - 42(6)-(\frac{-4(1.75)^3}{3} + \frac{31(1.75)^2}{2} - 42(1.75))$ = 51.17708