Question

Find the area enclosed between f(x)=0.3x2+7 and g(x)=x from x=−4 to x=8.

Answer 1

First, we sketch a picture to get a sense of the problem. g(x)=x is a diagonal line through (0,0) with slope = = 1. Since we are interested in the area between x = -4 and x = 8, we find the points on the line at these values. These are (-4, -4) and (8,8).

f(x) is a parabola. It's lowest point occurs when x = 0. It is the point (0,7). At x = -4 and x=8 it has the values 11.8 and 26.2 respectively. That is, it contains the points (-4, 11.8) and (8,26.2).

From these we make a rough sketch (see attachment). This is a sketch and mine is very incorrect when it comes to scale but what matters here is which of the curves is on top, which is below and whether they intersect anywhere in the interval, so my rough sketch is good enough. From the sketch we see that f(x) is always above (greater than) g(x).

To find the area between the curves over the given interval we integrate their difference and since f(x) is strictly greater than g(x) we subtract as follows: f(x) - g(x). The limits of integration are the values -4 and 8 (the x-values between which we are looking for the area.

Now let's integrate:
$\int\limits^{8}_ {-4}f(x)-g(x) \, dx = \int\limits^{8}_ {-4}.3 x^{2} +7-x \, dx$
The integral yields: $[tex](\frac{.3 (8)^{3} }{3} +7(8)- \frac{ (8)^{2} }{2}) -(\frac{.3 (-4)^{3} }{3} +7(-4)- \frac{ (-4)^{2} }{2}) = 117.6$ [/tex]
We evaluate this for 8 and for -4 subtracting the second FROM the first to get: