Find the area enclosed between

Question

Find the area enclosed between f(x)=0.8x^2+9 and g(x)=x from x=-4 to x=4.

 

Answer 1

If you sketch the graphs of the given functions, you will find that the function f(x) is on top. The area between the two curves is the definite integral of the difference of the top and bottom graphs over the given intervals.

That is:

 _{∫(f(x) - g(x) dx over the intervals [-4,4]}

 

It is also obvious that these two graphs do not intersect.

 

The difference of the two functions yields:

(0.8x²+9) - x

 

Integrating this difference and placing the intervals, we have:

 

_-4 [0.8x³/3 +9x - x²/2]⁴ = [0.8*4³ + 9(4) -4²/2] - [0.8(-4)³ + 9(-4) - (-4)²/2]

 

= 51.2/3 + 36 -8 - [(-51.2)/3 + (-36) - 8] = 106.133 square units

 

Answer 2

To begin, you might sketch the graph, which is a parabola opening upwardly and having vertex at (0,9).

This has symmetry w.r.t. y-axis, so you need only to calculate definite integral from x=0 to x=4, and then double the result.

 

Antiderivative is 0.8x³/3 + 9x (we'll ignore +C) so this 'half-area' is (4/15)(4³) + 9(4) = 796/15 or about 53.066...

 

Now double it!