What is the area of the region bounded between the graphs of y= -x^2 + 8x and y =x^2 + 2x?
2 Answers
The area is
Explanation:
First step is always to find intersection points. These will be the endpoints of the area we must find.
Since both functions have
#{(y = -x^2 + 8x), (y = x^2 + 2x):}#
readily.
#x^2 +2x= -x^2 + 8x#
#2x^2 - 6x = 0#
#2x(x - 3) = 0#
#x = 0 or 3#
If we test a random point in this interval into both of the functions, we find that
#A = A_"highest curve" - A_"lowest curve"#
#A = int_0^3-x^2 +8x- (x^2 + 2x)dx#
#A = int_0^3 -x^2 + 8x - x^2 - 2x dx#
#A = int_0^3 -2x^2 + 6x dx#
The integral is given by
#A = [-2/3x^3 + 3x^2]_0^3#
#A = -2/3(3)^3 + 3(3)^2 - 0#
#A = -2/3(27) + 3(9)#
#A = -18 + 27#
#A = 9#
Hopefully this helps!
Explanation:
If you observe the graph of these two functions together, the area sought is between the two points of intersection. Therefore if we equate the functions together, we can then just find the area between the single function and the x axis.
The points of intersection are:
So our integral is:
Plugging in the upper and lower bounds:
Area
Graph of the two separate function with area 9 sq units.
Graph of single function with area 9 sq units.