Please try yourself first.
sketch a graph.
Find where the two curves intersect.
Then integrate the difference between intersection points.
Note, sketch carefully between x = 0 and x = 5. I think you may find two regions.
The total area enclosed by the graphs of
y=6x^2–x^3+x and y=x^2+5x is???
2 answers
Find the points where the curves intersect.
y1 = 6x^2 –x^3 + x = y2 = x^2 + 5x
x^3 - 5x^2 +4x = 0
x(x^2 -5x +4) = 0
x(x-4)(x-1) = 0
One curve (y2) is higher from x=0 to x=1, the other (y1) is higher from x=1 to x=4.
Add the integral from 0 to 1 of (y2 - y1) to the integral from 1 to 4 of (y1 - y2). You need to reverse the sign of the integrand to avoid having one enclosed ares subtracted from the other.
y1 = 6x^2 –x^3 + x = y2 = x^2 + 5x
x^3 - 5x^2 +4x = 0
x(x^2 -5x +4) = 0
x(x-4)(x-1) = 0
One curve (y2) is higher from x=0 to x=1, the other (y1) is higher from x=1 to x=4.
Add the integral from 0 to 1 of (y2 - y1) to the integral from 1 to 4 of (y1 - y2). You need to reverse the sign of the integrand to avoid having one enclosed ares subtracted from the other.
2 answers