so area
= integral (x^2 - mx + 1) dx from 0 to 5
= x^3 /3 - (m/2)x^2 + x from 0 to 5
= (125/3 - 25m/2 + 5) - (0-0+0)
but this equals 11.3024
times 6
250 - 75m + 30 = 67.8144
-75m = -212.1856
m = 2.8291
Suppose the total area under the curve y=x^2-mx+1 from 0 to 5 is 11.3024. Find m.
I know that this means that i need to find the integral of the absolute value of x^2-mx+1 from 0 to 5 and set it equal to 11.3024, but I am not sure where to go from there.
3 answers
I am trying to find the total area under the curve, not just the integral...
You know the roots are
x = (m±√(m^2-4))/2
So, if m<=2 there are not two real roots, so the total area is the same as the integral.
If m>2, then you need to divide the integral into 3 parts, with the roots as the limits of integration.
Reiny's solution of m=2.8291 indicates to me that you will need to break up the integral, since the curve dips below the x-axis.
x = (m±√(m^2-4))/2
So, if m<=2 there are not two real roots, so the total area is the same as the integral.
If m>2, then you need to divide the integral into 3 parts, with the roots as the limits of integration.
Reiny's solution of m=2.8291 indicates to me that you will need to break up the integral, since the curve dips below the x-axis.
3 answers