a. f ' = 6(x-3)^2
now simply se this equal to 6
b. the integration limits are 0 and the x-intercept (set f = 0 simple enough)
the area is a naegative of the integral of f(x) on this interval
Hello guys! Im having trouble with this question i would appreciate the help.
curve y=( 2x-3)^3
a) find the x coordinates of the two points on the curve which have gradient 6
b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.
Thank you in advance!
a) Take the derivative a set it equal to 6.
solve for x when
b) You are integrating
To find the upper limit of integration you need to set your function equal to zero, because that is the x value of your x intercept. Your function equals 0 when (2x-3) = 0, or when x = 3/2
So your integral becomes
You'll get a negative number for the area since the curve is below the x-axis on the interval which you're integrating, so your area will be the absolute value of the answer you get by evaluating that integral.