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!
I see Calculus26 already answered you, but I wrote this all out so I'll just post it anyway.
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.