Ap Calculus BC 2002 Form B Question 3?

a) We must start by finding the intersection points of the two curves.

#3/4x = 4x - x^3 + 1#

Solve using a graphing calculator to get

#x = 1.940#

Thus our bounds of integration will be from #x = 0# to #x = 1.940#. Therefore, letting #a = 1.940#, we get

#I = int_0^a 4x - x^3 + 1 - 3/4x dx ~~ 4.515#

Thus the area will be #4.515# square units.

b) Recall the formula for volume around the x-axis:

#V = pi int_b^c (f(x))^2 - (g(x))^2 dx#

Where #f(x)# is the upper function and #g(x)# the lower. Thus in our case

#V = pi int_0^a (4x -x^3 + 1)^2 - (3/4x)^2 dx#

Once again using a calculator to evaluate we get

#V = 57.463# cubic units

c) The perimeter is given by adding the arc length of the linear function on #[0, 1.940]#, the arc length of the cubic function on #[0, 1.940]# and #y_2(0) - y_1(0) = 4(0) - 0^3 + 1 - 3/2(0) = 1#. We must recall the arc length formula:

#A = int_b^c sqrt(1 + (dy/dx)^2)dx#

#P = int_0^a sqrt(1 + (4x - 3x^2)^2)dx + int_0^a sqrt(1 + (3/2)^2) dx + 1#

#P = 7.528# units

The last couple of steps would have not been required on the exam because it states NOT TO EVALUATE the arc length.

Hopefully this helps!