# Ap Calculus BC 2002 Form B Question 3?

## Apr 29, 2018

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

$\frac{3}{4} x = 4 x - {x}^{3} + 1$

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

$I = {\int}_{0}^{a} 4 x - {x}^{3} + 1 - \frac{3}{4} x \mathrm{dx} \approx 4.515$

Thus the area will be $4.515$ square units.

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

$V = \pi {\int}_{b}^{c} {\left(f \left(x\right)\right)}^{2} - {\left(g \left(x\right)\right)}^{2} \mathrm{dx}$

Where $f \left(x\right)$ is the upper function and $g \left(x\right)$ the lower. Thus in our case

$V = \pi {\int}_{0}^{a} {\left(4 x - {x}^{3} + 1\right)}^{2} - {\left(\frac{3}{4} x\right)}^{2} \mathrm{dx}$

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 $\left[0 , 1.940\right]$, the arc length of the cubic function on $\left[0 , 1.940\right]$ and ${y}_{2} \left(0\right) - {y}_{1} \left(0\right) = 4 \left(0\right) - {0}^{3} + 1 - \frac{3}{2} \left(0\right) = 1$. We must recall the arc length formula:

$A = {\int}_{b}^{c} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

$P = {\int}_{0}^{a} \sqrt{1 + {\left(4 x - 3 {x}^{2}\right)}^{2}} \mathrm{dx} + {\int}_{0}^{a} \sqrt{1 + {\left(\frac{3}{2}\right)}^{2}} \mathrm{dx} + 1$

$P = 7.528$ 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!