Question 1

Which of the following correctly represents the area between $$y = 6 - 3x^2$ and y = 6x - 3$ for the values of x$ where the curves intersect?

Accepted Answer

$$\int_{-1}^{2} \left[(6 - 3x^2$) - (6x - 3)\right] \, dx$$

Question 2

Given the region bounded by $$y = 3 \cos x$$ and $$y = 3 \cos 2x$$ for $$0 \leq x \leq \pi$$, what is the correct expression for the area between the curves?

Accepted Answer

$$\int_{0}^{\pi} \left[3 \cos x - 3 \cos 2x\right] \, dx$$

Question 3

A region bounded by $$y = 2 - x$$, $$y = 2 - 2x$$, and $$y = 0 is $$revolved about $$x = 3. $$Which integral correctly represents the volume using the washer method?

Accepted Answer

$$\pi \int_{0}^{1} \left[(3 - (2 - x))^2$ - (3 - (2 - 2x))^2$\right] \, dx$$

Question 4

For the piecewise function $$f(x) = \begin{cases} 1 & 0 \leq x \leq 1 \ 2 - \sqrt{x} & 1 \u003c x \leq 4 \end{cases}$$, what is the total area under $$f(x)$$ from $$x = 0 to$$ $$x = 4$$?

Accepted Answer

$$1 + \int_{1}^{4} (2 - \sqrt{x}) \, dx$$

Question 5

When the region under $$f(x) in $$the first quadrant is revolved about the $$x-$$axis, which method and integral would you use to find the volume?

Accepted Answer

Disk method: $$\pi \int_{0}^{1} (1)^2$ \, dx + \pi \int_{1}^{4} (2 - \sqrt{x})^2$ \, dx$$

Question 6

When the region under $$f(x) is $$revolved about the $$y-$$axis, which integral represents the volume using the shell method?

Accepted Answer

$$2\pi \int_{0}^{1} x \cdot 1 \, dx + 2\pi \int_{1}^{4} x (2 - \sqrt{x}) \, dx$$

Question 7

What is the correct formula for the arc length of the curve $$y = \frac{x^3$}{6} + \frac{1}{2x} on$$ $$[1, 2]$$?

Accepted Answer

$$\int_{1}^{2} \sqrt{1 + \left(\frac{d}{dx}\left(\frac{x^3$}{6} + \frac{1}{2x}\right)\$$right)^2$$} \, dx$$

Question 8

A curve $$y = \frac{1}{4}(e$$^{2x}$$ + e$$^{-2x}$$) is $$revolved about the $$x-$$axis over $$[-2, 2]. $$Which integral represents the surface area generated?

Accepted Answer

$$2\pi \int_{-2}^{2} y \sqrt{1 + \left(\frac{dy}{dx}\$$right)^2$$} \, dx$$

Question 9

In a medical imaging scenario, the cross-section of a tumor is modeled by $$y = 6 - 3x^2$ and y = 6x - 3$. If the area between these curves is 32$, what does this area represent?

Accepted Answer

The physical cross-sectional area of the tumor in square units

Question 10

A chemical reaction produces a region bounded by $$y = 3 \cos x$$ and $$y = 3 \cos 2x$$ for $$0 \leq x \leq \pi. If $$the area is $$\frac{9\sqrt{3}}{2}$$, what does this value represent in a real-world context?

Accepted Answer

The total difference in concentration between two reactants over the interval

Question 11

If a region bounded by $$y = 2 - x$$, $$y = 2 - 2x$$, and $$y = 0 is $$revolved about $$x = 3$$, what does the resulting volume represent in a manufacturing context?

Accepted Answer

The volume of a hollow cylindrical part produced by rotating the region

Question 12

Which of the following is the correct formula for the surface area generated by revolving $$y = f(x)$$ about the $$x-$$axis from $$x = a to$$ $$x = b$$?

Accepted Answer

$$2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2$} \, dx$$

Question 13

A curve $$y = \frac{x^3$}{6} + \frac{1}{2x} is $$used to model the growth of a plant stem. If the length of the curve from $$x = 1 to$$ $$x = 2 is$$ $$\frac{17}{12}$$, what does this value represent?

Accepted Answer

The actual length of the plant stem between $$x = 1$$ and $$x = 2$$

Question 14

Which method is most appropriate for finding the volume of a solid generated by revolving a region about the $$y-$$axis?

Accepted Answer

Shell method

Question 15

In the context of calculus, what does the 'washer method' calculate?

Accepted Answer

The volume of a solid with a hole in the middle, generated by revolving a region around an axis

Question 16

If $$y = \frac{1}{4}(e$$^{2x}$$ + e$$^{-2x}$$) is $$revolved about the $$x-$$axis, which of the following is the correct formula for the surface area over $$[-2, 2]$$?

Accepted Answer

$$\frac{\pi}{8}(e^8$ + 16 - e$$^{-8}$$)$$

Question 17

A region is bounded by $$y = 2 - x$$, $$y = 2 - 2x$$, and $$y = 0. $$What is the best way to set up the limits of integration for the volume when revolving about $$x = 3$$?

Accepted Answer

Find the $$x-$$values where $$y = 0$$ for both lines and use those as the limits

Calculus Practice Exam Guidance: Areas, Volumes, Arc Length, and Surface Area

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