Q: The area of one leaf of the four-petaled rose $$r = \sin(2\theta) is $$given by which of the following?

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin(2\theta)]^2 \, d\theta$$

Q: Which of the following correctly sets up the integral for the area inside the cardioid $$r = 1 - \cos \theta$$?

$$\frac{1}{2} \int_{0}^{2\pi} (1 - \cos \$$theta)^2$$ \, d\theta$$

Question 1

Which of the following integrals correctly represents the area enclosed between the curves $$f(x) = x^2$ + 2$$ and $$g(x) = 2x + 5$$?

Accepted Answer

$$\int_{-1}^{3} \left[(2x + 5) - (x^2$ + 2)\right] \, dx$$

Question 2

The area of one leaf of the four-petaled rose $$r = \sin(2	heta) is $$given by which of the following?

Accepted Answer

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin(2	heta)]^2 \, d	heta$$

Question 3

A solid has a base bounded by $$x = 2 - y^2$ and x$$ = $$y^2 - 2$, with cross-sections perpendicular to the y$-axis that are squares. Which integral gives the volume of this solid?

Accepted Answer

$$\int_{-\sqrt{2}}^{\sqrt{2}} \left[(2 - y^2$) - (y^2$ - 2)\right]^2 \, dy$$

Question 4

Which of the following correctly sets up the integral for the area inside the cardioid $$r = 1 - \cos 	heta$$?

Accepted Answer

$$\frac{1}{2} \int_{0}^{2\pi} (1 - \cos \$$theta)^2$$ \, d	heta$$

Question 5

Given the region bounded by $$y = x^2$, y = 8$$ - $$x^2$$, and $$x = 0 in $$the first quadrant, which integral represents the volume generated by revolving this region about the $$y-$$axis?

Accepted Answer

$$2\pi \int_{0}^{2} x [(8 - x^2$) - x^2$] \, dx$$

Question 6

Find the arc length of $$f(x) = \frac{x^2$}{2} - \frac{\ln x}{4}$$ over $$[1, 3]. $$Which of the following integrals correctly represents this arc length?

Accepted Answer

$$\int_{1}^{3} \sqrt{1 + \left(x - \frac{1}{4x}\$$right)^2$$} \, dx$$

Question 7

Which of the following integrals correctly sets up the volume of the solid generated by revolving the region bounded by $$y = \ln x$$, $$y = 0$$, and $$x = e$$ about the line $$x = -2$$?

Accepted Answer

$$2\pi \int_{1}^{e} (x + 2) \ln x \, dx$$

Question 8

A region is bounded by $$x = y^2$ and x$$ = \sqrt{y}$$. What is the correct integral for the volume generated by rotating this region about the y$-axis?

Accepted Answer

$$2\pi \int_{0}^{1} y (\sqrt{y} - y^2$) \, dy$$

Question 9

Which of the following integrals correctly represents the arc length of $$y = 1 - e$$^{-x}$$ over $$[0, 1]$$?

Accepted Answer

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

Question 10

Suppose you are asked to find the area inside the circle $$r = 4 \cos 	heta$$ and to the right of the vertical line $$r = \sec 	heta. $$Which of the following is the correct setup for this area?

Accepted Answer

$$\frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} [4 \cos 	heta]^2 \, d	heta$$

Question 11

Given the region bounded by $$y = x^2$ - 1$$ and $$y = 3$$, with cross-sections perpendicular to the $$y-$$axis that are equilateral triangles, which integral gives the volume of the solid?

Accepted Answer

$$\int_{-1}^{3} \frac{\sqrt{3}}{4} \left([\sqrt{y + 1}] - [-\sqrt{y + 1}]\$$right)^2$$ \, dy$$

Question 12

Which of the following integrals correctly sets up the arc length of $$f(x) = x$$^{3/2}$$ over $$[0, 2]$$?

Accepted Answer

$$\int_{0}^{2} \sqrt{1 + \left(\frac{3}{2} x$$^{1/2}$$\$$right)^2$$} \, dx$$

Question 13

For the region bounded by $$y = x^2$ and y$$ = $$(x-2)^2$$ and the $$x-$$axis, which of the following best describes the area of the region?

Accepted Answer

$$\int_{0}^{2} \left|x^2$ - (x-2)^2$\right| \, dx$$

Question 14

Given the region bounded by $$f(x) = \sqrt{3} \cos x on$$ $$[0, \frac{\pi}{4}]$$, which integral represents the volume generated by rotating this region about the $$x-$$axis?

Accepted Answer

$$\pi \int_{0}^{\frac{\pi}{4}} [\sqrt{3} \cos x]^2 \, dx$$

Question 15

Which of the following is the correct setup for the volume of the solid generated by revolving the region bounded by $$x = y(4 - y)$$ and $$x = (y - 2)^2$ about the x$-axis?

Accepted Answer

$$\pi \int_{0}^{4} \left([y(4 - y)] - [(y - 2)^2$]\$$right)^2$$ \, dy$$

Question 16

For the region bounded by $$y = 2 - x^2$ and y$$ = $$x^2 - 2$, with cross-sections perpendicular to the x$-axis that are semicircles, which integral gives the volume of the solid?

Accepted Answer

$$\frac{\pi}{8} \int_{-\sqrt{2}}^{\sqrt{2}} \left[(2 - x^2$) - (x^2$ - 2)\right]^2 \, dx$$

Question 17

Which of the following is the correct setup for the volume of the solid generated by revolving the region bounded by $$y = x^2$, y = 8$$ - $$x^2$$, and $$x = 0 in $$the first quadrant about the line $$x = 3$$?

Accepted Answer

$$2\pi \int_{0}^{2} (3 - x)[(8 - x^2$) - x^2$] \, dx$$

Step-by-Step Calculus Guidance for Math 126 Exam 2 Review

Guida di studio - Domande di pratica

Study guide - Flashcard