# pre-calculus questions

• Jan 26th 2008, 07:35 PM
daneeyah
pre-calculus questions
I have an online quiz, that I am currently taking and should submit pretty soon.
The answer in bold is the one I chose. If I am wrong, please explain why :confused:
Thanks for any input, I appreciate it.
---

1) Let R be the region between the curves y=x^2
and y=2x-(x^2).

To find the area of R, we need to

A) integrate [(x^2)-(2x-x^2)] from 0 to 1
B) integrate [(2x-x^2)-(x^2)] from 0 to 1
C) integrate [(x^2)-(2x-x^2)] from 0 to 2
D) integrate [(2x-x^2)-(x^2)] from 0 to 2

2)Let R be the region bounded by the curves y=(x^2)+3 and y=x, for x in [-1,1].

The area of R is equal to

A) 2
B) 17/3
C) 20/3

3) Let R be the region bounded by the curves x=(y^2) and y=x+5, for y in [-1,2].

Sketch the region, then answer the following question.

The most convenient way to find the area of R is to slice the region...

A) horizontally
B) vertically
• Jan 26th 2008, 07:49 PM
Jhevon
Quote:

Originally Posted by daneeyah
I have an online quiz, that I am currently taking and should submit pretty soon.
The answer in bold is the one I chose. If I am wrong, please explain why :confused:
Thanks for any input, I appreciate it.
---

1) Let R be the region between the curves y=x^2
and y=2x-(x^2).

To find the area of R, we need to

A) integrate [(x^2)-(2x-x^2)] from 0 to 1
B) integrate [(2x-x^2)-(x^2)] from 0 to 1
C) integrate [(x^2)-(2x-x^2)] from 0 to 2
D) integrate [(2x-x^2)-(x^2)] from 0 to 2

correct
• Jan 26th 2008, 07:51 PM
daneeyah
Thank you I appreciate it :)

I am pretty sure #2 is correct, but #3 I am a bit unsure about.
Thanks again for your time, I appreciate it.
• Jan 26th 2008, 07:52 PM
Jhevon
Quote:

Originally Posted by daneeyah
2)Let R be the region bounded by the curves y=(x^2)+3 and y=x, for x in [-1,1].

The area of R is equal to

A) 2
B) 17/3
C) 20/3

yes
• Jan 26th 2008, 07:56 PM
Jhevon
Quote:

Originally Posted by daneeyah
3) Let R be the region bounded by the curves x=(y^2) and y=x+5, for y in [-1,2].

Sketch the region, then answer the following question.

The most convenient way to find the area of R is to slice the region...

A) horizontally
B) vertically

no. did you graph it?

it is almost obvious from the graph that the best way to find the area here is to write both functions in terms of y and do the required integral. in that case, we need rectangles whose lengths are parallel to the x-axis (you need to know how we define the integral based on Riemann sums to know what rectangles we are talking about here)
• Jan 26th 2008, 08:00 PM
daneeyah
Quote:

Originally Posted by Jhevon
no. did you graph it?

it is almost obvious from the graph that the best way to find the area here is to write both functions in terms of y and do the required integral. in that case, we need rectangles whose lengths are parallel to the x-axis (you need to know how we define the integral based on Riemann sums to know what rectangles we are talking about here)

I just graphed it again.. I see how horizontal would be better.
Thanks again.. I just submitted my quiz and got a 100. ;)
• Jan 26th 2008, 08:03 PM
mr fantastic
Quote:

Originally Posted by daneeyah
I have an online quiz, that I am currently taking and should submit pretty soon.
The answer in bold is the one I chose. If I am wrong, please explain why :confused:
Thanks for any input, I appreciate it.
---

1) Let R be the region between the curves y=x^2
and y=2x-(x^2).

To find the area of R, we need to

A) integrate [(x^2)-(2x-x^2)] from 0 to 1
B) integrate [(2x-x^2)-(x^2)] from 0 to 1 Mr F says: Correct.
C) integrate [(x^2)-(2x-x^2)] from 0 to 2
D) integrate [(2x-x^2)-(x^2)] from 0 to 2

2)Let R be the region bounded by the curves y=(x^2)+3 and y=x, for x in [-1,1].

The area of R is equal to

A) 2
B) 17/3
C) 20/3 Mr F says: Correct.

3) Let R be the region bounded by the curves x=(y^2) and y=x+5, for y in [-1,2].

Sketch the region, then answer the following question.

The most convenient way to find the area of R is to slice the region...

A) horizontally
B) vertically Mr F says: Horizontal is most convenient for me. But you know what they say .... one man's meat is another man's poison ..... I think the question is quite stupid, actually.

..
• Jan 26th 2008, 08:04 PM
Jhevon
Quote:

Originally Posted by daneeyah
I just graphed it again.. I see how horizontal would be better.
Thanks again..

you're welcome :)

Quote:

I just submitted my quiz and got a 100. ;)
(Clapping) very nice!
• Jan 26th 2008, 08:28 PM
mr fantastic
Quote:

Originally Posted by Jhevon
you're welcome :)

(Clapping) very nice!

Yes, well done Jhevon! (Rofl) ;)