Calculus Help

[MrCodeDude]

Tomorrow is a test for Calc and usually my teacher holds these "study sessions" where we go in the evening to his class and we just do a bunch of problems to help us study. Tonight he couldn't do it, but we're still having a test. 🙁 Anyway, I have a bunch of even problems assigned and there are only odd answers in the back of the book. Anyway, ATOT has bailed me out in the past, I figured it's worth it to ask again 😉

Find the area between the two intersecting lines.

FIRST PROBLEM
x = y^2 + 1, x = y+3

I solved for y for both equations and got:

y = sqrt[x-1] , y = x-3

Setting both equations equal to each other, you find out that they intersect at x = 2 and 5.

Integral from 2 to 5 of [ sqrt[x-1] - (x-3) ]

Is that the correct integral? Cause if you graph both equations, you need the negative sqrt[x-1].

 

[Capone]

first equation should be sqrt[x-1]

edit nm you wrote it right in the integral

 

[MrCodeDude]

Originally posted by: Capone
first equation should be sqrt[x-1]

edit nm you wrote it right in the integral

😉 Yeah, meant x-1 🙂

 

[MrCodeDude]

NM, I heard you can just switch the x's and y's.. ?

 

[dude8604]

why don't you graph the x = function and do the integral in terms of dy.

 

[Capone]

Had to get out my calculator, yeah you do need the - sqrt too
because it should be y = + or - sqrt[x-1]

edit: post above was my next step

 

[PowerMacG5]

Originally posted by: dude8604
why don't you graph the x = function and do the integral in terms of dy.

What he said.

 

[MrCodeDude]

Is the answer 9/2 ?

Or 27 / 6?

 

[Capone]

9/2 = 27/6

 

[MrCodeDude]

Originally posted by: Capone
9/2 = 27/6

LOL. If Calculus doesn't kill me, arithmetic will 🙁