- #1
gokugreene
- 47
- 0
Hey guys, I am curious if I am setting this up right. Could you take a look and make sure I am on the right path?
I have three questions.
1) I am trying to find the area bound by [tex]y=x^2[/tex] and [tex]y=4x+5[/tex]
Upper function [tex]y=4x+5[/tex]
Lower function [tex]y=x^2[/tex]
For my integral I have [tex]\int^{5}_{-1}4x-5-x^2[/tex]
I then get [tex]\left[\frac{-x^{3}}{3}+2x^{2}+5x\right]^{5}_{-1}[/tex]
and my answer is 110/3 units squared
Is my answer correct? Am I doing it right? I just figured out you can do it this way. My teacher has been teaching us to use squares, triangles, and other geometric methods (not talking about riemann sums) rather than just pure integration...
Ok, second question..
Find the area bounded by y=x-2, y=4, the x-axis, and the y-axis.
Would I set this problem up like [tex]\int^{6}_{2}x-2-\int^{2}_{0}x-2[/tex] this?
Last question..
I need to find the derivative of [tex]ln(x-1)^{2}[/tex]
I am confused as to whether they are saying to square x-1 or the natural log.
Thanks guys
I have three questions.
1) I am trying to find the area bound by [tex]y=x^2[/tex] and [tex]y=4x+5[/tex]
Upper function [tex]y=4x+5[/tex]
Lower function [tex]y=x^2[/tex]
For my integral I have [tex]\int^{5}_{-1}4x-5-x^2[/tex]
I then get [tex]\left[\frac{-x^{3}}{3}+2x^{2}+5x\right]^{5}_{-1}[/tex]
and my answer is 110/3 units squared
Is my answer correct? Am I doing it right? I just figured out you can do it this way. My teacher has been teaching us to use squares, triangles, and other geometric methods (not talking about riemann sums) rather than just pure integration...
Ok, second question..
Find the area bounded by y=x-2, y=4, the x-axis, and the y-axis.
Would I set this problem up like [tex]\int^{6}_{2}x-2-\int^{2}_{0}x-2[/tex] this?
Last question..
I need to find the derivative of [tex]ln(x-1)^{2}[/tex]
I am confused as to whether they are saying to square x-1 or the natural log.
Thanks guys