#2a) I don't get the process that cound the x-coordinate and point of tangency between f(x) and g(x)
#7c) Since the area between the two functions from x=-2 to x=0 is revolved around the x-axis, why is it that the function is subtracted from the chord when you use washers and not the other way around?
#9b) I'm having difficulty understanding how they differentiated from(e^x-e^(-x)x to [xe+xe^x+e^(-x)-e^x].
#10b) How did you determine that 1/b is the slope of l and why does b=1/n?
#18b) How was it determined that A(w) would equal to we^(-w^2)?
Thanks
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5 comments:
for 10b) i answered it in Priya's post, which is titled "HW#9 Area and Volume 10b"
for 18b)the area is that because it's just length times height which is we^(-w^2) if you make it w^2 or (e^(-w^2))^2 i'm not sure if the problem will work out but i just did we^(-w^2) (sorry, i don't really have a solid answer for this)
for 7c) I got the integral to be the chord squared subtracted from the function squared, and i multiplied it by pi. i did it like that because from x=-4 to x=0, the chord is under the function. but my interval was from -4 to 0 not -2 to 0. did you square root the function? that's what i did. and my answer came out to be 8pi/3, which matches the answer in the answer section. sorry if i'm wrong though.
for 9b) the way you can integrate (x)[e^x-e^(-x)] is use the uv-vdu
u=x du=1
dv=[e^x-e^(-x)] v=e^x+e^(-x)
uv-vdu=
x(e^x+e^(-x))-*integral of [e^x+e^(-x)]dx
=[x(e^x+e^(-x))-(e^x-e^(-x))]
then you plug in 2 and 0 and multiply by 2pi. hope it helps
for 18b the Area of a rectangle is length(x) times height (y)=xy
x=w
y=h(w)=e^(-w^2)
A(w)=xy= w[e^(-w^2)]
Thanks for your help
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