find the area of the region in the first quadrant that is bounded above by the graph of f(x) = sqrt(x) and below by the x-axis and the line g(x) = x - 2 using integration with respect to y. show steps/explain.
draw it, always find the points of intersection once you have all that it is easier to approach the problem
how do i find y upper and lower
ya, i've got a rough sketch like that
looks like 2 and 0
for the y-axis
well... u have an exact graph.. how do i find those values otherwise
find the intercsection
set the two formulas equal
and find both coordinates
\[y^2=y+2\implies y^2-y-2=(y+1)(y-2)=0\]and we don't care about the negative intersect since we're in QI
I solved it for y because it's easier than solving \[\sqrt x=x-2\], and also because we are looking for the y-value anyway
how'd you get y^2 = y + 2
oh... i think i got it
\[\sqrt x=y\implies x=y^2\]\[x-2=y\implies y+2=x\]setting\[x=x\]we get\[y^2=y+2\]
now how do we decide which is our upper and which is our lower do you know?
0 lower, 2 upper... cause 2 is bigger than 0 and it looks right?
yes, but which function minus which function in the integrand?
right most - left most. so here it's y+2 - y^2
yep
well that's all happy integrating!
how come i dont have to split it into two separate pieces like in the "with respect to x" version
because I neglected to mention that part :P
lol
oh, no, it;s because there is no "chink" in the "right-left" like there is in the "top-bottom" way along x
it might be right.. i was just wondering
define chink, lol
sharp corner-type pointy thing ^^^ oxford english standard
haha
meh, well i guess i'll just have to eye-ball it
|dw:1344234978887:dw|very rough sketch as we gio along x we are adding up little rectangles that stretch up and down
... eye-balling it probably not good for bridge building :)
yeah, i gotcha
|dw:1344235038918:dw|but when we get to the point we have to use a different bottom....
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