compute the area enclosed by the graphs of y = x + 6 , y = x^3 , and y = -x/2
whats your graph of the bounded region ...
sketched it out, and looks like left bounds are the intersection of the linear functions, and right bounds are the intersection of x + x and x^3
so, -4 to 2
just as an aside, how would you solve x + 6 = x^3
complete the cube maybe, but i just wolfed it
or it's obvious that it's two
wasnt obvious to me till now lol
or to me until I saw the wolf graph of the three
ok, now is our boundary consistent, unchanging, along our interval?
no, it changes at 0
then what are your thoughts as to how we work this?
calc from -4 to 0, add it to from 0 to 2
very good, so whats our setup, what do you need help with
reassurance, will post the final answer, these are much easier than some of the antiderivatives from previous chapters
got 24 for A1
\[\int_{-4}^{0}x+6+\frac12x~dx+\int_{0}^{2}x+6-x^3~dx\] \[\left.\frac12x^2+6x+\frac14x^2\right|_{-4}^{0}~+~\left. \frac12x^2+6x-\frac14x^4\right|_0^{2}\] \[\left.-\frac12(-4)^2-6(-4)-\frac14(-4)^2\right.~+~\left. \frac12(2)^2+6(2)-\frac14(2)^4\right.\]
-8+24-4 = 12 for A1
-3/4 * -64 - 24 = A1 ...is the area formula I cam up with after plugging in
show me your antiderivative
oops, cubed -4, should be squared
got the same as you, (3/4)x^2 + 6x
good :)
0 - (3/4(-4)^2 - 24 = ...
-6(-4) = +24
oops, darn parenthesis
12
[3/4 (0)^2 + 6(0)] - [3/4 (-4)^2 +6(-4)] [0 + 0] - [12 -24] [0] - [-12] = 12
A2 = 10
correct
AA = 22
very good
woot, getting funner
yeah, these are to build your confidence, theyll tear it down later ;)
not me, probably done after this...bigger and better things to learn that computers can't do for us
but yeah, this should be rudamentary for someone pursuing any sciences
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