The factor I desire to carry out this is to calculation the area between this curve, the y-axis, y=5 and y=1.

You are watching: X in terms of y

I found the answer to it is in 5 making use of a roundabout method.

But my very first attempt to be to shot and express x in terms of y.

Basic algebra provided me one expression v an $x^3$.


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I think that this will get really confusing if you incorporate with respect come y.

Here is what I suggest you do.

$y = 2x + \frac 8x^2 -5$

Evaluated in ~ $y = 1$ and also $y = 5$

$1 = 2x + \frac 8x^2 -5\\x = 2$

$5 = 2x + \frac 8x^2 -5\\x = 1$

$\int_0^1 (5-1) \;dx + \int_1^2 2x + \frac 8x^2 -5 - 1\;dx$


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Basic algebra is correct.

From$y=2x+\frac8x^2-5$,multiplying by $x^2$we get$yx^2=2x^3+8-5x^2$,or$2x^3-(5-y)x^2+8= 0$.

This is,unfortunately,a cubicwhich can be resolved by thetraditionally confusing formula.

Its derivative is$6x^2-2(5-y)x= 0$which has roots$x=0$and$x=\dfrac5-y3$.

Note thatat $x=0$the duty is $8$and at$\dfrac5-y3$the value is$2(\frac5-y3)^3-(5-y)(\frac5-y3)^2+8=(\frac5-y3)^3(2-3)+8=-(\frac5-y3)^3+8=(\fracy-53)^3+8$.From this,we can determine whichvalues of $y$give one equation with$1$ or $3$ genuine roots.


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marty cohenmarty cohen
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To find the area, there isn"t a certain reason to uncover $x$ in regards to $y$. A graph of the role shows that the area is relatively simple to find.

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You just have actually to discover the area under your duty from $1$ come $2$ (you can use the basic Theorem of Calculus here), and also manipulate it to gain your desired area.


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CalconymCalconym
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