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Find the total area of the finite domains bounded between the curve $y=x^3−4x$ and the line $x+2y=2$.

I have sketched a graph of the curve and the line and found the points they intersect they $x$-axis and $y$-axis by equating them to 0. Afterwards, I tried to equate both equations together to find the point they intersect and I managed to reduced them down to $2x^3-7x-2=0$. Factorised it and I got $(x-2)(2x^2+4x+1)=0$. Then I tried to find the roots of the equation. Hmm and I got stuck...

Parcly Taxel
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Thomas King
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1 Answers1

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The steps that you performed so far are correct. Now, you need to solve for $(x-2)(2x^2+4x+1)=0$. Then, you just need to combine the two roots of the quadratic and the root of $(x-2)$. You can solve for the quadratic using the quadratic formula $x=\begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$. Next, you could then using the points of intersection as your bounds of integration to determine the areas.

Jesse Meng
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  • Of course he should take care in computing the area. The way the question is posed I would presume all area is to be counted as positive. The integrals will have one positive area and one negative area. – Dan Sp. Mar 11 '18 at 04:43
  • Yes, of course. He needs to integrate the absolute value of the differences for each interval. – Jesse Meng Mar 11 '18 at 04:48
  • Hi guys, thanks for your ur reply. so i used the quadratic formula and got the roots of the eqn. x=2, x=(-4+sqrt(8))/4, x=(-4-sqrt(8))/4. I used them as the limits for integration and got 5.14461 for the bigger area and 1.41422 for the smaller area. Thus the total area i got is 6.56(2s.f). Just to be sure, i could add them, tgt right? Even though one of the domain bounded is in the negative area. – Thomas King Mar 11 '18 at 13:25
  • Yes, you want to add up the total areas. It does not make sense for areas to cancel with each other. – Jesse Meng Mar 13 '18 at 04:49