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I am currently trying to do this question but I am having trouble getting the correct answer of 25.

I am also having trouble understanding whether my limits are right and whether my method of doing is correct.

I am trying to take the big regions (with the triangles) - regions of triangle to get the shaded area.

This is my current working $(\int_0^4\int_0^\sqrt xx\:dydx\;-\;\int_1^4\int_0^\sqrt xx\:dydx\;)+\;(\int_0^9\int_{-x}^0x\:dydx\;-\;\int_1^9\int_{-x}^0\:x\:dydx)$

enter image description here

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    $\begingroup$ If you are trying to get the area, the integrand should be $1,$ not $x$ $\endgroup$ – saulspatz Mar 16 at 14:00
  • $\begingroup$ it still doesnt get me to 25 though, then what is the $\int\:\int\:x\;DA$ for? $\endgroup$ – deviljones Mar 16 at 14:04
  • $\begingroup$ It should be $\int\int\mathrm{dA},$ not $\int\int x\mathrm{dA},$ but I'm not sure you have the limits of integration right. I'll check. $\endgroup$ – saulspatz Mar 16 at 14:08
  • $\begingroup$ The equation of the straight lines containing the hypotenuses of the right-angled triangles in the first and fourth quadrants is $y=\frac23\cdot(x-1)$ and $y=\frac38\cdot(1-x)$ respectively. The integral should be:$$\left(\int_0^4\int_0^\sqrt xdy~dx-\int_1^4\int_0^{\frac23\cdot(x-1)}dy~dx\right)+\left(\int_0^9\int^0_{-\sqrt x}dy~dx-\int_1^9\int_{\frac38\cdot(1-x)}^0dy~dx)\right)$$ $\endgroup$ – Shubham Johri Mar 16 at 14:18
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    $\begingroup$ Anywho I don't think the question wants you to find the area of the shaded region, which is $25/3$ and not $25$, as you mentioned in your post. It probably wants you to find $\int\int_AxdA$ as mentioned in the picture, which indeed evaluates to $25$. You just need to change the integrand from $1$ to $x$ in each of the above integrals with identical limits, i.e.$$\left(\int_0^4\int_0^\sqrt xxdy~dx-\int_1^4\int_0^{\frac23\cdot(x-1)}xdy~dx\right)+\left(\int_0^9\int^0_{-\sqrt x}xdy~dx-\int_1^9\int_{\frac38\cdot(1-x)}^0xdy~dx)\right)$$ $\endgroup$ – Shubham Johri Mar 16 at 14:32
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First of all, to get the area, the integrand should be $1$ or if you like $\mathrm{dA},$ not $x$ (or $x\mathrm{dA}).$ Also, you have the limits of integration incorrect. In the region above the $x-$axis, the limits in the first integral are correct, but the inner limits in the second integral are wrong. $y$ runs from $0$ to the line segment joining joining the points $(1,0)$ and $(4,2)$ You need to use this line to determine the $y$ value.

Below the $x-$ axis you have the limits incorrect in both integrals. In the first integral the lower limit should be $-\sqrt{x}$ not $-x,$ though I suspect this is a typo, and in the second integral, you've made the same sort of error in the inner limits as you did above the $x-$axis.

I think, with these pointers, you'll be able to correct your mistakes. Good luck.

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  • $\begingroup$ I don't think the question was about finding the area, rather $\int\int_AxdA$. I think the OP made a mistake in the body of the post. $\endgroup$ – Shubham Johri Mar 16 at 14:33
  • $\begingroup$ @ShubhamJohri Could be. I haven't actually computed any integrals. If you have, then I'm sure you are correct. $\endgroup$ – saulspatz Mar 16 at 14:36
  • $\begingroup$ yes it was a typo, apologies on my part, nevertheless, i was still wrong for the other limits, didnt know I have to find the equations of the 2 straight lines by finding the gradient and expressing it in y=mx+c form $\endgroup$ – deviljones Mar 16 at 14:38
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    $\begingroup$ It's just a matter of finding the equations of the curves that bound the regions in question. Each of those curves puts restrictions on the points $(x,y)$ that lie in the shaded region. $\endgroup$ – saulspatz Mar 16 at 14:42

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