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Conflicted on whether I did this correctly. Any suggestions are appreciated! enter image description here

Eight 8
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3 Answers3

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Your initial calculation of $\frac{1}{3}+\frac{49}{3}=\frac{50}{3}$ was correct for the two green areas

Your revised calculation of $5$ is correct for the purple area

What I do not know is whether the question is asking for the two green areas of $\frac{50}{3}$, or just the darker green area of $\frac{49}{3}$, or the purple area of $5$

enter image description here

Henry
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  • Thanks! I updated my answer. Does that look better? – Eight 8 Aug 16 '20 at 04:18
  • @Eight8 - All your calculations of areas have been correct, both initial and revised. I am not certain precisely which area you actually being asked for – Henry Aug 16 '20 at 09:19
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I think you've shaded the wrong region. Here is a plot of the three curves in question:

enter image description here

The only region that is bounded by all three curves and no others, is the one between $x = 1$ and $x = 4$, and from $y = 1$ to $y = 16$. Note you are not asked for the region below $y = \sqrt{x}$.

Therefore, the desired area is $$\int_{x=1}^4 x^2 - \sqrt{x} \, dx.$$ I leave the computation as an exercise.

heropup
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If you want to find the area of the purple shaded region, that is not what you have.

That area would be $\int_0^1 x^2 \ dx + \int_1^4 \sqrt x$

But it is not clear that you in fact do want the that shaded area. That would would include the boundary $y= 0$ which is not mentioned in the instructions.

My picture looks like:

enter image description here

Eventually the parabola will hit the line, but I truncated the picture.

Which is what your work shows.

You have an unexplained sign change at the last step. It is not wrong, but I would get in front of it, rather than flipping a sign with no explanation.

$\int_0^4 |\sqrt x - x^2| \ dx = \int_0^1 \sqrt x - x^2 \ dx + \int_1^4 x^2-\sqrt x \ dx$

Doug M
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