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find the area between:

$$\begin{array}{} y=4\\ y=2\sqrt{x}\\ y=3-x \end{array}$$

I have found the intersection points between the curves to find the interval but even while doing it with respect to x or to y, my answer is always wrong. please help

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    $\begingroup$ You should post your answer so that we can point out the mistake, if any. $\endgroup$ – Justpassingby Dec 9 '15 at 19:22
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    $\begingroup$ If you show us your work then we might be able to tell you where you’re going wrong. $\endgroup$ – amd Dec 9 '15 at 19:22
  • $\begingroup$ the problem is not so clear have you made a picture? $\endgroup$ – Dr. Sonnhard Graubner Dec 9 '15 at 19:30
  • $\begingroup$ It looks straight forward to me. Exactly WHAT did you do? What answer did you get? What answer do you think you should have gotten? $\endgroup$ – user247327 Dec 9 '15 at 19:32
  • $\begingroup$ help yourself with wolframalpha.com/input/… $\endgroup$ – janmarqz Dec 9 '15 at 19:34
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Hints

  1. Take a look at this picture

enter image description here

  1. You should compute

$$\text{Area}=\color{blue}{\text{Blue}}+\color{green}{\text{Green}}=\int_{-1}^{1} (4-(3-x)) dx + \int_{1}^{4} (4-2\sqrt x) dx$$

  1. The final answer is $\frac{14}{3}$
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    $\begingroup$ +1 for plots. Answer with pictures beats answers without pictures all day every day. $\endgroup$ – Kaster Dec 9 '15 at 19:43
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HINT: Draw a diagram indicating the required area bounded by $y=4$, $y=2\sqrt x$ & $y=3-x$. Divide the bounded region into two smaller regions, then one should get area bounded $$=\frac{1}{2}(2\times 2)+\left(3\times 4-\int_{1}^{4}2\sqrt x\ dx\right) $$

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