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Find the area between the two curves $y = x^2 + ax$ and $y = 2x$ between $x=1$ and $x=2$, given that $a>2$

I found the antiderivative of $x^2+ax-2x$ and included the area between $x=1$ and $x=2$ which is $\dfrac{3a-4}{6}-\dfrac{6a-4}{3}$. I don't understand what $(a>2)$ means in the problem.

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  • $\begingroup$ $(a>2)$ is not part of the second equation, if that is part of the confusion? it is a separate part. $\endgroup$ – Dre Sep 12 '13 at 2:27
  • $\begingroup$ Oh. When I was editing the question, I implemented Dre's suggestion without thinking. If that is the confusion, please edit it back to the way it was... $\endgroup$ – apnorton Sep 12 '13 at 2:31
  • $\begingroup$ the (a>2) is written right next to the "y=2x" part of the question. I don't why it's there or what how to incorporate it $\endgroup$ – user94418 Sep 12 '13 at 2:37
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Hint: the area between two curves $f(x)$ and $g(x)$ can be found by the formula $$A=\int_c^d{[f(x)-g(x)]dx}$$

It seems the reason it's asking you for $a>2$ is it will cancel the $2x$ if $a=2$. Perhaps you should plot the graphs for different values of $a$.

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  • $\begingroup$ I have used that however I still have "a" in the answer. I don't know what to do with it. $\endgroup$ – user94418 Sep 12 '13 at 2:35
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    $\begingroup$ The $a$ is just an arbitrary scalar. If it equals 2 your $x^2+ax-2x$ just becomes $x^2$ and they probably are trying you to work with multiple terms and arbitrary scalers. $\endgroup$ – Eleven-Eleven Sep 12 '13 at 2:43
  • $\begingroup$ Unfortunately it is asking me for a final answer on the website. I tried entering it with the "a" still in the answer and it doesn't accept it. How can I get a single answer when "a" is anything from greather than 2 to infinity? $\endgroup$ – user94418 Sep 12 '13 at 2:53
  • $\begingroup$ Did you integrate properly? I got $\frac{7}{3}+\frac{3(a-2)}{2}$ $\endgroup$ – Eleven-Eleven Sep 12 '13 at 2:58
  • $\begingroup$ Having $a$ is completely acceptable in the answer if it gives you something like $a>2$. I suspect you are missing something critical or you made a calculation error. $\endgroup$ – Eleven-Eleven Sep 12 '13 at 3:00

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