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If $ f(x) = x^2 $ and $ g(x)=a ~, a \in \mathbb{R}, ~ a > 0 $, find the area between the parabola and the line that equals $4/3 $.

I know the integral is $$ A = 2\int_0^{\sqrt{a}} (a - x^2)dx = \dfrac{4} {3} \rightarrow a=1 $$

The thing is, according to my workbook, the answer is $ a = (4)^{1/3} $

I just dont see why the answer is that, even though I see that when $a=1$ the given area is not satisficied.

Thanks is advance.

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    $\begingroup$ Please edit your question, it does not make sense. What is $u$? How can a line be "equal" to $\frac 43 u^2$? $\endgroup$ – Ritam_Dasgupta May 24 at 16:52
  • $\begingroup$ As remarked by Ritam_Dasgupta it is impossible to understand what you mean: a straight line has equation $y=mx+c$ ; (may be is it $y=x+\frac43 u^2$ ?); why haven't answered his question an hour later ? $\endgroup$ – Jean Marie May 24 at 17:48
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    $\begingroup$ $ u^2 $ just mean square units, for instance, square meters, it is just a way to define generic area units. For straight line, I'm sorry, I meant a horizontal line, denote as $ g(x) = a $, so the slope is 0. It is the same as to say $ y = a $ $\endgroup$ – Daniel May 24 at 20:32
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It turned out to be that my workbook was wrong. The correct answer was $ a = 1 $, as I guessed in the first place. Sorry for the inconvenience.

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