# Area between curve and straight line with parameter

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.

• Please edit your question, it does not make sense. What is $u$? How can a line be "equal" to $\frac 43 u^2$? – Ritam_Dasgupta May 24 at 16:52
• 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 ? – Jean Marie May 24 at 17:48
• $u^2$ 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 $y = a$ – Daniel May 24 at 20:32
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.