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How to solve this question ?

The integral would be 0.3x^3+xc^2 + c 0.3(64)^3 +(64)c^2 +c =0 c= 78643.2/( -64+c) am I right ?

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Draw a picture. You will get a familiar parabola, with axis the $y$-axis. Note that the region we want the area of is below the $x$-axis, and extends from $x=-c$ to $x=c$. By symmetry, the area is $$2\int_0^c (0-(x^2-c^2))\,dx.$$ So we are integrating $c^2-x^2$, and then doubling. Calculation gives $\frac{4}{3}c^3$.

We want this to be $64$. It is now not hard to solve for $c$.

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The problem should be restated because the area "between" the function $y=x^2-c^2$ and $y=0$ is always equal to $\infty$.

I would restated the problem as follows

Find the exact positive value of $c$ if the area of the region bounded by $y=x^2-c^2$ and the x-axis is 64.

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Next, make some sketches. Clearly the region bounded by the curve and the x-axis is the region in blue. The area between the 2 curves is the |pink area|+|blue area|.

The area bounded would be $$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg|$$ $$= \bigg|\frac{x^3}{3}-c^2x\bigg|_{-c}^c\bigg| = \bigg|\bigg((\frac{c^3}{3}-c^3) - (-\frac{c^3}{3}-(-c^3))\bigg)\bigg|$$ $$= \bigg|-\frac{2}{3}c^3- \frac{2}{3}c^3\bigg|$$ $$= \bigg|-\frac{4}{3}c^3\bigg| = \frac{4}{3}c^3$$

But $$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg| = 64$$ so $$\frac{4}{3}c^3 = 64$$

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