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I am trying to find the area under the curve $f(x)= C - x^4$ bounded by the $x$ and $y$ axes, where $x$ is any real positive number.

I know to find the area under the curve I need to evaluate the integral of $f(x)$. For example to find the area under the curve on the positive $x$ axis I take the integral of $C -x^4$ over $[0, c^{\frac{1}{4}}]$. However when you find the area using that formula, you find the area under the $y$ axis as well.

Any help on how to find the area of $C - x^4$ when $x \ge 0$, $y \ge 0$, $C \ge 0$ would be greatly appreciated.

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  • $\begingroup$ You seem to be a little confused about what kind of area an integral measures, because you already found the right answer without recognizing that it is the right answer. If you count all the area under the curve, not just the area between the curve and the $x$-axis, the area is unbounded ("infinite" if you prefer that term), certainly not the finite real value that your integral will produce. In other words, the integral is doing exactly what you want already. $\endgroup$ – David K Sep 30 '15 at 14:49
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When $c$ is larger than zero the area enclosed by $[0,c^{1/4}]$ is above the $x$ axis.

To prove this consider the minimum of $c-x^2=f(x)$ continuous on the interval $[0,c^{1/4}]$,

We notice that when $c>0$, $f'(x)=0$ occurs at a local maximum only, when $x=0$. Thus, we find the values of the endpoints to find the minimum. $x=0$ is already checked, so our minimum has to be at $x=c^{1/4}$. Substituting values we see our minumum is $f(x)=c-(c^{1/4})^4=0$.

Thus by integrating in this region you find the area enclosed by $c-x^4$ and the $x$ and $y$ axis only.

$$\int_{0} ^ {c^{1/4}} (c-x^4) dx$$

$$=c(c^{1/4})-\frac{(c^{1/4})^5}{5}-0$$

$$=(4/5)c^{5/4}$$

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