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