Area bounded by $f(x)=1/e^{x^2}$ and two lines $x=a, x=3a$

Question

The area is bounded by $$f(x)=e^{-x^2}$$ and two lines $$x=k$$ and $$x=3k$$ Find an expression for the area including $k$, and determine the value for $k$ which will maximize the area.

I general, you find the area between curves by finding their point(s) on intersection, taking the difference between them and taking the definite integral at the said point(s) - which gives us the integral.

Should I use that the same procedure here?

Answer 1

The area is simply $$\int_k^{3k}e^{-x^2}\,dx$$ But the primitive of the integrand isn't elementary. See https://en.wikipedia.org/wiki/Error_function. In any case, calculating explicitly the area, isn't required. Only the derivative is required.

Answer 2

While $$F(k):=\int_k^{3k}f(x)\,\mathrm dx$$ cannot be expressed elementary, you only need to find $k$ that maximizes $F(k)$ and not the explicit maximal value of $F$. For finding the maximizing $k$, note that $$F'(k)=3f(3k)-f(k) $$ (why?) and solve for $F'(k)=0$.

Answer 3

As Martin said, the area is given by $$A(k):=\int_k^{3k}e^{-x^2}\,dx.$$ To compute the maximum with respect to $k$, we use differentiation under the integral sign:

$$A'(k)=3e^{-(3k)^2}-e^{-k^2}=3(e^{-k^2})^9-e^{-k^2}.$$ The function $A(k)$ will attain an extremum at $k_0$ iff $A'(k_0)=0$. Setting $y:=e^{-k^2}$ this happens when $$3y^9-y=0\iff y=0\quad \lor \quad y=\pm\frac{1}{\sqrt[8]{3}}.$$ But we still need to solve for $k$, and the only possible value of $y$ which admits a solution is $+\frac{1}{\sqrt[8]{3}}$, so: $$e^{-k^2}=\frac{1}{\sqrt[8]{3}}\iff-k^2=-\frac{\ln(3)}{8}\iff k=\pm\sqrt{\frac{\ln(3)}{8}}.$$ Of course for the maximum we choose the $+$ sign.