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I know how to solve it by the use of integrals, but I wonder why solving it using a graph would bring incorrect result? I mean, why would it be wrong to just consider roots of these functions and solve it drawing solely on them?

Consider two functions:
$f (x) = a − x^2 $
$g(x) = x^4 − a $ For precisely which values of $a > 0$ is the area of the region bounded by the $x$-axis and the curve $y=f(x)$ bigger than the area of the region bounded by the $x$-axis and the curve $y=g(x)$?

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First of all, the question is not entirely well-posed: are you intended to calculate the area along the entire infinite $x$-axis?(!) Some subinterval ($[-\sqrt{a},\sqrt{a}]$ for the $f$ and $[-a^{1/4}, a^{1/4}]$ for $g$, say)? Something else?

In any case, knowing the roots of $f$ and $g$ doesn't help you. It's like if I asked you, "I have two rectangles. One rectangle has width $n$, and the other has width $m$, but I refuse to tell you anything about the heights of the rectangles. Which rectangle has greater area?" You cannot say. Similarly you cannot say anything about the area under a curve if you just know the roots of the curve... you also need to know the heights of the curve at every $x$ to compute the area.

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  • $\begingroup$ "Area bounded by the $x$ axis and the curve" has a natural interpretation, which is that its the region between the $x$-intercepts. Then if you calculate the areas in terms of $a$, you get that the area of the first region is $\frac{4}{3}a^{3/2}$, and the area of the second region is $\frac{8}{5}a^{5/4}$ (assuming I computed them correcty). And so the question becomes "For what values of $a>0$ is $\frac{4}{3}a^{3/2} > \frac{8}{5}a^{5/4}$?" which can certainly be answered. I don't understand your argument in the second paragraph. $\endgroup$ – Arturo Magidin Oct 25 '17 at 6:11
  • $\begingroup$ @ArturoMagidin "has a natural interpretation" perhaps. I'm willing to concede that for functions with exactly two $x$-intercepts, that's the natural interpretation, and allude to that interpretation in my post (setting aside the question of whether area is being computed signed or unsigned). "I don't understand your argument in the second paragraph." The OP question is not how to solve the problem using integrals, but rather, whether one can compare the integrals without computing them, just from inspecting the location of the roots of the two functions. I answered "no." $\endgroup$ – user7530 Oct 25 '17 at 6:39
  • $\begingroup$ Oh, I see. Carry on. It's late and I should go to bed. Sorry. $\endgroup$ – Arturo Magidin Oct 25 '17 at 6:51

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