Area enclosed between $x$-axis from $a$ to $b$ and the curve $f(x)$ is finite when

  1. $a=0, \quad b=\infty,\quad f(x)=e^{-5x^5}$
  2. $a=-\infty,\quad b=\infty,\quad f(x)=e^{-5x^5}$
  3. $a=-7,\quad b=\infty,\quad f(x)=1/x^4$
  4. $a=-7,\quad b=7,\quad f(x)=1/x^4$

How can this be solved?

  • $\begingroup$ yes how do you know $\endgroup$ – sam Jul 16 '13 at 10:18


$$\begin{align*}\bullet&\;\;\;\int\limits_0^\infty e^{-5x^5}dx\le\int_0^\infty e^{-x}\,dx=\lim_{b\to \infty}\left(-\frac1{e^b}+1\right)\ldots\\ \bullet&\;\;\int\limits_{-\infty}^\infty e^{-5x^5}dx=\int\limits_{-\infty}^0 e^{-5x^5}dx+\int\limits_0^\infty e^{-5x^5}dx\stackrel{\text{subst.:}\;x=-u}=\int\limits_0^\infty\left(e^{5x^5}+e^{-5x^5}\right)dx\;\ldots\\ \bullet&\;\;\int\limits_{-7}^\infty\frac{dx}{x^4}=\int\limits_{-7}^0\frac{dx}{x^4}+\ldots\;:\;\;\text{but}\;\;\int\limits_{-7}^0\frac{dx}{x^4}=-\frac13\lim_{\epsilon\to 0}\left(\frac1{\epsilon^3}+\frac1{7^3}\right)=\ldots\end{align*}$$

  • 1
    $\begingroup$ Classified solutions +1 $\endgroup$ – mrs Jul 16 '13 at 10:23

As $$\lim_{x\to +\infty}x^5\exp(-5x^5)=0<\infty$$ so according to Comparison test; 1. is convergent and you can evaluate that via definition. But about 4. since $$\lim_{x\to 0^+}(x-0)^4\times \frac{1}{x^4}=1\neq 0$$ and so Comparison test again agrees $$\int_0^1\frac{dx}{x^4}<\int_{-7}^{\infty}\frac{dx}{x^4}$$ is divergent and so we don't have a finite area in this case.

  • $\begingroup$ why not be option 2 $\endgroup$ – sam Jul 16 '13 at 9:55
  • 1
    $\begingroup$ @santoshkumar: I think for 2 if you consider $\int_{-\infty}^0$ and set $-x=t$ then you get your answer. Try it $\endgroup$ – mrs Jul 16 '13 at 9:57
  • $\begingroup$ $^++_+^{+_{+_+}}+ _+^+\quad$ $\endgroup$ – amWhy Jul 17 '13 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.