2
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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?

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  • $\begingroup$ yes how do you know $\endgroup$ – sam Jul 16 '13 at 10:18
4
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Hints:

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

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

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

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