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765

Part of 2011 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 765

Source: 1994 Yokohama City University entrance exam/Medicine

11/16/2011
Define two functions g(x), f(x) (x≥0)g(x),\ f(x)\ (x\geq 0) by g(x)=∫0xe−t2dt, f(x)=∫01e−(1+s2)x1+s2ds.g(x)=\int_0^x e^{-t^2}dt,\ f(x)=\int_0^1 \frac{e^{-(1+s^2)x}}{1+s^2}ds.
Now we know that f′(x)=−∫01e−(1+s2)xds.f'(x)=-\int_0^1 e^{-(1+s^2)x}ds.
(1) Find f(0).f(0).
(2) Show that f(x)≤π4e−x (x≥0).f(x)\leq \frac{\pi}{4}e^{-x}\ (x\geq 0).
(3) Let h(x)={g(x)}2h(x)=\{g(\sqrt{x})\}^2. Show that f′(x)=−h′(x).f'(x)=-h'(x).
(4) Find lim⁡x→+∞g(x)\lim_{x\rightarrow +\infty} g(x)
Please solve the problem without using Double Integral or Jacobian for those Japanese High School Students who don't study them.
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