Answer: (a) For all positive integers \(n\), \(I_n=nI_{n-1}-1\).
(b) \(I_4=24\mathrm{e}-65\).
(c) \(\dfrac{65}{24}\lt \mathrm{e}\lt \dfrac{11}{4}\).
Given \(I_n=\int_0^1(1-x)^n\mathrm{e}^x\,dx\), integrate by parts with \(u=(1-x)^n\) and \(dv=\mathrm{e}^x\,dx\). Then \(du=-n(1-x)^{n-1}\,dx\) and \(v=\mathrm{e}^x\).
So
\(I_n=\left[(1-x)^n\mathrm{e}^x\right]_0^1-\int_0^1 \mathrm{e}^x\big(-n(1-x)^{n-1}\big)\,dx\)
\(=\left[(1-x)^n\mathrm{e}^x\right]_0^1+n\int_0^1(1-x)^{n-1}\mathrm{e}^x\,dx\).
Now \(\left[(1-x)^n\mathrm{e}^x\right]_0^1=0-1=-1\), so
\(I_n=-1+nI_{n-1}=nI_{n-1}-1\).
Next,
\(I_0=\int_0^1 \mathrm{e}^x\,dx=\left[\mathrm{e}^x\right]_0^1=\mathrm{e}-1\).
Using the recurrence:
\(I_1=I_0-1=\mathrm{e}-2\),
\(I_2=2I_1-1=2\mathrm{e}-5\),
\(I_3=3I_2-1=6\mathrm{e}-16\),
\(I_4=4I_3-1=24\mathrm{e}-65\).
The area enclosed by the \(x\)-axis, the \(y\)-axis and the curve \(y=(1-x)^4\mathrm{e}^x\) for \(0\le x\le1\) is
\(\int_0^1(1-x)^4\mathrm{e}^x\,dx=I_4=24\mathrm{e}-65\).
On \(0\le x\le1\), the curve lies above the \(x\)-axis, so the area is positive. Also, since the region lies inside the rectangle \(0\le x\le1\), \(0\le y\le1\), its area is less than \(1\). Hence
\(0\lt 24\mathrm{e}-65\lt 1\).
Therefore
\(65\lt 24\mathrm{e}\lt 66\),
so
\(\dfrac{65}{24}\lt \mathrm{e}\lt \dfrac{66}{24}=\dfrac{11}{4}\).
