Answer: Using integration by parts,
\(I_n=\int_0^1(1-x)^n e^x\,dx=\left[(1-x)^n e^x\right]_0^1+\int_0^1 n(1-x)^{n-1}e^x\,dx.\)
The boundary term is \(0-1=-1\), so
\(I_n=nI_{n-1}-1.\)
Also,
\(I_0=\int_0^1 e^x\,dx=\left[e^x\right]_0^1=e-1.\)
Hence
\(I_1=e-2,\quad I_2=2e-5,\quad I_3=6e-16,\quad I_4=24e-65.\)
Therefore
\(I_4=24e-65.\)
The area enclosed by the axes and the curve \(y=(1-x)^4e^x\) for \(0\le x\le 1\) is
\(\int_0^1(1-x)^4e^x\,dx=I_4=24e-65.\)
Since this area is positive and the whole region lies inside the unit square,
\(0\lt 24e-65\lt 1.\)
So
\(65\lt 24e\lt 66,\)
and therefore
\(\frac{65}{24}\lt e\lt \frac{66}{24}=\frac{11}{4}.\)
Start with
\(I_n=\int_0^1(1-x)^n e^x\,dx.\)
Integrate by parts with \(u=(1-x)^n\) and \(dv=e^x\,dx\). Then
\(du=-n(1-x)^{n-1}\,dx,\qquad v=e^x.\)
So
\(I_n=\left[(1-x)^n e^x\right]_0^1-\int_0^1 e^x\big(-n(1-x)^{n-1}\big)\,dx\)
which simplifies to
\(I_n=\left[(1-x)^n e^x\right]_0^1+\int_0^1 n(1-x)^{n-1}e^x\,dx.\)
Now
\(\left[(1-x)^n e^x\right]_0^1=(1-1)^n e^1-(1-0)^n e^0=0-1=-1,\)
and
\(\int_0^1 n(1-x)^{n-1}e^x\,dx=nI_{n-1}.\)
Therefore, for all positive integers \(n\),
\(I_n=nI_{n-1}-1.\)
Next find \(I_4\). First,
\(I_0=\int_0^1 e^x\,dx=\left[e^x\right]_0^1=e-1.\)
Apply the recurrence repeatedly:
\(I_1=1\cdot I_0-1=(e-1)-1=e-2,\)
\(I_2=2I_1-1=2(e-2)-1=2e-5,\)
\(I_3=3I_2-1=3(2e-5)-1=6e-16,\)
\(I_4=4I_3-1=4(6e-16)-1=24e-65.\)
So the exact value is
\(I_4=24e-65.\)
Now consider the region bounded by the \(x\)-axis, the \(y\)-axis, and the curve \(y=(1-x)^4e^x\) for \(0\le x\le 1\). Its area is
\(A=\int_0^1(1-x)^4e^x\,dx=I_4=24e-65.\)
On \(0\le x\le 1\), the curve is above the \(x\)-axis, so \(A\gt 0\).
Also, for \(0\le x\le 1\), we have \((1-x)^4\le 1\) and \(e^x\le e\lt 3\), but more importantly the curve starts at \((0,1)\) and decreases to \((1,0)\), so the enclosed region lies inside the unit square \(0\le x\le 1\), \(0\le y\le 1\). Hence
\(A\lt 1.\)
Therefore
\(0\lt 24e-65\lt 1.\)
Add 65 and divide by 24:
\(65\lt 24e\lt 66\)
so
\(\frac{65}{24}\lt e\lt \frac{66}{24}=\frac{11}{4}.\)
