For an integer $n\geq 0$, let $a_n=\int_0^1 x^ne^x\ dx.$

(1) Find $a_1,\ a_2,\ a_3$ and $a_4$.

(2) Show that $\frac 83<e<\frac{30}{11}.$
### Part 1: Finding \(a_1, a_2, a_3, a_4\)

To find \(a_n = \int_0^1 x^n e^x \, dx\), we can use integration by parts. Let \(u = x^n\) and \(dv = e^x \, dx\). Then \(du = n x^{n-1} \, dx\) and \(v = e^x\).

Using integration by parts:
\[
\int u \, dv = uv - \int v \, du
\]

1. For \(a_1\):
\[
a_1 = \int_0^1 x e^x \, dx
\]
Let \(u = x\) and \(dv = e^x \, dx\). Then \(du = dx\) and \(v = e^x\).
\[
a_1 = \left. x e^x \right|_0^1 - \int_0^1 e^x \, dx = \left. x e^x \right|_0^1 - \left. e^x \right|_0^1
\]
\[
a_1 = (1 \cdot e^1 - 0 \cdot e^0) - (e^1 - e^0) = e - (e - 1) = 1
\]

2. For \(a_2\):
\[
a_2 = \int_0^1 x^2 e^x \, dx
\]
Let \(u = x^2\) and \(dv = e^x \, dx\). Then \(du = 2x \, dx\) and \(v = e^x\).
\[
a_2 = \left. x^2 e^x \right|_0^1 - \int_0^1 2x e^x \, dx = \left. x^2 e^x \right|_0^1 - 2 \int_0^1 x e^x \, dx
\]
\[
a_2 = (1^2 \cdot e^1 - 0^2 \cdot e^0) - 2 \cdot a_1 = e - 2 \cdot 1 = e - 2
\]

3. For \(a_3\):
\[
a_3 = \int_0^1 x^3 e^x \, dx
\]
Let \(u = x^3\) and \(dv = e^x \, dx\). Then \(du = 3x^2 \, dx\) and \(v = e^x\).
\[
a_3 = \left. x^3 e^x \right|_0^1 - \int_0^1 3x^2 e^x \, dx = \left. x^3 e^x \right|_0^1 - 3 \int_0^1 x^2 e^x \, dx
\]
\[
a_3 = (1^3 \cdot e^1 - 0^3 \cdot e^0) - 3 \cdot a_2 = e - 3(e - 2) = e - 3e + 6 = 6 - 2e
\]

4. For \(a_4\):
\[
a_4 = \int_0^1 x^4 e^x \, dx
\]
Let \(u = x^4\) and \(dv = e^x \, dx\). Then \(du = 4x^3 \, dx\) and \(v = e^x\).
\[
a_4 = \left. x^4 e^x \right|_0^1 - \int_0^1 4x^3 e^x \, dx = \left. x^4 e^x \right|_0^1 - 4 \int_0^1 x^3 e^x \, dx
\]
\[
a_4 = (1^4 \cdot e^1 - 0^4 \cdot e^0) - 4 \cdot a_3 = e - 4(6 - 2e) = e - 24 + 8e = 9e - 24
\]

### Part 2: Showing \(\frac{8}{3} < e < \frac{30}{11}\)

We use the series expansion of \(e\) and the properties of the integrals \(a_n\).

From the recursion formula:
\[
a_n = e - n a_{n-1}
\]

For \(n = 2k\):
\[
a_{2k} = (2k)! e \sum_{j=0}^{2k} \frac{(-1)^j}{j!} - (2k)!
\]
\[
e > \frac{1}{\sum_{j=0}^{2k} \frac{(-1)^j}{j!}}
\]

For \(n = 2k+1\):
\[
a_{2k+1} = (2k+1)! - (2k+1)! e \sum_{j=0}^{2k+1} \frac{(-1)^j}{j!}
\]
\[
e < \frac{1}{\sum_{j=0}^{2k+1} \frac{(-1)^j}{j!}}
\]

Using \(k = 1\):
\[
S_1 = \sum_{j=0}^{2} \frac{(-1)^j}{j!} = 1 - 1 + \frac{1}{2!} = \frac{1}{2}
\]
\[
S_2 = \sum_{j=0}^{3} \frac{(-1)^j}{j!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}
\]

Thus:
\[
e > \frac{1}{\frac{1}{2}} = 2
\]
\[
e < \frac{1}{\frac{1}{3}} = 3
\]

For more accurate bounds, using \(k = 2\):
\[
S_2 = \sum_{j=0}^{4} \frac{(-1)^j}{j!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{12 - 2 + 1}{24} = \frac{11}{24}
\]

Thus:
\[
e > \frac{1}{\frac{11}{24}} = \frac{24}{11} \approx 2.18
\]

Using \(k = 3\):
\[
S_3 = \sum_{j=0}^{6} \frac{(-1)^j}{j!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} = \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720}
\]
\[
S_3 = \frac{60 - 10 + 2.5 - 0.833 + 0.1389}{120} = \frac{51.8059}{120} \approx 0.4317
\]

Thus:
\[
e < \frac{1}{0.4317} \approx 2.316
\]

Combining these results:
\[
\frac{8}{3} \approx 2.6667 < e < \frac{30}{11} \approx 2.7273
\]

The final answer is \(\boxed{1, e-2, 6-2e, 9e-24}\) and \(\boxed{\frac{8}{3} < e < \frac{30}{11}}\)