Calculate the integral: integral from e to (e^e) x³

Calculate the integral: $\int_e^{e^e} x^3 e^{x-4} \, dx$.

We will use substitution to transform this integral into a more standard form.

Step 1: Simplify and use substitution

First, rewrite the integrand:

\[x^3 e^{x-4} = x^3 e^x e^{-4} = e^{-4} x^3 e^x\]

Since $e^{-4}$ is a constant, we have:

\[\int_e^{e^e} x^3 e^{x-4} \, dx = e^{-4} \int_e^{e^e} x^3 e^x \, dx\]

Step 2: Repeated integration by parts for $\int x^3 e^x\,dx$

Recall: $\int u \, dv = uv - \int v \, du$

First application: Let $u_1 = x^3$, $dv_1 = e^x \, dx$, so $du_1 = 3x^2 \, dx$, $v_1 = e^x$:

\[\int x^3 e^x \, dx = x^3 e^x - 3\int x^2 e^x \, dx\]

Second application: For $\int x^2 e^x \, dx$, let $u_2 = x^2$, $dv_2 = e^x \, dx$, so $du_2 = 2x \, dx$, $v_2 = e^x$:

\[\int x^2 e^x \, dx = x^2 e^x - 2\int x e^x \, dx\]

Third application: For $\int x e^x \, dx$, let $u_3 = x$, $dv_3 = e^x \, dx$, so $du_3 = dx$, $v_3 = e^x$:

\[\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x\]

Step 3: Combine the results

Working backwards:

\[\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) = e^x(x^2 - 2x + 2)\]

\[\int x^3 e^x \, dx = x^3 e^x - 3[e^x(x^2 - 2x + 2)] = e^x(x^3 - 3x^2 + 6x - 6)\]

Step 4: Apply the Newton-Leibniz formula

\[\int_e^{e^e} x^3 e^{x-4} \, dx = e^{-4} \left[e^x(x^3 - 3x^2 + 6x - 6)\right]_e^{e^e}\]

Step 5: Evaluate at the upper limit $x = e^e$

\[F(e^e) = e^{-4} \cdot e^{e^e} \cdot [(e^e)^3 - 3(e^e)^2 + 6e^e - 6] = e^{e^e - 4} \cdot [e^{3e} - 3e^{2e} + 6e^e - 6]\]

Step 6: Evaluate at the lower limit $x = e$

\[F(e) = e^{-4} \cdot e^e \cdot [e^3 - 3e^2 + 6e - 6] = e^{e-4} \cdot [e^3 - 3e^2 + 6e - 6]\]

Step 7: Calculate the definite integral

\[\int_e^{e^e} x^3 e^{x-4} \, dx = F(e^e) - F(e) = e^{e^e - 4}[e^{3e} - 3e^{2e} + 6e^e - 6] - e^{e-4}[e^3 - 3e^2 + 6e - 6]\]

This can be factored slightly:

\[= e^{-4}\left\{e^{e^e}[e^{3e} - 3e^{2e} + 6e^e - 6] - e^e[e^3 - 3e^2 + 6e - 6]\right\}\]

Note on the result:

This integral does not simplify to a particularly elegant closed form. The exact answer involves exponential terms with very large exponents (since $e^e \approx 15.15$).

Pattern for $\int x^n e^x \, dx$:

In general, for $\int x^n e^x \, dx$, repeated integration by parts gives:

\[\int x^n e^x \, dx = e^x \sum_{k=0}^{n} (-1)^{n-k} \frac{n!}{k!} x^k + C\]

For $n = 3$:

\[\int x^3 e^x \, dx = e^x(x^3 - 3x^2 + 6x - 6) + C\]

which matches our result.

Final answer:

\[\boxed{\int_e^{e^e} x^3 e^{x-4} \, dx = e^{-4}\left[e^x(x^3 - 3x^2 + 6x - 6)\right]_e^{e^e}}\]

Or more explicitly:

\[\boxed{\int_e^{e^e} x^3 e^{x-4} \, dx = e^{e^e-4}(e^{3e} - 3e^{2e} + 6e^e - 6) - e^{e-4}(e^3 - 3e^2 + 6e - 6)}\]