Find the value of the integral $\int_{0}^{1} x^{7} e^{-x^{4}} d x$

Step 1 ：First, we recognize that this is an improper integral of the form \(\int_{a}^{b} f(x) dx\), where \(f(x) = x^{7} e^{-x^{4}}\).

Step 2 ：We can solve this integral using the method of substitution. Let's set \(u = x^{4}\), then \(du = 4x^{3} dx\).

Step 3 ：We need to adjust the \(x^{7}\) in the integral to match the \(x^{3}\) in \(du\). We can rewrite \(x^{7}\) as \(x^{4}x^{3}\), which is \(ux^{3}\).

Step 4 ：Substitute \(u\) and \(du\) into the integral, we get \(\frac{1}{4}\int_{0}^{1} u e^{-u} du\).

Step 5 ：This is now a standard integral of the form \(\int u e^{-u} du\), which can be solved using integration by parts. The formula for integration by parts is \(\int u dv = uv - \int v du\).

Step 6 ：Let's set \(v = e^{-u}\) and \(dv = -e^{-u} du\). Then the integral becomes \(-\frac{1}{4}\int u dv\).

Step 7 ：Applying the formula for integration by parts, we get \(-\frac{1}{4} [uv - \int v du]\), which simplifies to \(-\frac{1}{4} [ue^{-u} - \int e^{-u} du]\).

Step 8 ：The integral \(\int e^{-u} du\) is simply \(-e^{-u}\). Substituting this back in, we get \(-\frac{1}{4} [ue^{-u} + e^{-u}]\).

Step 9 ：Evaluate this from 0 to 1, we get \(-\frac{1}{4} [(1e^{-1} + e^{-1}) - (0e^{0} + e^{0})]\), which simplifies to \(-\frac{1}{4} [2e^{-1} - 1]\).

Step 10 ：Simplify this to get the final answer, \(\boxed{-\frac{1}{2e} + \frac{1}{4}}\).