Question

Derive a reduction formula and compute the integral \(\int_{-1}^{0} x^{n} e^{x} d x,(n\) is a positive integer \()\).

Short answer

Question: Determine the definite integral \(\int_{-1}^{0} x^{n} e^{x} dx\) for positive integer values of \(n\). Answer: \(\int_{-1}^{0} x^{n} e^{x} dx = 1 - \sum_{k=0}^{n}(-1)^{k} \binom{n}{k} e^{-1}\)

Step-by-step solution

Perform integration by parts

Let's choose \(u=x^n\) and \(dv=e^x dx\). Thus, \(du=nx^{n-1}dx\) and \(v=e^x\). Then the integration by parts formula, \(\int u dv = u v - \int v du\), gives us: \(\int x^{n} e^{x} dx = x^n e^x - n\int x^{n-1} e^x dx\).

Look for a pattern for the reduction formula

Observe that: \(I_n = \int x^{n} e^{x} dx = x^n e^x - nI_{n-1}\), where \(I_n = \int x^{n} e^{x} dx\). The pattern for the reduction formula can be expressed as: \(I_n = x^n e^x - nx^{n-1} e^x + n(n-1) I_{n-2}\). By continuing this process, we can see a pattern that leads to the reduction formula: \(I_n = e^x\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}x^{n-k}\).

Apply the reduction formula to the given definite integral

Now we apply this reduction formula to compute the desired integral: \(\int_{-1}^{0} x^{n} e^{x} dx = \left[e^x\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}x^{n-k}\right]_{-1}^{0}\) \(= \left[e^0\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}0^{n-k} - e^{-1}\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}(-1)^{n-k}\right]\) \(= 1 - \sum_{k=0}^{n}(-1)^{k} \binom{n}{k} e^{-1}\). This is the final expression for the integral \(\int_{-1}^{0} x^{n} e^{x} dx\).