Question

Consider the integral \(\int x^{n} e^{x} d x .\) Use integration by parts to evaluate the integral if (a) \(n=1\) (b) \(n=2\) (c) \(n=3\) (d) Conjecture the value of the integral for any positive integer \(n\) (e) Writing to Learn Give a convincing argument that your conjecture in part (d) is true.

Short answer

The results for the integral \(\int{x^n e^{x}}dx\) when \(n=1, 2, 3\) are \(xe^{x} - e^{x} + C\), \(x^{2} e^{x} - 2 (x e^{x} - e^{x}) + C\) and \(x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (x e^{x} - e^{x})) + C\). The conjecture for any positive integer \(n\) is that \(\int{x^n e^{x}}dx = n \int{x^{n-1} e^{x}}dx\), leading to a final step of \(e^{x} + C\) once power of \(x\) reaches 0. The conjecture is proven to be true by recursive applications of integration by parts.

Step-by-step solution

Base Case (n=1)

To start with, let's evaluate the integral when n=1. We are going to use the formula for integration by parts: \(\int u dv = uv - \int v du\). Let \(u = x\) and \(dv = e^x dx\), as we can differentiare \(x\) and it nullifies, while integrating \(e^x dx\) results in \(e^x\). Substituting these values, we get: \( \int x e^{x} d x = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} + C\).

Case when (n=2)

When \(n = 2\), let's use the formula for integration by parts and let \(u = x^2\) and \(dv = e^x dx\). Now we can apply integration by parts to evaluate the integral. Following same steps, we get: \(\int x^{2} e^{x} dx = x^{2} e^{x} - 2\int x e^{x} dx\). The integral \(\int x e^{x} dx\) was calculated in the preceding step which we can reuse, the result comes out to be \(x^{2} e^{x} - 2 (x e^{x} - e^{x}) + C\).

Case when (n=3)

In this case, \(u = x^3\) and \(dv = e^x dx\). We then find out \(\int x^{3} e^{x} dx = x^{3} e^{x} - 3\int x^{2} e^{x} dx = x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (x e^{x} - e^{x})) + C\). From these steps, we can start to observe a pattern emerge for the ith integral.

Conjecture for any positive integer (n)

From the pattern observed in the above steps: The conjectured general formula for the nth integral is: \(\int x^{n} e^{x} dx = n \int x^{n-1} e^{x}dx\), eventually leading to a final step of \(\int e^{x} dx = e^{x} + C\).

Proof of the conjecture

Our conjecture is basically the result of recursively applying the integration by parts, where at each step we reduce the power of x by 1, so eventually after \(n\) iteration the power of \(x\) will become 0 and the integral \( \int e^{x} dx \) can be directly written as \(e^{x} + C\), so the conjecture stands true.