Question

Prove that \(\int f^{\prime}(x) F(x) d x=f^{\prime}(x) F(x)-f(x) F^{\prime}(x)+\int f(x) F^{\prime \prime}(x) d x\) and generally \(\int \mathrm{f}^{(\mathrm{n})}(\mathrm{x}) \mathrm{F}(\mathrm{x}) \mathrm{dx}=\mathrm{f}^{\mathrm{f}-1}(\mathrm{x}) \mathrm{F}(\mathrm{x})\) \(-f^{(n-2)}(x) F^{\prime}(x)+\ldots+(-1)^{n} \int f(x) F^{(n)}(x) d x\).

Short answer

Question: Prove that for any positive integer n, the integral of the nth derivative of f(x) times F(x) is given by the following formula: \(\int f^{(n)}(x) F(x) dx = f^{(n-1)}(x) F(x) - f^{(n-2)}(x) F^{\prime}(x) + \ldots +(-1)^{n} \int f(x) F^{(n)}(x) dx\)

Step-by-step solution

Show the formula for the first-order derivative

First, we need to show that \(\int f^{\prime}(x) F(x) dx = f^{\prime}(x) F(x) - f(x) F^{\prime}(x) + \int f(x) F^{\prime \prime}(x) dx\). To do this, use integration by parts. Let \(u = F(x)\) and \(dv = f^{\prime}(x)dx\). Then differentiation and integration will give: \(du = F^{\prime}(x) dx, v = f(x)\). Now applying the formula for integration by parts \(\int udv = uv - \int vdu\), we have: \(\int F(x) f^{\prime}(x)dx = F(x)f(x) - \int f(x)F^{\prime}(x) dx\) Now we need to add and subtract the term \(f(x)F^{\prime}(x)\) on both sides of the equation to get the desired formula: \(f^{\prime}(x)F(x) - f(x)F^{\prime}(x) + \int f(x)F^{\prime}(x) dx = f^{\prime}(x)F(x) - f(x)F^{\prime}(x) + \int f(x)F^{\prime \prime}(x) dx\)

Show the formula for general n

We will now prove the formula for general n by induction: For \(n=1\), we have already shown the formula in step 1. Assume the formula holds for \(n=m\), i.e., \(\int f^{(m)}(x) F(x) dx = f^{(m-1)}(x) F(x) - f^{(m-2)}(x) F^{\prime}(x) + \ldots +(-1)^{m} \int f(x) F^{(m)}(x) dx\) Now, differentiating both sides of the equation with respect to x, we get: \(f^{(m+1)}(x) F(x) + f^{(m)}(x) F^{\prime}(x) = f^{(m)}(x) F(x) - f^{(m-1)}(x) F^{\prime}(x) + \ldots +(-1)^{m}f(x) F^{(m)}(x) + (-1)^{m}f(x) F^{(m+1)}(x)\) Now, using integration by parts on the left side with \(u = F(x)\) and \(dv = f^{(m+1)}(x) dx\), we have: \(\int f^{(m+1)}(x) F(x) dx = f^{(m)}(x) F(x) - \int f^{(m)}(x) F^{\prime}(x) dx\) Now substitute the right side of the induction assumption equation into the equation above and rearrange: \(\int f^{(m+1)}(x) F(x) dx = f^{(m)}(x) F(x) - f^{(m-1)}(x) F^{\prime}(x) + \ldots +(-1)^{m+1} \int f(x) F^{(m+1)}(x) dx\) The result is the formula for \(n=m+1\). Hence, the formula is proven for general n by induction.