Question

Suppose \(\mathrm{f}\) and \(\mathrm{g}\) are continuous and \(f(a)=f(b)=0\) Prove \(\int_{a}^{b} f(x) g(x) d x=-\int_{a}^{b} f^{\prime}(x) G(x) d x\) where \(\mathrm{G}(\mathrm{x})=\int_{\mathrm{a}}^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}\).

Short answer

Question: Prove the following identity: $$\int_{a}^{b} f(x) g(x) dx = -\int_{a}^{b} f'(x) G(x) dx,$$ where \(f(x)\) and \(g(x)\) are continuous functions, \(f(a) = f(b) = 0\), and \(G(x) = \int_{a}^{x} g(t) dt\). Answer: The identity can be proven using integration by parts, by choosing \(u(x) = f(x)\) and \(v'(x) = g(x)\) and applying the formula: $$\int_{a}^{b} u(x) v'(x) dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} u'(x) v(x) dx$$ After substituting the chosen functions and simplifying the equation using \(f(a) = f(b) = 0\), we get the desired identity: $$\int_{a}^{b} f(x) g(x) dx = -\int_{a}^{b} f'(x) G(x) dx$$

Step-by-step solution

Choose u(x) and v'(x) according to integration by parts strategy

First, we need to choose the functions \(u(x)\) and \(v'(x)\) such that one of them is \(f(x)g(x)\). An obvious choice is to set \(u(x) = f(x)\) and \(v'(x) = g(x)\). Now we find \(u'(x) = f'(x)\) and \(v(x) = G(x)\) by integrating \(v'(x)\) from \(a\) to \(x\).

Apply the integration by parts formula

First, we need to apply the integration by parts formula on the left side of the given identity: $$\int_{a}^{b} f(x) g(x) dx = \int_{a}^{b} u(x) v'(x) dx$$ Now using the integration by parts formula, we have: $$\int_{a}^{b} u(x) v'(x) dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} u'(x) v(x) dx$$ Substitute \(u(x) = f(x)\), \(u'(x) = f'(x)\), \(v'(x) = g(x)\), and \(v(x) = G(x)\) to get: $$\int_{a}^{b} f(x) g(x) dx = [f(x)G(x)]_{a}^{b} - \int_{a}^{b} f'(x) G(x) dx$$

Simplify the equation and prove the identity

Since \(f(a) = f(b) = 0\), the first term on the right side of the equation becomes zero: \([f(x)G(x)]_{a}^{b} = f(b)G(b) - f(a)G(a) = 0\). Now our equation becomes: $$\int_{a}^{b} f(x) g(x) dx = -\int_{a}^{b} f'(x) G(x) dx$$ This proves the given identity as required.