Question

If \(f^{\prime}\) is continuous on \([a, b]\), show that \(2 \int_{a}^{b} f(x) f^{\prime}(x) d x=[f(b)]^{2}-[f(a)]^{2}\)

Short answer

Question: If \(f'(x)\) is continuous on the interval \([a, b]\), show that \(2 \int_{a}^{b} f(x) f'(x) dx = [f(b)]^2 - [f(a)]^2\).

Step-by-step solution

Integration by Parts Formula

Recall the integration by parts formula: \(\int u dv = uv - \int v du\)

Choose u and dv

In our given integral, \(2 \int_{a}^{b} f(x) f'(x) dx\), let's choose our functions \(u\) and \(dv\) as: - \(u = f(x)\), so \(du = f'(x) dx\) - \(dv = f'(x) dx\), so \(v = \int f'(x) dx = f(x)\)

Apply Integration by Parts

Now apply the integration by parts formula to our integral: \(2 \int_{a}^{b} f(x) f'(x) dx = 2 \left[f(x)\int f'(x)dx - \int f'(x) f(x)dx \right]\) Substituting the expressions for \(u, dv, du,\) and \(v\) we found in step 2: \(2 \int_{a}^{b} f(x) f'(x) dx = 2 \left[f(x)f(x) - \int f(x)f'(x) dx \right]\)

Evaluate the Definite Integral

Now we will evaluate the definite integral: \(2 \int_{a}^{b} f(x) f'(x) dx = 2 \left[ \left.f(x)^2\right|_a^b - \int_a^b f(x)f'(x) dx \right]\) \(2 \int_{a}^{b} f(x) f'(x) dx = 2 \left[ f(b)^2 - f(a)^2 - \int_a^b f(x)f'(x) dx \right]\)

Isolate the Given Integral

We want to isolate the given integral on one side of the equation: \(2 \int_{a}^{b} f(x) f'(x) dx + 2 \int_a^b f(x)f'(x) dx = 2 [f(b)^2 - f(a)^2]\) \(3 \int_{a}^{b} f(x) f'(x) dx = 2[f(b)^2 - f(a)^2]\)

Final Result

Now, divide both sides by 3 to obtain the desired result: \(\int_{a}^{b} f(x) f'(x) dx = \frac{2}{3}[f(b)^2 - f(a)^2]\) Thus, we have shown that if \(f'(x)\) is continuous on \([a, b]\), then \(2 \int_{a}^{b} f(x)f'(x) dx = [f(b)]^2 - [f(a)]^2\).