Question

If \(f(0) = g(0) = 0\) and \(f''\) and \(g''\)are continuous , show that \(\int\limits_0^a {f\left( x \right)g''\left( x \right)dx = f\left( a \right)g'\left( a \right) - f'\left( a \right)g\left( a \right) + \int\limits_0^a {f''\left( x \right)g\left( x \right)dx} } \)

Short answer

Split the given integral into two forms then combine the result to get the required proof.

Step-by-step solution

Step 1: Substitution

It has been given that\(f\left( 0 \right) = g\left( 0 \right) = 0\), so let\(u = f\left( x \right)\)and\(dv = {g^\iota }^\iota \left( x \right)dx\)

• \(dv = {f^\iota }\left( x \right)dx,v = {g^\iota }\left( x \right)\)

Hence,

\(\int\limits_0^a {f\left( x \right){g^\iota }^\iota \left( x \right)dx = \left( {f\left( x \right){g^\iota }\left( x \right)} \right)_0^a - \int\limits_0^a {{f^\iota }\left( x \right){g^\iota }\left( x \right)dx} } \)

\( = f'(a)g'(a) - \int\limits_0^a {f'(x)g'(x)} dx\) 🡪 (1)

Substituting the value \(u = f''(x)\)

Now, let\(u = f'(x)\)

So,\(dv = g'(x)dx \Rightarrow du = f''(x)dx\)and\(v = g(x)\). So

\(\int\limits_0^a {f''(x)g'(x)dx = \left( {f'(x)g(x)} \right)_0^a} - \int\limits_0^a {f''(x)g(x)dx = f'(a)g(a) - \int\limits_0^a {f''(x)g(x)dx} } \) 🡪(2)

Combining both the equation

Now combining two results (1) and (2) we get\(\int_0^a {f(x)g''(x) = f(a)g'(a) - f'(a)g(a) + \int\limits_0^a {f''(x)g(x)dx} } \)

Hence, LHS=RHS (Proved)

Hence, it has been proved that when \(f(0) = g(0) = 0\) and \(f''\) and \(g''\)are continuous,

\(\int\limits_0^a {f(x)g'(x)dx = f(a)g'(a) - f'(a)g(a) + \int\limits_0^a {f''(x)g(x)dx} } \)

\(\)