An interesting integral

Question

$f(x)$ with continuous second derivative on $[0,1]$ satisfies $f(0)=0$, $f'(0)=0, f(1)=1, f'(1)=3.$ $$\int_{0}^{1}f(x)f'(x) (1+2(f'(x))^2+2f(x)f''(x))dx ? $$

I found the result of this integral. But I'm looking for different solutions.

Answer 1

\begin{eqnarray*} &&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) (1+2\left( f^{\prime }\left( x\right) \right) ^{2}+2f\left( x\right) f^{\prime \prime }\left( x)\right) dx \\ &=&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) dx+\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) (2\left( f^{\prime }\left( x\right) \right) ^{2}+2f\left( x\right) f^{\prime \prime }\left( x)\right) dx \\ &=&I_{1}+I_{2} \end{eqnarray*} Let's calculate the first integral. \begin{eqnarray*} I_{1} &=&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) dx \\ u &=&f\left( x\right) \ \text{find }du=f^{\prime }\left( x\right) dx \end{eqnarray*} Rewrite the given integral using the change of variables. \begin{eqnarray*} I_{1} &=&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) dx \\ &=&\int_{f\left( 0\right) }^{f\left( 1\right) }udu \\ &=&\left. \frac{u^{2}}{2}\right\vert _{f\left( 0\right) }^{f\left( 1\right) } \\ &=&\frac{f^{2}\left( 1\right) -f^{2}\left( 0\right) }{2} \\ &=&\frac{1}{2}. \end{eqnarray*} Similarly, the second integral \begin{eqnarray*} I_{2} &=&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) \left( 2\left( f^{\prime }\left( x\right) \right) ^{2}+2f\left( x\right) f^{\prime \prime }x)\right) dx \\ u &=&f\left( x\right) f^{\prime }\left( x\right) \ \text{find }du=\left( \left( f^{\prime }\left( x\right) \right) ^{2}+f\left( x\right) f^{\prime \prime }x)\right) dx \end{eqnarray*} Rewrite the given integral using the change of variables. \begin{eqnarray*} I_{2} &=&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) \left( 2\left( f^{\prime }\left( x\right) \right) ^{2}+2f\left( x\right) f^{\prime \prime }x)\right) dx \\ &=&\int\limits_{f\left( 0\right) f^{\prime }\left( 0\right) }^{f\left( 1\right) f^{\prime }\left( 1\right) }2udu \\ &=&\left. u^{2}\right\vert _{f\left( 0\right) f^{\prime }\left( 0\right) }^{f\left( 1\right) f^{\prime }\left( 1\right) } \\ &=&\left( f\left( 1\right) f^{\prime }\left( 1\right) \right) ^{2}-\left( f\left( 0\right) f^{\prime }\left( 0\right) \right) ^{2} \\ &=&9. \end{eqnarray*}

Thus, from $I_{1}$ and $I_{2}$ \begin{eqnarray*} &&\int_{0}^{1}f\left( x\right) f^{\prime }\left( x\right) (1+2\left( f^{\prime }\left( x\right) \right) ^{2}+2f\left( x\right) f^{\prime \prime }\left( x)\right) dx \\ &=&\frac{19}{2}. \end{eqnarray*}

Answer 2

By the chain rule we have the following identities:

\begin{align}\frac{d}{dx}f(x)f'(x) &= f'(x)^2+f(x)f''(x) \tag{1}\\ \frac{d}{dx}(f(x))^2 &= 2f(x)f'(x) \tag{2}\\ \frac{d}{dx}\left(\frac{d}{dx}(f(x)^2)\right)^2 &= 2\frac{d}{dx}(f(x))^2\cdot\frac{d^2}{dx^2}(f(x))^2 \tag{3}\end{align}

Hence we can rewrite the integral

\begin{align}&\int_0^1 f(x)f'(x)(1+2(f'(x))^2+2f(x)f''(x))dx \\ &\stackrel{(1)}{=} \int_0^1 f(x)f'(x)(1+2\frac{d}{dx}f(x)f'(x))dx \\ &\stackrel{(2)}{=} \frac{1}{2}\int_0^1 \frac{d}{dx}(f(x))^2(1+\frac{d^2}{dx^2}(f(x))^2)dx \\ &\stackrel{(3)}{=} \frac{1}{2}\int_0^1 \frac{d}{dx}(f(x))^2dx + \frac{1}{4}\int_0^1 \frac{d}{dx}\left(\frac{d}{dx}(f(x)^2)\right)^2dx \\ &= \frac{1}{2}(f(1)^2-f(0)^2 + \frac{1}{4}(2f(1)f'(1)-2f(0)f'(0))^2 = \frac{19}{2}\end{align}