\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let \(u = \sqrt {x}\), we have \(x = u^2\), and hence \(\Diff x = 2u \Diff u\).
When \(x = 0, u = 0\), and when \(x = 1, u = 1\).
Therefore, \begin {align*} \int _{0}^{1} f\left (\sqrt {x}\right )\Diff x & = \int _{0}^{1} f(u) \cdot 2u \Diff u \\ & = 2 \int _{0}^{1} u f(u) \Diff u \\ & = 2 \int _{0}^{1} x f(x) \Diff x. \end {align*}
We have \begin {align*} \int _{0}^{1} \left (g(x)\right )^2 \Diff x & = \int _{0}^{1} g\left (\sqrt {x}\right ) \Diff x - \frac {1}{3} \\ & = 2 \int _{0}^{1} x g(x) \Diff x - \int _{0}^{1} x^2 \Diff x, \end {align*}
and hence \begin {align*} \int _{0}^{1} \left [\left (g(x)\right )^2 - 2x g(x) + x^2\right ] \Diff x & = 0 \\ \int _{0}^{1} \left (g(x) - x\right )^2 \Diff x & = 0. \end {align*}
We have \(\left (g(x) - x\right )^2 \geq 0\) for all \(0 \leq x \leq 1\), and since \(g\) is continuous, we must have \(g(x) - x = 0\) for all \(0 \leq x \leq 1\) for the integral to evaluate to \(0\).
Hence, \(g(x) = x\) for \(0 \leq x \leq 1\).
Using integration by parts, we have \begin {align*} \int _{0}^{1} h(x) \Diff x & = \left [x h(x)\right ]_{0}^{1} - \int _{0}^{1} x \Diff h(x) \\ & = 1 \cdot h(1) - 0 \cdot h(0) - \int _{0}^{1} x h'(x) \Diff x \\ & = h(1) - \int _{0}^{1} x h'(x) \Diff x. \end {align*}
Hence, \begin {align*} \int _{0}^{1} \left (h'(x)\right )^2 \Diff x & = 2h(1) - 2 \int _{0}^{1} h(x) \Diff x - \frac {1}{3} \\ & = 2 h(1) - 2 \left [h(1) - \int _{0}^{1} x h'(x) \Diff x\right ] - \frac {1}{3} \\ & = 2 \int _{0}^{1} x h'(x) \Diff x - \frac {1}{3} \\ & = 2 \int _{0}^{1} x h'(x) \Diff x - \int _{0}^{1} x^2 \Diff x. \end {align*}
Therefore, similar to the previous part, we have \(h'(x) = x\) for \(0 \leq x \leq 1\), and hence \(h(x) = \frac {x^2}{2} + h(0) = \frac {x^2}{2}\) for \(0 \leq x \leq 1\).
First, we notice that \begin {align*} \int _{0}^{1} e^{-ax} \Diff x & = - \frac {1}{a} \left [e^{-ax}\right ]_{0}^{1} \\ & = - \frac {1}{a} \cdot \left [e^{-a} - 1\right ] \\ & = - \frac {e^{-a}}{a} + \frac {1}{a}. \end {align*}
Hence, \begin {align*} \int _{0}^{1} e^{ax} \left (k(x)\right )^2 \Diff x & = 2 \int _{0}^{1} k(x) \Diff x - \int _{0}^{1} e^{-ax} \Diff x + \frac {1}{a} - \frac {1}{a^2} - \frac {1}{4} \\ \int _{0}^{1} \left [e^{ax} \left (k(x)\right )^2 - 2 k(x) + e^{-ax}\right ] \Diff x & = - \frac {1}{a^2} + \frac {1}{a} - \frac {1}{4} \\ \int _{0}^{1} e^{-ax} \left [e^{2ax} \left (k(x)\right )^2 - 2 e^{ax} k(x) + 1\right ] \Diff x & = - \left (\frac {1}{a} - \frac {1}{2}\right )^2 \\ \int _{0}^{1} e^{-ax} \left (e^{ax} k(x)- 1\right )^2 \Diff x & = - \left (\frac {1}{a} - \frac {1}{2}\right )^2. \end {align*}
Since \(e^{-ax} > 0\), and \(\left (e^{ax} k(x) - 1\right )^2 \geq 0\), the integrand must be non-negative, and hence \[ \int _{0}^{1} e^{-ax} \left (e^{ax} k(x)- 1\right )^2 \Diff x \geq 0, \] meaning \[ \left (\frac {1}{a} - \frac {1}{2}\right )^2 \leq 0. \]
However, since this is a square, it is non-negative, and it must be the case that \(\left (\frac {1}{a} - \frac {1}{2}\right )^2 = 0\), giving \(\frac {1}{a} = \frac {1}{2}\), and hence \(a = 2\).
Therefore, \[ \int _{0}^{1} e^{-ax} \left (e^{ax} k(x)- 1\right )^2 \Diff x = 0, \] and since the integrand is continuous and non-negative over the interval, it must be zero everywhere for \(0 \leq x \leq 1\), and hence \[ e^{ax} k(x) - 1 = 0, \] giving \[ k(x) = e^{-ax} = e^{-2x} \] for \(0 \leq x \leq 1\).