Answer
See below
Work Step by Step
We are given:
$f_1(x)=1\\
f_2(x)=x\\
f_3(x)=x^4\\
a=0\\
b=1$
According to Gram-Schmidt we have:
$v_1=f_1(x)=1\\
v_2=f_2(x)-\frac{(f_2(x),v_1)}{||v_1||^2}v_1\\
=x-\frac{(x,1)}{||1||^2}1 \\
=x-\frac{\int ^1_0 x.1dx}{\int^1_0dx}.1\\
=x-\frac{ \frac{x^2}{2}|^1_0}{x|^1_0} \\
=x-\frac{\frac{1}{2}}{1}\\
=x-\frac{1}{2}\\
v_3=f_3(x)-\frac{(f_3(x),v_1)}{||v_1||^2}v_1-\frac{(f_3(x),v_2)}{||v_2||^2}v_2\\
=x^2-\frac{(x^2,1)}{||1||^2}1- \frac{(x^2,x-\frac{1}{2})}{||x-\frac{1}{2}||^2}(x-\frac{1}{2})\\
=x^2-\frac{\int ^1_0 x^2.1dx}{\int^1_0dx}-\frac{\int ^1_0 x^2(x-\frac{1}{2})dx}{\int^1_0(x-\frac{1}{2})^2dx}(x-\frac{1}{2})\\
=x^2-\frac{1}{3}-\frac{ \frac{x^4}{4}-\frac{x^3}{6}|^1_0}{\frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{4}|^1_0}(x-\frac{1}{2}) \\
=x^2-\frac{1}{3}-\frac{\frac{1}{12}}{\frac{1}{12}}(x-\frac{1}{2})\\
=x^2-\frac{1}{3}-x+\frac{1}{2} \\
=x^2-x+\frac{1}{6}$
Hence, an orthogonal basic spanned by the given factors is:
$\{1,x-\frac{1}{2},x^2-x+\frac{1}{6}\}$
