Answer
See below
Work Step by Step
We are given:
$f_1(x)=1\\
f_2(x)=x^2\\
f_3(x)=x^4\\
a=-1\\
b=1$
According to Gram-Schmidt we have:
$v_1=f_1(x)=1\\
v_2=f_2(x)-\frac{(f_2(x),v_1)}{||v_1||^2}v_1\\
=x^2-\frac{(x^2,1)}{||1||^2}1 \\
=x^2-\frac{\int ^1_{-1}x^2dx}{\int^1_{-1}dx}.1\\
=x^2-\frac{ \frac{x^3}{3}|^1_{-1}}{x|^1_{-1}} \\
=x^2-\frac{\frac{1}{3}-(-\frac{1}{3})}{1}\\
=x^2-\frac{1}{3}\\
v_3=f_3(x)-\frac{(f_3(x),v_1)}{||v_1||^2}v_1-\frac{(f_3(x),v_2)}{||v_2||^2}v_2\\
=x^4-\frac{(x^4,1)}{||1||^2}1- \frac{(x^4,x^2-\frac{1}{3})}{||x^2-\frac{1}{3}||^2}(x^2-\frac{1}{3})\\
=x^4-\frac{\int ^1_{-1} x^4.1dx}{\int^1_{-1}dx}-\frac{\int ^1_{-1} x^4(x^2-\frac{1}{3})dx}{\int^1_{-1}(x^2-\frac{1}{3})^2dx}(x-\frac{1}{3})\\
=x^4-\frac{\frac{x^5}{5}|^1_{-1}}{x|^1_{-1}}-\frac{ \frac{x^7}{7}-\frac{x^5}{15}|^1_0}{\frac{x^5}{5}-\frac{2x^3}{9}+\frac{x}{9}|^1_0}(x^2-\frac{1}{3}) \\
=x^4-\frac{1}{5}-\frac{\frac{16}{105}}{\frac{3}{45}}(x^2-\frac{1}{3})\\
=x^4-\frac{1}{5}-\frac{6}{7}(x^2-\frac{1}{3}) \\
=x^4-\frac{6}{7}x^2+\frac{3}{35}$
Hence, an orthogonal basic spanned by the given factors is:
$\{1,x^2-\frac{1}{3},x^4-\frac{6}{7}x^2+\frac{3}{35}\}$
