Answer
See below
Work Step by Step
Given $f_1(x)=1\\
f_2(x)=x\\
f_3(x)=\frac{1}{2}(3x^2-1)$
Obtain:
$=\int ^1_{-1}f_1(x)f_2(x)f_3(x)dx\\
=\int^1_{-1} x dx\\
=\frac{x^2}{2}|^1_{-1}\\
=\frac{1 }{2}-\frac{(-1)^2}{2}\\
=0$
$=\int ^1_{-1}f_1(x)f_3(x)dx\\
=\int^1_{-1} \frac{1}{2}(3x^2-1)dx\\
=(\frac{1}{2}(x^3-x))|^1_{-1}\\
=[\frac{1}{2}(1^3-1)]-[\frac{1}{2}(-1)^3-(-1)]\\
=0$
$=\int ^1_{-1}f_2(x)f_3(x)dx\\
=\int^1_{-1} \frac{1}{2}(3x^3-x) dx\\
=[\frac{1}{2}(\frac{3x^4}{4}-\frac{x^2}{2})]|^1_{-1}\\
=[\frac{1}{2}(\frac{3.1^4}{4}-\frac{1^2}{2})]-[\frac{1}{2}(\frac{3.(-1)^4}{4}-\frac{(-1)^2}{2})]\\
=0$
We have:
$||f_1||=\sqrt \\
=\sqrt \int^1_{-1}f_1(x)f_1(x)dx\\
=\sqrt \int^1_{-1}1dx\\
=\sqrt x|^1_{-1}\\
=\sqrt 1-(-1)\\
=\sqrt 2$
$||f_2||=\sqrt \\
=\sqrt \int^1_{-1}f_2(x)f_2(x)dx\\
=\sqrt \int^1_{-1} x^2 dx\\
=\sqrt \int^1_{-1}\frac{x^3}{3}dx|^1_{-1}\\
=\sqrt (\frac{1^3}{3}-\frac{(-1)^3}{3})\\
=\sqrt \frac{1}{3}-(-\frac{1}{3})\\
=\sqrt \frac{2}{3}\\
=\frac {\sqrt 6}{3}$
$||f_3||=\sqrt \\
=\sqrt \int^1_{-1}f_3(x)f_3(x)dx\\
=\sqrt \int^1_{-1} \frac{1}{4}(3x^2-1)^2)dx\\
=\sqrt \frac{1}{4}(9x^4-6x^2+1)dx\\
=\sqrt [\frac{1}{4}(\frac{9x^5}{5}-2x^3+x)]|^1_{-1}\\
=\sqrt [\frac{1}{4}(\frac{9.1^5}{5}-2.1^3+1)]-[\frac{1}{4}(\frac{9.(-1)^5}{5}-2.(-1)^3+(-1)]\\
=\sqrt [\frac{1}{4}(\frac{9}{5}-2+1)-\frac{1}{4}(\frac{-9}{5}+2-1)]\\
=\sqrt \frac{2}{5}\\
=\frac{\sqrt 10}{5}$
