Answer
See below
Work Step by Step
Given $f_1(x)=2x\\
f_2(x)=1+2x^2\\
f_3(x)=x^3-\frac{3}{5}x$
Obtain:
$=\int ^1_{-1}f_1(x)f_2(x)f_3(x)dx\\
=\int^1_{-1} 2x(1+2x^2)dx\\
=\int^1_{-1} 2x+4x^3dx\\
=x^2+x^4|^1_{-1}\\
=(1+1)-[(-1)^2+(-1)^4]\\
=0$
$=\int ^1_{-1}f_1(x)f_3(x)dx\\
=\int^1_{-1} 2x(x^3-\frac{3}{5}x)dx\\
=\int^1_{-1} (2x^4-\frac{6}{5}x^2)dx\\
=(2.\frac{x^5}{5}-\frac{6}{5}.\frac{x^3}{3})|^1_{-1}\\
=(2.\frac{1^5}{5}-\frac{6}{5}.\frac{1^3}{3})-[2.\frac{(-1)^5}{5}-\frac{6}{5}.\frac{(-1)^3}{3}]|^1_{-1}]\\
=0$
$=\int ^1_{-1}f_2(x)f_3(x)dx\\
=\int^1_{-1} (1+2x^2)(x^3-\frac{3}{5}x)dx\\
=\int^1_{-1}(x^3-\frac{3}{5}x+2x^5-\frac{6}{5}x^3)dx\\
=(-\frac{3}{5}.\frac{x^2}{2}-\frac{1}{5}.\frac{x^4}{4}+2.\frac{x^6}{6})|^1_{-1}\\
=[-\frac{3}{5}.\frac{1^2}{2}-\frac{1}{5}.\frac{1^4}{4}+2.\frac{1^6}{6}]-[-\frac{3}{5}.\frac{(-1)^2}{2}-\frac{1}{5}.\frac{(-1)^4}{4}+2.\frac{(-1)^6}{6}]\\
=0$
We have:
$||f_1||=\sqrt \\
=\sqrt \int^1_{-1}f_1(x)f_1(x)dx\\
=\sqrt \int^1_{-1}(2x)^2dx\\
=\sqrt 4.\frac{x^3}{3}|^1_{-1}\\
=\sqrt 4.\frac{1^3}{3}-4.\frac{(-1)^3}{3}\\
=\sqrt \frac{8}{3}$
$||f_2||=\sqrt \\
=\sqrt \int^1_{-1}f_2(x)f_2(x)dx\\
=\sqrt \int^1_{-1} (1+2x^2)^2 dx\\
=\sqrt \int^1_{-1} (1+4x^2+4x^4) dx\\
=\sqrt (x+\frac{4x^3}{3}+\frac{4x^5}{5})|^1_{-1}\\
=\sqrt (1+\frac{4.1^3}{3}+\frac{4.1^5}{5})-[(-1)+\frac{4.(-1)^3}{3}+\frac{4.(-1)^5}{5}]\\
=\sqrt \frac{94}{15}$
$||f_3||=\sqrt \\
=\sqrt \int^1_{-1}f_3(x)f_3(x)dx\\
=\sqrt \int^1_{-1} (x^3-\frac{3}{5})^2)dx\\
=\sqrt (x^6-\frac{6}{5}x^4+\frac{9}{25}x^2)dx\\
=\sqrt (\frac{x^7}{7}-\frac{6}{25}x^5+\frac{9}{75}x^3)|^1_{-1}\\
=\sqrt (\frac{1^7}{7}-\frac{6}{25}.1^5+\frac{9}{75}.1^3)-[\frac{(-1)^7}{7}-\frac{6}{25}(-1)^5+\frac{9}{75}(-1)^3]\\
=\sqrt \frac{4}{175}-(-\frac{4}{175})\\
=\sqrt \frac{8}{175}$
