University of Florida/Egm4313/s12.team11.gooding.k/R7

${\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)\ dx\!}$

${\displaystyle <f,g>=\int _{-2}^{10}x\cos(x)\ dx\!}$

Using integration by parts;

${\displaystyle <f,g>=[x\sin(x)+\cos(x)]_{-2}^{10}}$

                      ${\displaystyle <f,g>=-7.68\!}$

${\displaystyle \|f\|=<f,f>^{1/2}=\int _{a}^{b}f^{2}(x)\ dx\!}$

${\displaystyle =\int _{-2}^{10}[\cos(x)]^{2}\ dx\!}$

${\displaystyle =[.5(x+\sin(x)\cos(x)|_{-2}^{10}]^{1/2}}$

                       ${\displaystyle \|f\|=2.457\!}$

${\displaystyle \|g\|=\int _{a}^{b}g^{2}(x)\ dx\!}$

${\displaystyle =\int _{-2}^{10}x^{2}\ dx}$
${\displaystyle =[x^{3}/3]_{-2}^{10}}$

                       ${\displaystyle \|g\|={\frac {1008}{3}}\!}$

${\displaystyle cos(\theta )={\frac {<f,g>}{\|f\|\|g\|}}\!}$

${\displaystyle cos(\theta )={\frac {-7.68}{{\frac {1008}{3}}(2.457)}}}$

                      ${\displaystyle \theta =89.47}$

                      The two functions are nearly orthogonal.