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

Scalar Product
$$=\int_a^bf(x)g(x) \ dx\!$$

$$=\int_{-2}^{10}x\cos(x) \ dx\!$$

Using integration by parts;

$$=[x\sin(x)+\cos(x)]_{-2}^{10} $$

$$=-7.68\!$$

Magnitude
$$\| f \|=^{1/2}=\int_{a}^{b} f^2(x) \ dx\!$$

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

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

$$\| f \| =2.457\!$$

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

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

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

Angle Between Functions
$$cos(\theta)=\frac{}{\| f \| \| g \|}\!$$

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

$$\theta=89.47$$ The two functions are nearly orthogonal.