Use the inner product〈f,g〉=∫10f(x)g(x)dxin the vector space C0[0,1] of continuous functions on the domain [0,1] to find 〈f,g〉, ∥f∥, ∥g∥, and the angle αf,g between f(x) and g(x) forf(x)=−10x2−6 and g(x)=−9x−4.〈f,g〉= ,∥f∥= ,∥g∥= ,αf,g .

[tex]<f,g> = \int\limits^1_0 {10(-10x^{2} -6)(-9x-4)} \, dx \\ \\ =\int\limits^1_0 {900x^{3}+400x^{2} +540x+240 } \, dx\\ \\ = (225x^{4} + \frac{400x^{3} }{3} + 270x^{2} +240x)\\ = \frac{2605}{3}[/tex]

b) How

∥f∥ = <f,f> then:

∥f∥ = [tex]<f,f> = \int\limits^1_0 {10(-10x^{2} -6)(-10x^{2} -6)} \, dx \\ \\ =\int\limits^1_0 {1000x^{4}+1200x^{2} + 360} \, dx\\ \\ = (200x^{5} + 400x^{3} + 360x)\\ = 960[/tex]

c)

∥g∥ = <g,g>

∥g∥ = [tex]<g,g> = \int\limits^1_0 {10(-9x-4)(-9x-4)} \, dx \\ \\ =\int\limits^1_0 {810x^{2}+720x + 160} \, dx\\ \\ = (270x^{3} + 360x^{2} + 160x)\\ = 790[/tex]

d) Angle between f and g

<f,g> = ∥f∥∥g∥cosα

Thus

[tex]\alpha = cos^{-1}(\frac{2605/3}{(790)(960)} )\\\\\alpha = 90[/tex]