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If $\int_0^1 f(x)dx ≈ \alpha_0f(0) + \alpha_1f(1) + \beta_0f'(0) + \beta_1f'(1)$, find $\alpha_0 , \alpha_1 , \beta_0$ and $\beta_1$.

I have attempted this and when I work it out I don't find a value for $\beta_0$? Can someone tell me from my workings where I'm going wrong?

$f(x) = x^0, f(x) = x^1, f(x) = x^2,f(x) = x^3 $

$f'(x) =0, f'(x) =x, f'(x) = 2x, f'(x) = 3x^2 $

$\int_0^1 1 dx = 1, \int_0^1 x dx = 1/2, \int_0^1 x^2 dx = 1/3, \int_0^1 x^3 dx = 1/4 $

Then substituting into the formula above we get: $\alpha_0=1$ , $\alpha_0= 1/2$ , $\beta_1= .8833$ and no answer for $\beta_0$ ? Where am I going wrong?

EditPiAf
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IanWalter
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1 Answers1

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\begin{align} &f_0(x) = x^0, f_1(x) = x^1, ~ f_2(x) = x^2, ~ f_3(x) = x^3 \\ \implies&f_0'(x) =0, ~ ~ \boxed{f_1'(x) =1}, ~ f_2'(x) = 2x, ~f_3'(x) = 3x^2 \end{align} You had a mistake in the circled/boxed derivative.

Kyle
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