Suppose we want to find a formula for the weighted integral integral^1_0 Squarer

Question

Suppose we want to find a formula for the weighted integral integral^1_0 Squareroot xf(x)dx of the type integral^1_0 Squareroot^1_0 Squareroot xf (x)dx ~ a_1 f(0) + a_2 integral^1_0 f(x)dx, which can be useful if we already know the integral on the right and f(0). There are two approaches: interpolation - use a polynomial that satisfies certain data to substitute f in the integral and integrate - and the method of the undetermined coefficients. Use a linear interpolating polynomial p(x) of f such that P(0) = f(0) and integral^1_0 P(x)dx = integral^1_0 f(x)dx to determine a_1 and a_2. The method of the undetermined coefficients requires the formula to be exact for 1,x,x^2,... x^k-1, in which k is the number of coefficients we need to determine. Use this method to solve for a_1 and a_2.

Explanation / Answer

problem:

fx32dx{displaystyleint}fx^rac{3}{2},mathrm{d}x

Apply linearity:

=fx32dx=class{steps-node}{cssId{steps-node-1}{f}}{displaystyleint}x^rac{3}{2},mathrm{d}x

Now solving:

x32dx{displaystyleint}x^rac{3}{2},mathrm{d}x

Apply power rule:

xndx=xn+1n+1{displaystyleint}x^{mathtt{n}},mathrm{d}x=dfrac{x^{mathtt{n}+1}}{mathtt{n}+1} with n=32mathtt{n}=dfrac{3}{2}:

=2x525=dfrac{2x^rac{5}{2}}{5}

Plug in solved integrals:

fx32dxclass{steps-node}{cssId{steps-node-2}{f}}{displaystyleint}x^rac{3}{2},mathrm{d}x

=2fx525=dfrac{2fx^rac{5}{2}}{5}

The problem is solved:

fx32dx{displaystyleint}fx^rac{3}{2},mathrm{d}x

=2fx525+C=dfrac{2fx^rac{5}{2}}{5}+C

Antiderivative computed by Maxima:

g(x)dx=G(x)=oldsymbol{int{g(x)},mathrm{d}x = G(x) =}

2fx525+C

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