Suppose we want to find a formula for the weighted integral integral^1_0 Squarer
ID: 3008898 • Letter: S
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