# Nodes and weight of Gauss Quadrature

• MHB
• mathmari
In summary, Gauss quadrature can be calculated using a single node if f(x) is a function of only one variable, and it can be calculated using two nodes if f(x) is a function of two variables.

#### mathmari

Gold Member
MHB
Hey! :giggle:

Calculate the node $x_0$ and the weight $a_0$ of Gauss Quadrature so that $$\int_0^1w(x)f(x)\, dx\approx I_0(f)=a_0f(x_0)$$ where $w(x)=1+\sqrt{x}$.

I have done the following:

The Gauss quadrature formula with $(n + 1)=1$ node (i.e. $n=0$) integrates polynomials of degree $2n + 1=1$ exactly.
\begin{align*}\int_0^1w(x)\cdot 1\, dx=a_0 &\Rightarrow \int_0^1\left (1+\sqrt{x}\right )\cdot 1\, dx=a_0 \\ & \Rightarrow \int_0^1\left (1+\sqrt{x}\right )\, dx=a_0 \\ & \Rightarrow \int_0^1\left (1+x^{\frac{1}{2}}\right )\, dx=a_0 \\ & \Rightarrow \left [x+2x^{\frac{1}{2}+1}\right ]_0^1=a_0 \\ & \Rightarrow \left [x+2x^{\frac{3}{2}}\right ]_0^1=a_0 \\ & \Rightarrow 1+2 =a_0 \\ & \Rightarrow a_0 =3 \\ \int_0^1w(x)\cdot x\, dx=a_0\cdot x_0 &\Rightarrow \int_0^1\left (1+\sqrt{x}\right )\cdot x\, dx=3x_0 \\ & \Rightarrow \int_0^1\left (x+x\sqrt{x}\right )\, dx=3x_0 \\ & \Rightarrow \int_0^1\left (x+x^{\frac{3}{2}}\right )\, dx=3x_0 \\ & \Rightarrow \left [\frac{x^2}{2}+\frac{2}{3}x^{\frac{3}{2}+1}\right ]_0^1=3x_0 \\ & \Rightarrow \left [\frac{x^2}{2}+\frac{2}{3}x^{\frac{5}{2}}\right ]_0^1=3x_0 \\ & \Rightarrow \frac{1}{2}+\frac{2}{3} =3x_0 \\ & \Rightarrow \frac{7}{6} =3x_0 \\ & \Rightarrow x_0=\frac{7}{18} \end{align*}

Is that correct? Or if we say that it integrates polynomials of degree $1$ do we mean that $w(x)f(x)=1$ and $w(x)f(x)=x$ instead of $f(x)=1$ and $f(x)=x$ ?

:unsure:

mathmari said:
Or if we say that it integrates polynomials of degree $1$ do we mean that $w(x)f(x)=1$ and $w(x)f(x)=x$ instead of $f(x)=1$ and $f(x)=x$ ?

Hey mathmari!

I believe it is indeed intended that $f(x)=1$ and $f(x)=x$. (Nod)

mathmari said:
$$\int_0^1\left (1+x^{\frac{1}{2}}\right )\, dx=a_0 \Rightarrow \left [x+2x^{\frac{1}{2}+1}\right ]_0^1=a_0$$

If we take the derivative of $2x^{\frac{1}{2}+1}$, then we don't get $x^{\frac{1}{2}}$ do we?

mathmari said:
$$\int_0^1\left (x+x^{\frac{3}{2}}\right )\, dx=3x_0 \Rightarrow \left [\frac{x^2}{2}+\frac{2}{3}x^{\frac{3}{2}+1}\right ]_0^1$$

If we take the derivative of $\frac{2}{3}x^{\frac{3}{2}+1}$, then we don't get $x^{\frac{3}{2}}$ do we?

Klaas van Aarsen said:
If we take the derivative of $2x^{\frac{1}{2}+1}$, then we don't get $x^{\frac{1}{2}}$ do we?

If we take the derivative of $\frac{2}{3}x^{\frac{3}{2}+1}$, then we don't get $x^{\frac{3}{2}}$ do we?

Oh yes...

It should be:
\begin{align*}\int_0^1w(x)\cdot 1\, dx=a_0 &\Rightarrow \int_0^1\left (1+\sqrt{x}\right )\cdot 1\, dx=a_0 \\ & \Rightarrow \int_0^1\left (1+\sqrt{x}\right )\, dx=a_0 \\ & \Rightarrow \int_0^1\left (1+x^{\frac{1}{2}}\right )\, dx=a_0 \\ & \Rightarrow \left [x+\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}\right ]_0^1=a_0 \\ & \Rightarrow \left [x+\frac{2}{3}x^{\frac{3}{2}}\right ]_0^1=a_0 \\ & \Rightarrow 1+\frac{2}{3} =a_0 \\ & \Rightarrow a_0 =\frac{5}{3}\approx 1.6667 \\ \int_0^1w(x)\cdot x\, dx=a_0\cdot x_0 &\Rightarrow \int_0^1\left (1+\sqrt{x}\right )\cdot x\, dx=\frac{5}{3}x_0 \\ & \Rightarrow \int_0^1\left (x+x\sqrt{x}\right )\, dx=\frac{5}{3}x_0 \\ & \Rightarrow \int_0^1\left (x+x^{\frac{3}{2}}\right )\, dx=\frac{5}{3}x_0 \\ & \Rightarrow \left [\frac{x^2}{2}+\frac{1}{\frac{3}{2}+1}x^{\frac{3}{2}+1}\right ]_0^1=\frac{5}{3}x_0 \\ & \Rightarrow \left [\frac{x^2}{2}+\frac{2}{5}x^{\frac{5}{2}}\right ]_0^1=\frac{5}{3}x_0 \\ & \Rightarrow \frac{1}{2}+\frac{2}{5} =\frac{5}{3}x_0 \\ & \Rightarrow \frac{9}{10} =\frac{5}{3}x_0 \\ & \Rightarrow x_0=\frac{27}{50}=0,54 \end{align*}
:geek:

mathmari said:
It should be:
:geek:
It looks correct to me now. (Nod)

Klaas van Aarsen said:
It looks correct to me now. (Nod)

Great! Thank you!