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UPSC 2016 Maths Optional Paper 2 Q7c — Step-by-Step Solution

15 marks · Section B

Gaussian quadrature · Numerical Analysis · asked 3× in 13 yrs · Read the full method →

Question

For an integral $\displaystyle\int_{-1}^1 f(x)\,dx$, show that the two-point Gauss quadrature rule is given by $\displaystyle\int_{-1}^1 f(x)\,dx=f\!\left(\frac{1}{\sqrt3}\right)+f\!\left(-\frac{1}{\sqrt3}\right)$. Using this rule, estimate $\displaystyle\int_2^4 2xe^x\,dx$.

Technique

Derive Gauss–Legendre 2-point rule by enforcing exactness on $1,x,x^2,x^3$ (⇒ weights $1,1$, nodes $\pm1/\sqrt3$ = roots of $P_2$); rescale $[2,4]\to[-1,1]$ via $x=3+t$ (Jacobian $=1$); evaluate at the two nodes.

Solution

Setup. A two-point rule $\displaystyle\int_{-1}^1 f\,dx\approx w_1 f(x_1)+w_2 f(x_2)$ has four free parameters $w_1,w_2,x_1,x_2$; choosing them to integrate $1,x,x^2,x^3$ exactly gives a rule exact for all cubics (degree $2n-1=3$).

Step 1 — Derive the nodes and weights

Impose exactness on the monomials over $[-1,1]$:

By symmetry take $w_1=w_2=w$ and $x_2=-x_1$. Then equation 1 gives $w=1$; equations 2 and 4 are satisfied automatically; equation 3 gives $2x_1^2=\tfrac23$, so $x_1^2=\tfrac13$, i.e. $x_1=\dfrac{1}{\sqrt3},\ x_2=-\dfrac{1}{\sqrt3}$. Hence

Answer