UPSC 2024 Maths Optional Paper 2 Q7b-i — Step-by-Step Solution

7.5 marks · Section B

Simpson's 1/3 and 3/8 rules · Numerical Analysis · asked 5× in 13 yrs · Read the full method →

Question

Integrate $f(x)=5x^3-3x^2+2x+1$ from $x=-2$ to $x=4$ using Simpson’s $\tfrac{3}{8}$ rule with $h=1$ .

Seven points give 6 intervals = 2 groups of 3; apply the $3h/8\,[1,3,3,1]$ weights to each group.

Function values at $x=-2,-1,0,1,2,3,4$ :

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$f$	$-55$	$-9$	$1$	$5$	$33$	$115$	$281$

Group 1 ( $x=-2$ to $x=1$ ):

\frac{3}{8}[f(-2)+3f(-1)+3f(0)+f(1)]=\frac{3}{8}[-55-27+3+5]=\frac{3}{8}(-74)=-27.75.

Group 2 ( $x=1$ to $x=4$ ):

\frac{3}{8}[f(1)+3f(2)+3f(3)+f(4)]=\frac{3}{8}[5+99+345+281]=\frac{3}{8}(730)=273.75.

Total: $-27.75+273.75=246$ .

Note: Simpson’s $3/8$ rule is exact for cubics, so this equals the exact value.

\boxed{\;\int_{-2}^4 f(x)\,dx\approx 246\;\;(\text{Simpson's }\tfrac{3}{8}\text{ rule, exact for this cubic}).\;}