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

7.5 marks · Section B

Trapezoidal rule (composite; error) · Numerical Analysis · asked 3× in 13 yrs · Read the full method →

Question

Integrate $f(x)=5x^3-3x^2+2x+1$ from $x=-2$ to $x=4$ using the Trapezoidal rule with $h=1$ .

Composite Trapezoidal rule: endpoints weight 1, interior points weight 2, multiply by $h/2$ .

Function values (from part (i)): $f(-2)=-55$ , $f(-1)=-9$ , $f(0)=1$ , $f(1)=5$ , $f(2)=33$ , $f(3)=115$ , $f(4)=281$ .

Composite Trapezoidal rule with $h=1$ , $n=6$ :

\frac{h}{2}\bigl[f(x_0)+2\sum_{i=1}^{5}f(x_i)+f(x_6)\bigr]=\frac{1}{2}\bigl[-55+2(-9+1+5+33+115)+281\bigr].

Interior sum: $-9+1+5+33+115=145$ .

=\frac{1}{2}[-55+290+281]=\frac{516}{2}=258.

Comparison: Exact value $=246$ (computed in part (i)). The Trapezoidal rule overestimates by $12$ , as expected for a cubic with $O(h^2)$ error.

\boxed{\;\int_{-2}^4 f(x)\,dx\approx 258\;\;(\text{Trapezoidal rule; exact value is }246).\;}