(a) Use the trapezium rule with five ordinates to evaluate (int_{0} ^{1} frac{3}{1 + x^{2}} mathrm {d} x), correct to four significant figures.
(b) If (A = begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix}), find the image of the point (1, 2) under the linear transformation (A^{2} + A + 2I), where I is the (2 times 2) unit matrix.
Explanation
(a) (int _{0} ^{1} frac{3}{1 + x^{2}} mathrm {d} x)
| (x) | 0 | 0.25 | 0.50 | 0.75 | 1.0 |
| (x^{2}) | 0 | 0.06 | 0.25 | 0.56 | 1.0 |
| (1 + x^{2}) | 1 | 1.06 | 1.25 | 1.56 | 2.0 |
| (frac{3}{1 + x^{2}}) | 3 | 2.82 | 2.40 | 1.92 | 1.5 |
h = 0.25 ;
(y_{1} = 3, y_{2} = 2.82 , y_{3} = 2.40 , y_{4} = 1.92 , y_{5} = 1.5)
(y_{1} + y_{5} = 3 + 1.5 = 4.5)
(y_{2} + y_{3} + y_{4} = 2.82 + 2.40 + 1.92 = 7.14)
(2(y_{2} + y_{3} + y_{4}) = 14.28)
= (3[frac{1}{2} h ( y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})])
= (3[frac{1}{2} (0.25) (4.5 + 14.28)])
= (3(0.125)(18.78) = 2.3475)
(approxeq 2.348) (4 sig. figures).
(b) (A = begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix})
(A^{2} + A + 2I = begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix} begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix} + begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix} + 2 begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix})
= (begin{pmatrix} 4 + 3 & 2 + 2 \ 6 + 6 & 3 + 4 end{pmatrix} + begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix} + begin{pmatrix} 2 & 0 \ 0 & 2 end{pmatrix})
= (begin{pmatrix} 7 & 4 \ 12 & 7 end{pmatrix} + begin{pmatrix} 2 & 1 \ 3 & 2 end{pmatrix} + begin{pmatrix} 2 & 0 \ 0 & 2 end{pmatrix})
= (begin{pmatrix} 11 & 5 \ 15 & 11end{pmatrix})
(T : begin{pmatrix} 1 \ 2 end{pmatrix} to begin{pmatrix} 11 & 5 \ 15 & 11 end{pmatrix} begin{pmatrix} 1 \ 2 end{pmatrix})
=(begin{pmatrix} 11 + 10 \ 15 + 22 end{pmatrix} = begin{pmatrix} 21 \ 37 end{pmatrix})
The image of (1, 2) = (21, 37).