App. of trapezoidal rule

24) http://i729.photobucket.com/albums/ww291/kaka1307/0161-1.jpg ans: a) 20.64 bi)41.29 bii)no 27) http://i729.photobucket.com/albums/ww291/kaka1307/0162.jpg ans: a)103.2373 b)950 please give explanination on 27b.thx please show clear steps.thanks.

24a) h = (2 - 0)/4 = 0.5 So the trapezoidal rule gives a value: (0.5/2)[1 + e4 + 2(e0.25 + e + e2.25)] = 20.64 b i) Sub y = -x, then dx = -dy When x = 0, y = 0 and when x = -2, y = 2 Hence ∫(x = -2 → 0) ex^2 dx = -∫(y = 2 → 0) ey^2 dy = ∫(x = 0 → 2) ex^2 dx Hence I2 = 2I1 = 41.29 b ii) Let f(x) = ex^2, then f"(x) = 2ex^2(1+ 2x2) and max. |f"(x)| = 18e4 for both the intervals [0, 2] and [-2, 2] So when I2 is estimated with 8 sub-intervals, h is still 0.5 but the range is 4 which is double that of I1, so the error is even larger than estimating I1 with 4 sub-intervals. 27a) h = 4 and hence the estimation is: (4/2) [0 + 8e1.6 + 2(2e0.4 + 4e0.8 + 6e1.2)] = 103.2373 b) x = 100 + 4∫(t = 0 → 8) te0.2t dt + 200∫(t = 0 → 8) 1/(t + 1) dt = 100 + 4 x 103.2373 + 200 [ln (t + 1)] (t = 0 → 8) = 100 + 4 x 103.2373 + 200 ln 9 = 950 2011-02-01 17:17:09 補充： For 27a, h should be 2 and hence 4/2 becomes 2/2 = 1 