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Chapter 8: Q. 69 (page 702)

Use the results from Exercises 51鈥�60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61鈥�70 to within 0.001 of their values.
${鈭�}_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$

Short Answer

Expert verified

${鈭�}_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx 鈮� - 0.481$

Step by step solution

01

Step 1. Given information is:

${鈭�}_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx$

02

Step 2. Approximating the values

$From Q 59 . {鈭�}_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 3^{k}}{k!} (\frac{1 - {(0.5)}^{2 k + 3}}{2 k + 3}) This implies that to approximate the values of integrals, put the value of k {鈭�}_{0.5}^{1} x^{2} e^{- 3 x^{2}} dx = \frac{1}{3} (1 - 0 . 5^{4}) - \frac{3}{5} (1 - 0 . 5^{6}) + \frac{9}{2! 路 7} (1 - 0 . 5^{8}) = 0.3125 - 0.5906 + 0.6403 - 0.4995 鈮� - 0.481$

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