(a) Find the approximations T10 and M10 for 2 25e1/x dx. 1 (Round your answers to six decimal places.) T10 = M10 = (b) Estimate the errors in the approximations of part (a). (Round your answers to six decimal places.) |ET| ≤ 0.001667 Incorrect: Your answer is incorrect. |EM| ≤ (c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to within 0.0001? n = for Tn n = for Mn

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batolisis batolisis · Answer 1 · 2020-10-13T11:23:23+02:00

Answer:

A) T10 = 50.549391

M10 = 50.477547

B) ║ET10 ║= 0.169093

║EM10║= 0.084946

C)for ║ETN ║ N = 421

for ║EMN ║ N = 292

Step-by-step explanation:

A) F(x) = 25e[tex]^{\frac{1}{x} }[/tex]

where a = 1, b = 2 and bx = 1/10

attached below is the detailed solution

B) f'(x) = [tex]- \frac{25}{x^{2} } e^{\frac{1}{x} } ,[/tex] f"(x) = 25 [([tex](\frac{2x+1}{x^4} )e^{\frac{1}{x} }[/tex]

attached below is the detailed solution

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