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- 5. Prove that the equation has no solution in an ordered integral domain.General construction of Riemann integral of function f in ℝ³Prove that Coth-1x = 1/2 In (x + 1 / x - 1), x > 1 Hence or otherwise show that (i) Coth-1( 1 + 2tan2u ) = -In sinu (ii) the integral of e2coth-1 dx = x + 2In(x - 1) + k
- F function [0,136] continous and every t is part of [0,136] for F(t) > 0 . What is the value of this integral ?f(x)=integral from 0 to x of sin(3t^2) dt find the maclaurin ploynomial of degree 7 for f(x) use this polynomial to estimate the value of integral from 0 to .64 of sin(3x^2)dxa. Explain why 0 ≤ x²arctan(x) ≤ (pi*x²)/4 for all 0 ≤ x ≤ 1. b. Use the properties of the integrals to show that the value of the integral lower bound is 0, higher bound is 1 and the integral is x² arctan(x) dx lies on the interval [0,pi/12]