integral of 1 x By the first fundamental theorem of calculus since 1 x is continuous on 0 any function F x xx01 xdx with 0
We begin by noting some obvious facts Fact 1 F is continuous and strictly increasing Proof very straightforward Fact 2 ab a 1 tdt F b for all a b 0 Proof It can be proved by analysing Riemann sums that whenever a 0 and g is continuous on c b we have ab acg x a dx ab cg x dx If so why and why isn t this true for 1 x I m having difficulty understanding difference between the above functions in terms of Lebesgue Integration I d prefer if the above question could be answered by approximating both functions with simple functions and without using the equality of the Riemann Integral and the Lebesgue Integral
integral of 1 x
integral of 1 x
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Method 1 Integral Of 1 e x 1 YouTube
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Indefinite Integral Of 1 x 2 YouTube
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A note on the solution There s actually a much simpler solution I found after realizing the trig sub is redundant Here it is Start by integrating by parts with dv 1 I xe1 x xe1 x x2 dx xe1 x e1 x x dx We can substitute u 1 x du 1 x2 dx noting that x 1 u I xe1 x ueu u2 du xe1 x eu u du Usually when one asks for the integral of 1 x 1 x they ll be answered with ln x l n x but it s undefined for negative values Sometimes they ll be answered with ln x l n x which is defined in the entire real numbers line except 0 0 but that doesn t matter However this only works for real
Let s simplify the problem first by integrating from 1 to 1 There is a discontinuity at 0 so you write the integral as 1 11 xdx lim 0 11 xdx 1 1 xdx If you perform this calculation you obtain zero But that is not the only possible way to write the integral You can equally well write it as 1 The indefinite integral is a rational fraction and is typically solved using partial fractions decomposition You first factor the denominator x4 1 which has four complex roots the fourth roots of minus one let 0 1 2 and 3 Then you decompose 1 x4 1 a x 0 b x 1 c x 2 d x 3
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In your particular example your pattern of signs square roots and constants is best matched by arctangent We need to do a little algebra and a little substitution to get an exact match 1 x2 3 1 1 3x2 1 3 1 3 1 x 3 2 1 Now move the constant multiple 1 3 out front of your integral and set u x 3 so you get 1 1 x 1 x converges to 0 0 as x x The OP probably won t see this comment anyway as they have not logged in recently The posted answers are correct and another way to illustrate that the above reasoning posted in the question is not is to consider 1 x x 1 x x instead Again we are adding smaller and smaller numbers
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integral of 1 x - Let s simplify the problem first by integrating from 1 to 1 There is a discontinuity at 0 so you write the integral as 1 11 xdx lim 0 11 xdx 1 1 xdx If you perform this calculation you obtain zero But that is not the only possible way to write the integral You can equally well write it as 1