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What is the integral of 2 1-x dx with limits 0 to t?

Anonymous

∙ 12y ago

Updated: 12/13/2022

I think you meant the integral of 21-x dx, or as I write it ,/`21-x dx.

Use u-substitution where u = 1 - x. Note that the derivative of this equation is du = -1 dx, so we know dx = - du.

Rewriting the original integral in terms of u, integrating, then putting it back in terms of x gives us:

,/` 21-x dx

,/` 2u (-du)

- ,/` 2u du

- 2u / ln(2)

- 21-x / ln(2)

Note that the second to last line can be found in any integral table, and ln denotes natural log.

Now the bounds go from x = 0 to x = t, so:

[- 21-t / ln(2)] - [- 21-0 / ln(2)]

- 21-t / ln(2) + 2 / ln(2)

(-21-t + 2)/ln(2)

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∙ 12y ago

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Q: What is the integral of 2 1-x dx with limits 0 to t?

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Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.

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