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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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What is the integretion of modxdx?
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What is the definite integral from 0 to pi divided by 6 of sin cubed 2x cos 2x dx?
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What is a 2 step equations that equal 1?
# 1x+-1=0 # +1 | +1 # 1x = 1 # 1/1=1 # 1x+-1=0 # +1 | +1 # 1x = 1 # 1/1=1
Why derivative of mod x does not exist graphically?
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.
