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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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