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The way i started this is we know that $$\int_0^1 c(1-y)\,\mathrm{d}y=1$$

Do i just compute this to find c.


  • $\begingroup$ You still have f(y). You can omit it, because $f(y)=c(1-y)$ if $0<y<1$ $\endgroup$ Nov 23 '16 at 18:18
  • $\begingroup$ You're right i fixed that.Am i going about this question right, or am i way off ? $\endgroup$ Nov 23 '16 at 18:19
  • $\begingroup$ I changed my comment. $\endgroup$ Nov 23 '16 at 18:21
  • $\begingroup$ Alright removed f(y). $\endgroup$ Nov 23 '16 at 18:22
  • $\begingroup$ So is this the right way to solve this problem ? $\endgroup$ Nov 23 '16 at 18:24


$$\int_0^1 c(1-y)\,\mathrm{d}y=1$$

The constant can be factored out.

$$c\cdot \int_0^1 (1-y)\,\mathrm{d}y=1$$

And $$\int (1-y)\,\mathrm{d}y=y-\frac12 y^2$$

Now insert the limits.

  • $\begingroup$ Oh ok thanks so much. $\endgroup$ Nov 23 '16 at 18:35
  • $\begingroup$ You´re welcome. $\endgroup$ Nov 23 '16 at 18:36

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