# Statistics problem 8

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.

Thanks.

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

Hint:

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

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.

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