0
$\begingroup$

enter image description here

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.

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Oh ok thanks so much. $\endgroup$ Nov 23 '16 at 18:35
  • $\begingroup$ You´re welcome. $\endgroup$ Nov 23 '16 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.