So in this question we say if N is a known positive injured For what value of K. is the integral from one decay of X to the power of n minus one dx Equal to one over N. Well let's see if I have the integral from one. Check X X to the power of N -1 DX. What do I do to evaluate that definite interval? The first thing I do is I add one to my exported X to the N power and then I divide by that new exposure. So X to the end over it. And I'm going to evaluate this On the interval from 1 to K. And so that's giving me okay to the end power over in minus 1 to the end power over in equals one over it. Now, certainly one to the power of N is just one. So I'm getting K to the power of N over and -1 over N equals one over N. Now I'll add one over end to each side so that I'm getting K to the N over N equals two over N. Now I multiply both sides by N. That gives us thank you K. To the power of N equals two. And now my goal is to solve for K. Right? So I'm going to have to raise both sides to some power to get rid of that ends power I can raise both sides to the power of one over N. I'm going to raise both sides To the power of one over N. The reason is when you have a power to a power, what do you do to those exponents? You multiply those expose right End Times one over N is just one. And so here on the left, I'm getting just K to the first power and this is equal to chew to the power of one of ren answer choice. D. My final answer. Another way that you might do it, you might take the route of each side. But then when you take the root of something, that's the same as raising it to the power of one over and as well. And so K two the K equals two to the power of one over N. We got our anti derivative plugged in our limits of integration and solve for K. That was our strategy. Hopefully that was helpful. Have a good rest of your day.