Hello, can anyone help me?

$$\int\limits_{2}^{8}\sqrt({2x})dx=\sqrt{2}\int\limits_{2}^{8}(x ^{\frac{ 1 }{ 2 }})dx=\sqrt{2}(\frac{ x ^{1,5} }{ 1,5})$$ so, my questions are "Have I done everything right" And "How to find the answer?". p.s. there, of course, should be | with 8 on top of it and 2 on the bottom, but I didnt know how to "draw it"

Question

Hello, can anyone help me? 

$$\int\limits_{2}^{8}\sqrt({2x})dx=\sqrt{2}\int\limits_{2}^{8}(x ^{\frac{ 1 }{ 2 }})dx=\sqrt{2}(\frac{ x ^{1,5} }{ 1,5})$$ so, my questions are "Have I done everything right" And "How to find the answer?". p.s. there, of course, should be | with 8 on top of it and 2 on the bottom, but I didnt know how to "draw it"

turingtest · Answer

yes, it's right so far

kamille · Answer

so, I know  that I need to "put" 8 and 2 everywhere where is x, but! how to simplify (or I don't need to) $$\frac{ 8^{1,5} }{ 1,5 }$$

turingtest · Answer

it helps to remember that$$\large x^{a/b}=\sqrt[b]{x^a}$$and convert 1.5=3/2

kamille · Answer

$$\sqrt{2}(\frac{ \sqrt{8^{3}} }{ 1,5 }-\frac{ \sqrt{8} }{ 1,5 })=\sqrt{2}(\frac{ 8\sqrt{8} }{ 1,5 }-\frac{ \sqrt{8} }{ 1,5 })=\sqrt{2} * \frac{ 7\sqrt{8} }{ 1,5 }= 18\frac{ 2 }{ 3 }$$ Can you check now? @TuringTest

turingtest · Answer

yes, that is correct for the evaluation at 8, now you need to subtract the evaluation at 2

kamille · Answer

sorry? Haven't i done it? $$2^{\frac{ 3 }{ 2 }}=\sqrt{8}$$

turingtest · Answer

yeah, my bad, I got cut off
you have the right answer

kamille · Answer

Thank you a lot for your help!