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Kamille
Group Title
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"
 one year ago
 one year ago
Kamille Group Title
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"
 one year ago
 one year ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, it's right so far
 one year ago

Kamille Group TitleBest ResponseYou've already chosen the best response.0
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) \[\frac{ 8^{1,5} }{ 1,5 }\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it helps to remember that\[\large x^{a/b}=\sqrt[b]{x^a}\]and convert 1.5=3/2
 one year ago

Kamille Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{2}(\frac{ \sqrt{8^{3}} }{ 1,5 }\frac{ \sqrt{8} }{ 1,5 })=\sqrt{2}(\frac{ 8\sqrt{8} }{ 1,5 }\frac{ \sqrt{8} }{ 1,5 })=\sqrt{2} * \frac{ 7\sqrt{8} }{ 1,5 }= 18\frac{ 2 }{ 3 }\] Can you check now? @TuringTest
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, that is correct for the evaluation at 8, now you need to subtract the evaluation at 2
 one year ago

Kamille Group TitleBest ResponseYou've already chosen the best response.0
sorry? Haven't i done it? \[2^{\frac{ 3 }{ 2 }}=\sqrt{8}\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah, my bad, I got cut off you have the right answer
 one year ago

Kamille Group TitleBest ResponseYou've already chosen the best response.0
Thank you a lot for your help!
 one year ago
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