evaluate integral http://www.webassign.net/cgi-bin/symimage.cgi?expr=int_1%5E64%20%281%20%2B%202%20root3%28x%29%29%2Fsqrt%28x%29%20dx
\[\sqrt{x}=x ^{1/2}\]Move denominator to numerator; multiply through.
ok i am stuck on this: \[(1/2)/(1/2)+2^{5/6}/(5/6)\left(\begin{matrix}64 \\ 1\end{matrix}\right)\]
\[= \int\limits_{1}^{64} ( x^{-\frac{1}{2} } + 2 ( x^{\frac{1}{3} - \frac{1}{2} } ) )\]
\[= \int\limits_{1}^{64} ( x^{-\frac{1}{2}} + 2 x^{ -\frac{1}{6} } )dx \]
so its not ^5/6?
When you swing it to the top the 1/2 becomes negative.
right
so now what i do?
wait i add one to each of the numorator?
You start over. Integrate what elecengineer did. The negative is not showing on exponents.
yes but then i have to add one which becomes: \[x^{1/2} /(1/2)+ 2^{5/6}/(5/6)\]
You are writing rubbish. What is 2^(5/60?)
so well can you show me please..so i can pick up where im going wrong
\[\int\limits_{1}^{64}x ^{-1/2}+2x ^{-1/6}\]The negative not showing in exponent.
ok so whats the next step
Exponent work\[-(1/2)+(2/2)=(1/2)\]\[-(1/6)+(6/6)=(5/6)\]
Got it?
yes thats what i did
next step please
Well, you are confusing yourself. Above, you left of the x. Then instead of dividing, just flip it and multiply; makes everything cleaner.\[2x ^{1/2}+[2(6)/5]x ^{5/6}\]
ok so you riciporcal the 5/6 to 6/5 and mulitiple by 2
Yes, the 2 is already there as a coefficient.
ok now i understand
so now its 2x^(1/2)+(12/5)x^(5/6) right
did you get this? it is a definite integral
\[\int _1^{64} \frac{1}{\sqrt{x}}dx =2\sqrt{x}|_1^{64}=2(\sqrt{64}-\sqrt{1})=2\times 7=14\]
im doin it differentlt like: {2(8) + (12/5) (2^5)}-(22/5)
thats not right
\[\int_1^{64} \frac{2}{\sqrt[6]{x}}dx = \frac{12}{5}\sqrt[6]{x^5}|_1^{64}\]
\[=\frac{12}{5}(32-1)=74.4\]
so "final answer" should be \[14+74.4=88.4\] unless i made an arithmetic mistake
now can you explain why you are integrating and also converting decimals to fractions? what kind of math is this???
this is calculus... and my textbooks shows me to do this
oh ok. usually converting decimals to fractions comes first.
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