evaluate the integral
try wolfram
12/5 x^(5/2) [25,81]
Yeah. I think tonks method is accepetable.
Im not sure what you did :/
You can just input x inside the sqrt sign. making it integral of 6x^3/2
much lower than what you had, elaornelas
x = √x^2 x*√x = x^3/2
\[6\int\limits_{81}^{25} x ^{\frac{ 3 }{ 2 }}dx\]
hm I'm sorry I'm still kind of confused :/
i understand why you moved the 6, but not sure how to go about getting the solution from here
\[6\frac{ x ^{\frac{ 5 }{ 2 }} }{ \frac{ 5 }{ 2 } }] (81.25)\]
the integrand, 6x√x can be rewritten as 6x^(3/2) or 6x^1.5 then the power rule for anti-derivatives can be applied.
Just apply algebra. We know that xsqrt(x) = x^1 (x)^1/2 --> adding their exponents will yield to x^(1+(1/2)) or x^(3/2) or simply sqrt(x^3)
@elaornelas do you follow?
so so!
So what's next?
ok well let me show you what i have written down
\[-6\int\limits_{25}^{81}x(x)^{\frac{ 1 }{ 2 }}\]
Treat the x like it has an exponent of one and add the exponents (same base) to make it one term.
Why -6? Shouldn't be + And just simplify the expression \[x(x)^{\frac{ 1 }{ 2 }}\] this is actually same to \[x ^{\frac{ 3 }{ 2 }} = \sqrt{x ^{3}}\]
negative is fine because she switched the bounds
yes exactly @Tonks
oops, sorry I didn't see that. my bad :D
its fine! @Yttrium :)
So you get it now, why it became sqrt(x^3)??
@elaornelas do you get the x^(3/2) (same question)?
yes i do @Tonks so would this mean my result would be \[\frac{ -12x ^{\frac{ 5 }{ 2 }} }{ 5 }\]???
yes: -12/5 x^(5/2) [25,81]
Yeah :D
-12/5* 55924 blech!
yay! @Tonks @Yttrium :) OK so NOW….. is this the correct setup -[F(b)-F(a)]=__?
gotta go, good luck, you two!
Owyeah :D
ok well i got this real weird number :/ "134217.6"
uh oh is that wrong?
not sure :/ I'm confused @Yttrium
\[\int\limits 6x \sqrt{x } dx=\frac{12 x^{5/2}}{5} \]\[\left(\frac{12\ 25^{5/2}}{5}\right)-\left(\frac{12\ 81^{5/2}}{5}\right)=-\frac{671088}{5} \]
The fraction to 10 digits.\[N\left[-\frac{671088}{5},10\right]=-134217.6000 \]
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