OpenStudy (anonymous):

How would i go about solving a problem like this: For all x>0, find a function f and a number "a", such that :

6 years ago
OpenStudy (anonymous):

\[6+\int\limits_{a}^{x} (f(t))/(t^2) dt=2\sqrt{x}\]

6 years ago
OpenStudy (anonymous):

ok the derivative of the integral is the "integrand" right?

6 years ago
OpenStudy (anonymous):

yes

6 years ago
OpenStudy (anonymous):

oh wait let me think before i write....

6 years ago
OpenStudy (anonymous):

ok lets start with \[\int_a^x\frac{f(t)}{t^2}dt=2\sqrt{x}-6\]

6 years ago
OpenStudy (anonymous):

ok

6 years ago
OpenStudy (anonymous):

and this is \[F(x)-F(a)\]for some anti-derivative of \[\frac{f(x)}{x^2}\]

6 years ago
OpenStudy (anonymous):

but you already have the anti - derivative given. it is \[2\sqrt{x}\]

6 years ago
OpenStudy (anonymous):

so we put \[\frac{f(x)}{x^2}=2\sqrt{x}\] \[f(x)=2\sqrt{x}\times x^2=2x^{\frac{5}{2}}\]

6 years ago
OpenStudy (anonymous):

okay

6 years ago
OpenStudy (anonymous):

so now we only have to make sure that \[2\sqrt{a}=6\] making \[\sqrt{a}=3\] \[a=9\]

6 years ago
OpenStudy (anonymous):

hold on maybe i messed up damn damn damn

6 years ago
OpenStudy (anonymous):

okay

6 years ago
OpenStudy (anonymous):

let me start again. we have \[\int_a^x\frac{f(t)}{t^2}dt=2\sqrt{x}-6\]

6 years ago
OpenStudy (anonymous):

yea i messed up. take the derivative of both sides. we will adjust for the constant later. you know the derivative of the integral is the integrand so the left hand side is \[\frac{f(x)}{x^2}\] and the right hand side is \[\frac{1}{\sqrt{x}}\]

6 years ago
OpenStudy (anonymous):

got ya, applying the fundamental theorem of calculus

6 years ago
OpenStudy (anonymous):

NOW solve for \[f(x)\] to get \[f(x)=\frac{x^2}{\sqrt{x}}=x^{\frac{3}{2}}\]

6 years ago
OpenStudy (anonymous):

and that will do it. you know that \[f(x)=x^{\frac{3}{2}}\] and that \[a=9\] that part has not changed. why they wrote it in this goofy form is beyond me, but that is ok

6 years ago
OpenStudy (anonymous):

now do we still adust for the constant

6 years ago
OpenStudy (anonymous):

adjust*

6 years ago
OpenStudy (anonymous):

do we still adjust for the constant for no?

6 years ago
OpenStudy (anonymous):

you still there got one more tiny question? what was the whole point , why did you take the derivative of both sides?

6 years ago
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