Mathematics 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 : OpenStudy (anonymous):

$6+\int\limits_{a}^{x} (f(t))/(t^2) dt=2\sqrt{x}$ OpenStudy (anonymous):

ok the derivative of the integral is the "integrand" right? OpenStudy (anonymous):

yes OpenStudy (anonymous):

oh wait let me think before i write.... OpenStudy (anonymous):

ok lets start with $\int_a^x\frac{f(t)}{t^2}dt=2\sqrt{x}-6$ OpenStudy (anonymous):

ok OpenStudy (anonymous):

and this is $F(x)-F(a)$for some anti-derivative of $\frac{f(x)}{x^2}$ OpenStudy (anonymous):

but you already have the anti - derivative given. it is $2\sqrt{x}$ OpenStudy (anonymous):

so we put $\frac{f(x)}{x^2}=2\sqrt{x}$ $f(x)=2\sqrt{x}\times x^2=2x^{\frac{5}{2}}$ OpenStudy (anonymous):

okay OpenStudy (anonymous):

so now we only have to make sure that $2\sqrt{a}=6$ making $\sqrt{a}=3$ $a=9$ OpenStudy (anonymous):

hold on maybe i messed up damn damn damn OpenStudy (anonymous):

okay OpenStudy (anonymous):

let me start again. we have $\int_a^x\frac{f(t)}{t^2}dt=2\sqrt{x}-6$ 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 $\frac{1}{\sqrt{x}}$ OpenStudy (anonymous):

got ya, applying the fundamental theorem of calculus OpenStudy (anonymous):

NOW solve for $f(x)$ to get $f(x)=\frac{x^2}{\sqrt{x}}=x^{\frac{3}{2}}$ 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 OpenStudy (anonymous):

now do we still adust for the constant OpenStudy (anonymous): OpenStudy (anonymous):

do we still adjust for the constant for no? OpenStudy (anonymous):

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

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