Question

How would i go about solving a problem like this:

For all x>0, find a function f and a number "a", such that :

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

ok

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

okay

Answer 11

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

Answer 12

hold on maybe i messed up damn damn damn

Answer 13

okay

Answer 14

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

Answer 15

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
\[\frac{1}{\sqrt{x}}\]

Answer 16

got ya, applying the fundamental theorem of calculus

Answer 17

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

Answer 18

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

Answer 19

now do we still adust for the constant

Answer 20

adjust*

Answer 21

do we still adjust for the constant for no?

Answer 22

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