4
$\begingroup$

Find a function $f$ and a number $a$ such that: $$ 6+\int_{a}^{x}\frac{f(t)}{t^2}\:\mathrm{d}t=2\sqrt{x} $$ For all $x>0$

From Fundamental Theorem of Calculus section. Having some trouble with this. Any help?

$\endgroup$
1
  • $\begingroup$ What have you tried? The fact that this is in the Fundamental Theorem of Calculus section is a strong indicator as to what you should do. $\endgroup$ Commented Aug 13, 2013 at 16:37

2 Answers 2

7
$\begingroup$

Hint ::

Just differentiate, get an expression for $f$ and then substitute back to obtain the value of $a$.

EDIT :

To differentiate the integral, use the property : $$\dfrac d{dx}\large\int^{g(x)}_{f(x)}h(t)dt=h(g(x)).g'(x)-h(f(x)).f'(x)$$

$\endgroup$
0
6
$\begingroup$

$$ \int_a^x \frac{f(t)}{t^2} \ dt = 2\sqrt{x} - 6 $$

$$\mathrm{ differentiate \ using \ leibniz \ rule } $$

$$\frac{f(x)}{x^2} = \frac{1}{\sqrt{x} } \Rightarrow f(x) = x\sqrt{x}$$

$$\int_a^x \frac{\sqrt{t}}{t} \ dt = 2\sqrt{x} - 6 $$

$$\int_a^x t^{-\frac{1}{2}} \ dt = \left | 2\sqrt{t} \right|_a^x = 2\sqrt{x} - 2\sqrt{a} = 2\sqrt{x} - 6$$

$$ \Rightarrow 2\sqrt{a} = 6 \Rightarrow a = 9 $$

$\endgroup$
6
  • $\begingroup$ sorry @Vijay but it looks like floor function . $\endgroup$
    – what'sup
    Commented Aug 13, 2013 at 16:53
  • $\begingroup$ I get it up until the 3rd line. Where did $$\int_a^x \frac{\sqrt{t}}{t} \ dt $$ come from? $\endgroup$
    – user5826
    Commented Aug 13, 2013 at 17:08
  • $\begingroup$ $$ \frac{f(t)}{t^2} $$ $\endgroup$
    – what'sup
    Commented Aug 13, 2013 at 17:11
  • $\begingroup$ @Juan Because $f(t)/t^2 = 1/\sqrt t $ $\endgroup$ Commented Aug 13, 2013 at 17:12
  • $\begingroup$ @Juan Did you understand ? $\endgroup$
    – what'sup
    Commented Aug 13, 2013 at 17:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .