# 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 $x>0$

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?

• 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. Commented Aug 13, 2013 at 16:37

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 : $$\dfrac d{dx}\large\int^{g(x)}_{f(x)}h(t)dt=h(g(x)).g'(x)-h(f(x)).f'(x)$$

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

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

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

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

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

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

• sorry @Vijay but it looks like floor function . Commented Aug 13, 2013 at 16:53
• I get it up until the 3rd line. Where did $$\int_a^x \frac{\sqrt{t}}{t} \ dt$$ come from? Commented Aug 13, 2013 at 17:08
• $$\frac{f(t)}{t^2}$$ Commented Aug 13, 2013 at 17:11
• @Juan Because $f(t)/t^2 = 1/\sqrt t$ Commented Aug 13, 2013 at 17:12
• @Juan Did you understand ? Commented Aug 13, 2013 at 17:13