Question

If \(\displaystyle \int_{9}^{x}\frac{f(y)}{y^2}\,dy=2\sqrt{x}-6\), then \(f(x)=\)

• A. \(\sqrt{x}\)
• B. \(x\sqrt{x}\)
• C. \(x^2\sqrt{x}\)
• D. \(x+\sqrt{x}\)

Hint:

If \[ \int_a^x g(y)\,dy=F(x), \] then by the Fundamental Theorem of Calculus, \[ g(x)=F'(x). \] This is very useful for finding unknown functions inside definite integrals.

Answer 1

The Correct Option is B

Step 1: Use the Fundamental Theorem of Calculus.
Given, \[ \int_{9}^{x}\frac{f(y)}{y^2}\,dy=2\sqrt{x}-6 \] Differentiate both sides with respect to \(x\).
By the Fundamental Theorem of Calculus, \[ \frac{d}{dx}\left(\int_{9}^{x}\frac{f(y)}{y^2}\,dy\right) = \frac{f(x)}{x^2} \]

Step 2: Differentiate the right hand side.
\[ \frac{d}{dx}(2\sqrt{x}-6) = 2\cdot \frac{1}{2\sqrt{x}}-0 \] \[ = \frac{1}{\sqrt{x}} \]

Step 3: Equate both derivatives.
Therefore, \[ \frac{f(x)}{x^2}=\frac{1}{\sqrt{x}} \] Multiplying both sides by \(x^2\), \[ f(x)=\frac{x^2}{\sqrt{x}} \] \[ f(x)=x^{2-\frac12} \] \[ f(x)=x^{3/2} \] \[ f(x)=x\sqrt{x} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{x\sqrt{x}} \]