Question

Let $f:(0,\infty) \rightarrow R$ and $F(x) = \int_{0}^{x}f(t)dt$. If $F(x^2) = x^2(1+x)$ then $f(4)$ equals:

Answer 1

4

Answer 2

To find the value of $f(4)$, we will use the given information and the Fundamental Theorem of Calculus.

Given:

1. $F(x) = \int_{0}^{x}f(t)dt$
2. $F(x^2) = x^2(1+x)$

Step 1: Relate $F(x)$ and $f(x)$ using the Fundamental Theorem of Calculus. According to the Fundamental Theorem of Calculus, if $F(x) = \int_{a}^{x}f(t)dt$, then $F'(x) = f(x)$. Therefore, from the first given condition, we have: $F'(x) = f(x)$

Step 2: Differentiate the second given equation with respect to $x$. The second equation is $F(x^2) = x^2(1+x)$. Differentiate both sides with respect to $x$: $\frac{d}{dx}(F(x^2)) = \frac{d}{dx}(x^2(1+x))$

For the left side, use the chain rule: $\frac{d}{dx}F(g(x)) = F'(g(x)) \cdot g'(x)$. Here $g(x) = x^2$, so $g'(x) = 2x$. $\frac{d}{dx}(F(x^2)) = F'(x^2) \cdot 2x$

For the right side, first expand $x^2(1+x)$ to $x^2+x^3$, then differentiate: $\frac{d}{dx}(x^2+x^3) = 2x + 3x^2$

Equating the derivatives of both sides: $F'(x^2) \cdot 2x = 2x + 3x^2$

Step 3: Substitute $F'(x^2)$ with $f(x^2)$. From Step 1, we know $F'(x) = f(x)$. Replacing $x$ with $x^2$, we get $F'(x^2) = f(x^2)$. Substitute this into the equation from Step 2: $f(x^2) \cdot 2x = 2x + 3x^2$

Step 4: Solve for $f(x^2)$. Divide both sides by $2x$. Since the domain of $f$ is $(0,\infty)$, $x$ must be positive, so $2x \neq 0$. $f(x^2) = \frac{2x + 3x^2}{2x}$ $f(x^2) = \frac{2x}{2x} + \frac{3x^2}{2x}$ $f(x^2) = 1 + \frac{3x}{2}$

Step 5: Find $f(4)$. We need to find $f(4)$. From the expression $f(x^2) = 1 + \frac{3x}{2}$, we need $x^2 = 4$. Since $x \in (0,\infty)$, we take the positive square root, so $x = 2$. Substitute $x=2$ into the expression for $f(x^2)$: $f(4) = 1 + \frac{3(2)}{2}$ $f(4) = 1 + 3$ $f(4) = 4$

The final answer is $4$.