Applying the product rule to 2(x+2)² f(x) gives 2[2(x+2)f(x) + (x+2)² f'(x)]. Differentiating -3(x+2)² gives -6(x+2). By the Fundamental Theorem of Calculus, the derivative of 10 ∫_0^x (t+2)f(t)dt is 10(x+2)f(x).

Applying the product rule to 2(x+2)² f(x) gives 2[2(x+2)f(x) + (x+2)² f'(x)]. Differentiating -3(x+2)² gives -6(x+2). By the Fundamental Theorem of Calculus, the derivative of 10 ∫_0^x (t+2)f(t)dt is 10(x+2)f(x).

We expand and simplify the equation. Since x ≥ 0, x+2 is never zero, so we can divide the entire equation by 2(x+2) to simplify it further.

We can rewrite the differential equation as ((x+2)f'(x) - 3f(x))/((x+2)^4) = (3)/((x+2)^4). The left side is the derivative of (f(x))/((x+2)³). Integrating both sides gives (f(x))/((x+2)³) = -(1)/((x+2)³) + C.

We can rewrite the differential equation as ((x+2)f'(x) - 3f(x))/((x+2)^4) = (3)/((x+2)^4). The left side is the derivative of (f(x))/((x+2)³). Integrating both sides gives (f(x))/((x+2)³) = -(1)/((x+2)³) + C.

Let $f$ be a differentiable function such that $2(x + 2)^2 f(x) - 3(x + 2)^2 = 10 \int_0^x (t + 2) f(t) dt$ , $x \geq 0$ . Then $f(2)$ is equal to

Answer: $19$

Step-by-step solution

Step 1: Rewrite the given equation

The problem provides an equation involving a differentiable function $f(x)$ and a definite integral. Our goal is to find the value of $f(2)$ . First, let's rewrite the given equation to make it easier to differentiate.

Step 2: Differentiate both sides with respect to x

To eliminate the integral, we differentiate both sides of the equation with respect to $x$ . We will use the product rule for differentiation on the left side and the Fundamental Theorem of Calculus on the right side.

Step 3: Apply differentiation rules

Applying the product rule to $2(x+2)^2 f(x)$ gives $2[2(x+2)f(x) + (x+2)^2 f'(x)]$ . Differentiating $-3(x+2)^2$ gives $-6(x+2)$ . By the Fundamental Theorem of Calculus, the derivative of $10 \int_0^x (t+2)f(t)dt$ is $10(x+2)f(x)$ .

Step 4: Simplify the differential equation

We expand and simplify the equation. Since $x \geq 0$ , $x+2$ is never zero, so we can divide the entire equation by $2(x+2)$ to simplify it further.

Step 5: Rearrange into a linear first-order differential equation

Rearranging the terms, we get a first-order linear differential equation: $(x+2)f'(x) - 3f(x) = 3$ . This equation can be solved using an integrating factor or by recognizing it as a derivative of a product.

Step 6: Solve the differential equation

We can rewrite the differential equation as $\frac{(x+2)f'(x) - 3f(x)}{(x+2)^4} = \frac{3}{(x+2)^4}$ . The left side is the derivative of $\frac{f(x)}{(x+2)^3}$ . Integrating both sides gives $\frac{f(x)}{(x+2)^3} = -\frac{1}{(x+2)^3} + C$ .

Step 7: Find the constant C and f(2)

From the integrated form, $f(x) = -1 + C(x+2)^3$ . To find $C$ , we use the original equation at $x=0$ . Substituting $x=0$ into the original equation gives $8f(0) - 12 = 0$ , so $f(0) = 3/2$ . Plugging this into our solution for $f(x)$ yields $C = 5/16$ . Therefore, $f(x) = -1 + \frac{5}{16}(x+2)^3$ . Finally, we calculate $f(2)$ by substituting $x=2$ into this expression.

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Let fff be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt2(x + 2)^2 f(x) - 3(x + 2)^2 = 10 \int_0^x (t + 2) f(t) dt2(x+2)2f(x)−3(x+2)2=10∫0x​(t+2)f(t)dt, x≥0x \geq 0x≥0. Then f(2)f(2)f(2) is equal to