Question:

Step 1: Differentiate both sides of the given integral equation with respect to $x$. We are given that $\displaystyle\int_{x}^{x+1} f(t)\, dt = e^x$ for every $x \geq 0$. To find a relationship between function values, we differentiate both sides with respect to $x$. Using the Leibniz rule for differentiating integrals with variable limits: $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t)\, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$$ Step 2: Apply the differentiation formula to our specific integral. In our case, $a(x) = x$ and $b(x) = x+1$, so $a'(x) = 1$ and $b'(x) = 1$. Therefore: $$\frac{d}{dx} \int_{x}^{x+1} f(t)\, dt = f(x+1) \cdot 1 - f(x) \cdot 1 = f(x+1) - f(x)$$ Step 3: Differentiate the right-hand side of the original equation. The derivative of $e^x$ with respect to $x$ is simply $e^x$. Therefore, differentiating both sides of the given equation yields: $$f(x+1) - f(x) = e^x$$ This is our key functional equation. Step 4: Apply the functional equation at $x = 0$. Substituting $x = 0$ into $f(x+1) - f(x) = e^x$: $$f(1) - f(0) = e^0 = 1$$ Step 5: Apply the functional equation at $x = 1$. Substituting $x = 1$ into $f(x+1) - f(x) = e^x$: $$f(2) - f(1) = e^1 = e$$ Step 6: Find $f(2) - f(0)$ by adding the two equations. To find $f(2) - f(0)$, we add the results from Steps 4 and 5: $$f(2) - f(0) = [f(2) - f(1)] + [f(1) - f(0)] = e + 1$$ The middle term $f(1)$ cancels out, giving us: $$f(2) - f(0) = e + 1$$ **Final Answer:** The value of $f(2) - f(0)$ equals $e + 1$, which corresponds to **Option 3**.