University Entrance Last Question

$\begin{aligned} \int_x^{x+1} f(t) \ dt & = e^x \\ \int_0^{x+1} f(t) \ dt - \int_0^x f(t) \ dt & = e^x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ f(x+1) - f(x) & = e^x & \small \color{#3D99F6} \text{Putting }x = 1 \\ f(2) - f(1) & = e \quad ...(1) & \small \color{#3D99F6} \text{Putting }x = 1=0 \\ f(1) - f(0) & = 1 \quad ...(2) & \small \color{#3D99F6} \text{Adding the two equations }(1) + (2) \\ \implies f(2) - f(0) & = \boxed{e+1} \end{aligned}$

Let $f(t) = Ae^{t}$ so that the above definite integral computes to:

$\int_{x}^{x+1} Ae^{t} dt = Ae^{t}|_{x}^{x+1} = Ae^{x+1} - Ae^{x} = Ae^{x}(e - 1) = e^x$

or $A = \frac{1}{e-1} \Rightarrow f(x) = \frac{e^x}{e-1}.$ Thus, $f(2) - f(0) = \frac{e^{2} - 1}{e-1} = \frac{(e+1)(e-1)}{e-1} = \boxed{e+1}.$

The given integral from x to x+1 plus the same integral from x+1 to x+2 equals exp(x) + exp(x+1).

f(2)-f(0) = exp(0) + exp(1) = 1+e, because we set x = 0.