University Entrance Last Question

Calculus Level 2

Suppose that a continuous function f ( x ) f(x) satisfies the relation x x + 1 f ( t ) d t = e x \displaystyle \int_x^{x+1} f(t)\,dt = e^x for every x 0 x\ge 0 . Find f ( 2 ) f ( 0 ) f(2)-f(0) .

e 1 e-1 e + 1 e+1 e 2 + 1 e^2+1 1 1

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3 solutions

Chew-Seong Cheong
May 25, 2019

x x + 1 f ( t ) d t = e x 0 x + 1 f ( t ) d t 0 x f ( t ) d t = e x Differentiate both sides w.r.t. x f ( x + 1 ) f ( x ) = e x Putting x = 1 f ( 2 ) f ( 1 ) = e . . . ( 1 ) Putting x = 1 = 0 f ( 1 ) f ( 0 ) = 1 . . . ( 2 ) Adding the two equations ( 1 ) + ( 2 ) f ( 2 ) f ( 0 ) = e + 1 \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}

Dang it, I've never seen this type of problem before. Thanks!

Elijah Wenn - 2 years ago

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You are welcome

Chew-Seong Cheong - 2 years ago
Tom Engelsman
May 24, 2019

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

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

Kris Hauchecorne
May 27, 2019

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.

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