Question

Let f(x) = |x — 2] and g(x) = f(f(x)) , x ∈ [0,4], then $\int\limits^3_0$ (g(x) — f(x)) dx =

Answer 1

It had given f(x) = |x - 2| and g(x) = f(f(x)) , x ∈ [0, 4]

we have to find $\int\limits^3_0{(g(x)-f(x))}\,dx$

solution : here for x > 2, f(x) = x - 2

for 0 < x < 2 , f(x) = -(x - 2)

so g(x) = ||x - 2| - 2|

for x ≤ 0 , g(x) = (2 - x) - 2 = - x

for 0 < x < 2, g(x) = 2 - (2 - x) = x

for 2 ≤ x < 4, g(x) = 2 - (x - 2) = 4 - x

for x ≥ 4, g(x) = (x - 2) - 2 = x - 4

now $\int\limits^3_0{(g(x)-f(x))}\,dx=\int\limits^2_0{(g(x)-f(x))}\,dx+\int\limits^3_2{(f(x)-g(x))}\,dx$

= $\int\limits^2_0{g(x)}\,dx+\int\limits^3_2{g(x)}\,dx-\int\limits^2_0{f(x)}\,dx-\int\limits^3_2{f(x)}\,dx$

= $\int\limits^2_0{x}\,dx+\int\limits^3_2{(4-x)}\,dx-\int\limits^2_0{-(x-2)}\,dx-\int\limits^3_2{(x-2)}\,dx$

= [x²/2]²₀ + [4x - x²/2]³₂ + [x²/2 -2x]²₀ - [x²/2 - 2x]³₂

= 2 + (12 - 9/2) - (8 - 2) + (2 - 4) - [(9/2 - 6) - (2 - 4)]

= 2 + 15/2 - 6 - 2 - [-3/2 + 2]

= 15/2 - 6 + 3/2 - 2

= 9 - 8

= 1

Therefore the value of $\int\limits^3_0{(g(x)-f(x))}\,dx$ is 1