Find an expression for g(x)?

Question

let

f(x)=

0 if x<0

x if 0≤ x ≤ 1

2-x if 1< x ≤ 2

0 if x>2

and g(x)= integrate from 0 to x f(t) dt

find an expression for g(x) similar to the one for f(x).

where is f defferentiable?

where is g differentiable

please help having to teach this to myself teacher no help

Answer 1

We have:

f(x)=

0 if x&lt;0

x if 0≤ x ≤ 1

2-x if 1&lt; x ≤ 2

0 if x&gt;2

and

g(x) = ∫ [0 to x] f(x) dx.

First, I observe that as f(x) is continuous, taking limits (e.g., lim x→0⁻) for the integration is unnecessary.

If x &lt; 0,

g(x) = ∫ [0 to x] 0 dx = 0.

If 0 ≤ x ≤ 1,

g(x) = ∫ [0 to x] x dx = x²/2 |[0,x] = x²/2 - 0²/2 = x²/2.

If 1 &lt; x ≤ 2,

g(x) = ∫ [0 to x] f(x) dx =

∫ [0 to 1] f(x) dx + ∫ [1 to x] f(x) dx =  

1²/2 + ∫ [1 to x] (2-x) dx =  

1/2 + (2x - x²/2) |[1,x] =

1/2 + (2x - x²/2) - (2(1) - (1)²/2) =

1/2 + (2x - x²/2) - (3/2) =

(2x - x²/2) - 1.

If 2 &lt; x,

g(x) = ∫ [0 to x] f(x) dx =

∫ [0 to 2] f(x) dx + ∫ [2 to x] f(x) dx =  

(2(2) - (2)²/2) - 1 + ∫ [1 to x] 0 dx =  

1 + 0 =

1.

To summarize,

g(x)=

0 if x&lt;0

x²/2 if 0 ≤ x ≤ 1

(2x - x²/2) - 1 if 1&lt; x ≤ 2

1 if x&gt;2

(You will eventually learn, if you continue math, that g is called a (probability) distribution function, although the integral for such is from -∞ to +∞.)

f is differentiable everywhere save at x=0, x=1, and x=2, as the lim x→0⁻ (f(x)-f(0))/(x-0) = 0 and lim x→0⁺ (f(x)-f(0))/(x-0) = 1, the lim x→1⁻ (f(x)-f(1))/(x-1) = 1 and lim x→1⁺ (f(x)-f(1))/(x-1) = -1, and the lim x→2⁻ (f(x)-f(2))/(x-2) = -1 and lim x→2⁺ (f(x)-f(2))/(x-2) = 0.

g is differentiable everywhere, as its derivative is the continuous function f.