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] f(t)dt on [0, x]
i) Find an expression for g(x) similar to the one for f(x).
ii) Where is f differentiable? Where is g differentiable?
I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).
But from that, how do I figure out an expression for g(x) from the piecewise function?
Any and all help is appreciated here.
Here's part (a).
Did you make a sketch?
a) Find an expression for similar to the one for
gives the area under the graph from
The graph looks like this:Code:| | 1+ * | *:::* | *:::::::* | *:::::::::| * - * * * - - - * - + - * * * - - 0 1 x 2
Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.
For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).
For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is .
For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is . The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is .
The total area is . AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.
For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.