# Sketch an approximate graph of the area function A(x)= ʃ(superscript x)(subscript 0) f(t) dt restricted to the interval [0,4].based on the graph f(t) (which can be found at the following link...

Sketch an approximate graph of the area function A(x)= ʃ(superscript x)(subscript 0) f(t) dt restricted to the interval [0,4].

based on the graph f(t) (which can be found at the following link http://s19.postimage.org/bjz86zm0j/calc_hw_23_3_graph.png )

Sketch an approximate graph of the area function A(x)= ʃ(superscript x)(subscript 0) f(t) dt restricted to the interval [0,4].

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Since the integral `A(x)=int_0^x f(t)dt` is the area under the graph of f(t) starting at 0 and moving to the right, we can get an approximate graph of A(x) by dividing the function into triangles and adding each triangle to the area as we go. Also, A(0)=0 since there is no area under the graph at the start.

The first section to consider is the triangle from 0 to 1, which has height 1. This complete triangle has area `1/2`. The triangle area also increases quickly from zero to slowly at 1 (like an upside down parabola).

The next section is from 1 to 2 since this is a negative area. It is the same as the first section but when x=2, we are back to zero total area.

The next section is from 2 to 2.75. This has negative area, and goes from subtracting area quickly to only a little bit. The area ends up being at -0.375.

The last section is from x=2.75 to x=4. This has positive area, and ultimately adds 2.5 to the area, for a final value of A(4)=2.125. The crossing line appears to be about x=3 or so.

This means the approximate graph for A(x) is: