Based on the graph  f(t) (attached in the link bellow) fill out approximate yhe values in the table of A(x)= ʃ(superscript x)(subscript 0) f(t) dt.  Based on the graph  f(t) attached to this...

Based on the graph  f(t) (attached in the link bellow) fill out approximate yhe values in the table of A(x)= ʃ(superscript x)(subscript 0) f(t) dt.

 

Based on the graph  f(t) attached to this link (http://s19.postimage.org/bjz86zm0j/calc_hw_23_3_graph.png  ),

(a) fill out approximate values in the table of:

A(x)= ʃ(superscript x)(subscript 0) f(t) dt. 

(The link to this table is: http://s19.postimage.org/mv1vvcsvn/calc_hw_23_3_table.png )

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lfryerda | High School Teacher | (Level 2) Educator

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The only exact value of `A(x)` that we can calculate is `A(0)` since there is no area.  For the rest of the table we need to use approximations to find the area using rectangles or triangles as necessary.

For `A(1)`, we see that the area under the curve is approximately a rectangle, that goes from x=0 to x=1 and has a height of 1 and base of 1, so `A(1)=0.5`.

The next cell in the table `A(2)` has a positive area and an equal amount of negative area below the x-axis, so `A(2)=0`.

For `A(2.75)`, we need to add an additional negative triangular area with base 0.75 and negative height 1, so `A(2.75)=-0.375`.

`A(3)` adds a small positive area to `A(2.75)`.  In this case the triangle has a base of 0.25 and height of 0.75 to get a total area of `A(3)=-0.375+0.1875=-0.1875`.

Finally `A(4)`, is adding a trapezoidal area to `A(3)`.  The left height is 0.75, the right height is 5 and the base is 1.  This means that `A(4)=-0.1875+2.875=2.6875`.

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