# (a) Sketch a graph of A(x)= ʃ(superscript x)(subscript 1) (t^2) dt by first sketching a graph of t^2 and by counting small squares. (a) Sketch a graph of A(x)= ʃ(superscript x)(subscript 1)...

(a) Sketch a graph of A(x)= ʃ(superscript x)(subscript 1) (t^2) dt by first sketching a graph of t^2 and by counting small squares.

(a) Sketch a graph of A(x)= ʃ(superscript x)(subscript 1) (t^2) dt by first sketching a graph of t^2 and by counting small squares. (b) Note that A(1)= 0 (Why?). (c) Also, note that the FTC gives us A’(x)= x^2 Why? Based on this information how can we obtain an algebraic formula for A(x)?

*print*Print*list*Cite

Consider the integral `A(x)=int_1^x t^2 dt`. Calculating the integral is the same as finding the area from t to x between the graph of `t^2` and the horizontal axis. The graph of `t^2` is given by:

Supposing that x is a small number (say x=4) then we can count the squares under the graph (in this case about 4, with each square have 5 units, so the total area is about 20 square units).

However, there is an easier way to find A(x). Notice that A(1)=0 since there is no area under the curve from 1 to 1.

Also `A'(x)=x^2` by the FTC, so to find A(x), we need to find another function that has the same derivative as `x^2`. Any function `1/3 x^3+C` has the same derivative, were C is any constant. Since A(1)=0, then we can find C by substituting for 1 in the function.

This means that `0=1/3(1)+C`

which means that `C=-1/3`.

**Therefore, the integral is `A(x)=1/3 x^3-1/3`.**