# find the area function A(x) for the function y=t with a=0. please include graph with solution.

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### 1 Answer

By Fundamental Theorem of Integral Calculus ,the area function A(x)

is

`A(x)=int_a^xy(t)dt`

`=int_a^xtdt`

Now

`inttdt=t^2/2+c` where c is an integrating constant.We determine C with the given condition a=0.

Put

`t^2/2+c=0`

`0/2+c=0` (put t=0=a)

c=0

Thus we have

`A(x)=int_0^xtdt`

`=x^2/2`

Thus area function is

`A(x)=x^2/2`

Its graph is