# find the area function A(x) for the functiony=2t+1 with a=0. please include graph with solution

### 1 Answer | Add Yours

By fundamental theorem of integral calculus ,the area function is

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

`=int_a^x(2t+1)dt`

`` But we have

`int(2t+1)dt=t^2+t+c` ,where c is constant to be determined under given condition a=0

Thus we have (we assume point has zero area)

so put

`t^2+t+c=0`

`c=0` (when you put t=0)

Thus area function is

`A(x)=int_0^x(2t+1)dt`

`=x^2+x`

Thus area function is

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

and its graph is

**Sources:**