# using the formula for the area of triangles (or trapezoids) find the function: A(x)= ʃ (superscript x)(subscript -1) (3- 3t)dt. Then calculate A’(x).Graph f(t)= 3 – 3t. Assume that -1<...

using the formula for the area of triangles (or trapezoids) find the function: A(x)= ʃ (superscript x)(subscript -1) (3- 3t)dt. Then calculate A’(x).

Graph f(t)= 3 – 3t. Assume that -1< x< 1 and using the formula for the area of triangles (or trapezoids) find the function: A(x)= ʃ (superscript x)(subscript -1) (3- 3t)dt. Then calculate A’(x).

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The graph of `f(t)=3-3t` is:

The integral `A(x)=int_{-1}^x (3-3t)dt` is found by getting the area between the line and the horizontal axis, where the area is negative to the right of x=1.

If `x<=1`, then the area is the trapezoid from -1 to x, which is has base x+1, one height of 6 and the other height of 3-3x which has area `A(x)=1/2 (x+1)(6+3-3x)=3/2(x+1)(3-x)=3/2(-x^2+2x+3)`.

If `x>1`, then the area is the initial triangle of area 6, and then subtract the remaining triangle from 1 to x with height 3-3x and a base of x-1. This gives and area `A(x)=6+1/2(x-1)3(1-x)`. This simplifies to

`A(x)=6-3/2(x-1)^2`

`=3/2(4-x^2+2x-1)`

`=3/2(-x^2+2x+3)`

**Since the formula is the same in both cases, we just have the area as **

**`A(x)=3/2(-x^2+2x+3)` **

**The derivative of the area is the original integrand, which is**

**`A'(x)=3-3x` **