# Find the area of the region between the curves y=x+1, y=9-x^2, x=-1 and x=2can you show me the steps cuz i tried it dozens of times and i always get the wrong answer

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You should notice that the curve `y=9-x^2` is found above the line `y = x + 1` , over the interval [-1,2], hence, you need to evaluate the definite integral of the following function such that:

`int_(-1)^2 (9-x^2-x-1) dx = int_(-1)^2 (8 - x - x^2) dx`

`int_(-1)^2 (8 - x - x^2) dx = int_(-1)^2 8dx - int_(-1)^2 xdx - int_(-1)^2 x^2dx`

Using the formula `int_a^b x^n dx = (x^(n+1))/(n+1)|_a^b` yields:

`int_(-1)^2 (8 - x - x^2) dx = (8x - x^2/2 - x^3/3)|_(-1)^2`

`int_(-1)^2 (8 - x - x^2) dx = (8*2 - 4/2 - 8/3 - 8*(-1) + 1/2 + ((-1)^3)/3)`

`int_(-1)^2 (8 - x - x^2) dx = 16 - 2 - 8/3 + 8 + 1/2- 1/3`

`int_(-1)^2 (8 - x - x^2) dx = 22 + 1/2 - 9/3`

`int_(-1)^2 (8 - x - x^2) dx = 19 + 1/2`

`int_(-1)^2 (8 - x - x^2) dx = 39/2`

**Hence, evaluating the area of the region between the indicated curves yields `int_(-1)^2 (9-x^2-x-1) dx = 39/2` .**

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