Question

What is `int_-1^2xdx`?

Answer 1

`1 1/2`

Explanation:

Standard form: `int(x^n).dx = 1/(n+1) x^(n+1)`

Note that `x = x^1`

`int_-1^2 x .dx = [ 1/2 x^2]_(-1) ^2`

`[1/2 (2)^2] - [ 1/2 (-1)^2]`

`2 - 1/2 = 1 1/2`

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