Question

Question #f8c66

Answer 1

`int_(-1)^1 xe^(x^8) dx = 0`

Explanation:

I assume "with" boundaries of -1 and 1

ie `int_-1^1 xe^(x^8) dx`

`int_-1^1 xe^(x^8) dx = 1/8int_-1^1 8xe^(x^8) dx`
`:. int_(-1)^1 xe^(x^8) dx = 1/8[e^(x^8)]_(-1)^1`

`:. int_(-1)^1 xe^(x^8) dx = 1/8{e^1 - e^1}`
`:. int_(-1)^1 xe^(x^8) dx = 0`

graph{xe^(x^8) [-10, 10, -5, 5]}

NB If you wanted the Area bounded by the curve then by symmetry we would want:

`A = 2int_0^1 xe^(x^8) dx = 2 * 1/8[e^(x^8)]_0^1`
`:. A = 1/4{e^1 - e^0}`
`:. A = 1/4(e-1)`