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What is the definite integral `int_0^8 e^x - x^2 + 1 dx`

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xataexa | Student, Kindergarten | Salutatorian

Posted September 7, 2013 at 2:25 PM via web

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What is the definite integral `int_0^8 e^x - x^2 + 1 dx`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 7, 2013 at 2:31 PM (Answer #1)

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The value of `int_0^8 e^x - x^2 + 1 dx` has to be determined.

`int_0^8 e^x - x^2 + 1 dx`

= `int_0^8 e^x dx - int_0^8 x^2 dx + int_0^8 1 dx`

= `(e^x)_0^8 - (x^3/3)_0^8 + x_0^8`

= `e^8 - e^0 - (8^3/3) + 0 + 8 - 0`

= `e^8 - 1 - 8^3/3 + 8`

=` e^8 - 491/3`

The definite integral `int_0^8 e^x - x^2 + 1 dx = e^8 - 491/3`

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