Homework Help

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

carolinesmiths's profile pic

Posted via web

dislike 0 like

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

1 Answer | Add Yours

justaguide's profile pic

Posted (Answer #1)

dislike 0 like

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`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes