Verified answer

The integral of the function is [(x^3)/3 -2e^x]

just insert the limits to get the definite integral:

[1/3 - 2e] - [0 -2] = 1/3 -2e + 2 = 2.3333 -2e

which I'll leave you to simplify any further as you wish.

Note that "Stumped" is wrong in adding an integration constant (c) to a definite integral. The integration constant cancels when you insert the limits and subtract.

Spilt up the integral to x^2 and 2e^x. taking the integral of x^2 gives you x^3/3 and the integral of 2e^x is 2e^x. combining them your governing equations is {(x^3/3) - (2e^x)}. Now run this thru the fun. Theo. of Calc. F(b)-F(a) where b=1 and a=0.

{[1^3/3 - 2e^1] - [0^3/3 - 2e^0]} doing the algebra gives you:

(1/3 - 2e^1) - (0/3 - 2e^0)

1/3 - 2e - 0/3 + 2 (remember that anything raised to the zero power is equal to 1)

giving you 7/3 - 2e + c.

The integral from low limit 0 to 1 of [x² - 2e^x] dx =

[(x^3)/3 - 2e^x] evaluated at 1, then subtract [(x^3)/3 - 2e^x] evaluated at zero.

I get 2&(1/3) - 2e.