What is the definite integral of x^2*e^x between the limits x = 0 and x = 1?

We'll integrate by parts the function,
twice.

We'll use the
formula:

Int udv = u*v-Int
vdu

We'll put u = x^2 => du =
2x

We'll put dv = e^x => v =
e^x

We'll apply the
formula:

Int x^2*e^x dx = x^2*e^x - Int 2x*e^x
dx

We'll repeat the procedure again for the term Int
2x*e^xdx:

We'll put u = 2x => du =
2

We'll put dv = e^x => v =
e^x

We'll apply the
formula:

Int 2x*e^xdx = 2x*e^x - 2Int e^x
dx

Int 2x*e^xdx = 2x*e^x -
2e^x

The integral of the given function
is:

Int x^2*e^x dx = x^2*e^x - 2x*e^x +
2e^x

We'll apply Leibniz
Newton:

F(1) = e - 2e + 2e =
e

F(0) = 2

Int x^2*e^x dx =
F(1) - F(0) = e - 2

The definite integral of
the given function is: Int x^2*e^x dx = e - 2.