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

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The definite integral of x^2*e^x is required between x = 0 and x = 1.

Let's find the indefinite integral first. This requires the use of integration by parts.

Int [ x^2*e^x dx]

Let u = x^2 => du = 2x* dx

dv = e^x dx => v = e^x

=> x^2*e^x - Int [ 2x*e^x dx]

Again for Int[x*e^x dx] we need to use integration by parts

let u = x => du = dx

e^x dx = dv => v = e^x

=> x^2*e^x - 2*( x*e^x - Int [ e^x dx])

=> x^2*e^x - 2*( x*e^x - e^x)

=> x^2*e^x - 2*x*e^x + 2*e^x

Now for x = 0

x^2*e^x - 2*x*e^x + 2*e^x = 2*1

For x = 1

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

=> e

**The required definite integral is e - 2**

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.**