# Find the region enclosed by the curves: `y=e^(x)` , `y=xe^(x)` , x=0``

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### 1 Answer

You need to find the limits of integration, hence, you need to solve the equation `e^x = x*e^x` , such that:

`e^x = x*e^x => e^x - x*e^x = 0`

Factoring out` e^x` yields:

`e^x(1 - x) = 0`

Since `e^x > 0` , hence `1 - x = 0 => x = 1`

You need to evaluate the functions `y = e^x` and `y = x*e^x` at `x = 0` and `x = 1/2` , such that:

`x = 0 => y = e^0 => y = 1`

`x = 0 => y = 0*e^0 = 0`

`x = 1/2 => y = e^(1/2) => y = sqrt e `

`x = 1/2 => y = (sqrt e)/2`

Hence `e^x > x*e^x ` for `x in [0,1]` , thus, evaluating the area of the region enclosed by the given curves, yields:

`int_0^1 (e^x - x*e^x)dx`

Using the property of linearity of integral, you need to split the integral in two, such that:

`int_0^1 (e^x)dx - int_0^1 (x*e^x)dx`

You need to solve `int_0^1 (x*e^x)dx` using integration by parts, such that:

`int udv = uv - int vdu`

`u = x => du = dx`

`dv = e^x => v = e^x`

`int_0^1 (x*e^x)dx= x*e^x|_0^1 - int_0^1 e^x dx`

`int_0^1 (e^x - x*e^x)dx = int_0^1 (e^x)dx -x*e^x|_0^1 + int_0^1 e^x dx`

`int_0^1 (e^x - x*e^x)dx = 2e^x|_0^1 - x*e^x|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 (e^x - x*e^x)dx = 2e^1 - 2e^0 - 1*e^1 + 0*e^0`

`int_0^1 (e^x - x*e^x)dx = 2e - 2 - e`

`int_0^1 (e^x - x*e^x)dx = e - 2`

**Hence, evaluating the area of the region enclosed by the given curves, yields **`int_0^1 (e^x - x*e^x)dx = e - 2.`