Question

Sketch the region enclosed by the given curves and

find its area.

13. \(y = {e^x},y = x{e^x}, x = 0\)

Short answer

The Area\( = \int_0^1 {{e^x}} - x{e^x}dx = e - 2\)

Step-by-step solution

Area enclosed Two Curve

Area enclosed between two curve \(y = f(x)\) and \(y = g(x)\) over \((a,b)\) such that \(f(x) \ge g(x)\) is

\(A = \int_a^b f (x) - g(x)dx\)

From the graph we can see that \({e^x} \ge x{e^x}\) on the Interval \((0,1)\) Therefore the Area between the two curve is

\(A = _0^1{e^x} - x{e^x}dx\)

\(A = _0^1(1 - x){e^x}dx\)

\(A = \;\;\;(1 - x){e^x}dx_0^1\)

Final Proof

Do Integration by parts as shown below :

\(A = (1 - x)\;\;\;{e^x}dx - \frac{{d(1 - x)}}{{dx}} \cdot {e^x}dx\;\;\;dxe_0^1\)

\(A = (1 - x){e^x} + {e^x}dxe_0^1{e^x}{e^1} + (1 - x){e^x} + {e^0}{\rm{ }}\)

\(A = \;\;\;(2 - x){e^1} - (2 - 0){e^0}\)

\(A = e - 2\)