Question

Find the area enclosed by the curve \(x = 1 + {e^t},y = t - {t^2}\) and the x-axis.

Short answer

The area enclosed of the given curve is given as: \(A = 3 - e\)

Step-by-step solution

Draw the graph of the given expression

The parametric equation of curve is:

\(\begin{array}{l}x = 1 + {e^t}\\y = t - {t^2}\end{array}\)

The graph of the ellipse when is shown below:

Find the area

The area enclosed by the curve\(x = f(t),y = g(t)\)as \(t\)increases from\(\alpha \)to\(\beta \)is:

\(A = \int\limits_\alpha ^\beta {g(t)f'(t)} dt\)

Here\(f(t) = 1 + {e^t},g(t) = t - {t^2}\)

Find\(f'(t)\)

\(\begin{array}{l}f(t) = 1 + {e^t}\\f'(t) = {e^t}\end{array}\)

Now we need to find the interval of t.

Since the curve has intersections with x-axis so we need to let y equal to zero to get the values of t.

\(\begin{array}{l}t - {t^2} = 0\\t(1 - t) = 0\\t = 0,1\end{array}\)

The interval will be: \(\left( {0,1} \right)\)

\(\begin{array}{l}A = \int\limits_\alpha ^\beta {g(t)f'(t)} dt\\A = \int\limits_0^1 {t{e^t}} dt - \int\limits_0^1 {{t^2}{e^t}} dt\\u = t,du = dt,v = {e^t},dv = {e^t}dt\\\int\limits_0^1 {t{e^t}} dt = \left( {t{e^t}} \right)_0^1 - \int\limits_0^1 {{e^t}} dt\\\int\limits_0^1 {t{e^t}} dt = e - (e - 1)\\\int\limits_0^1 {t{e^t}} dt = 1\\u = {t^2},du = 2tdt,v = {e^t},dv = {e^t}dt\\\int\limits_0^1 {{t^2}{e^t}} dt = \left( {{t^2}{e^t}} \right)_0^1 - 2\int\limits_0^1 {t{e^t}} dt\\\int\limits_0^1 {{t^2}{e^t}} dt = (e - 0) - 2(1)\\\int\limits_0^1 {{t^2}{e^t}} dt = e - 2\end{array}\)

The area of the curve is:

\(\begin{array}{l}A = \int\limits_0^1 {t{e^t}} dt - \int\limits_0^1 {{t^2}{e^t}} dt\\A = 1 - (e - 2)\\A = 3 - e\end{array}\)

Hence, the area enclosed of the given curve is given as: \(A = 3 - e\).