Find the area between the following curves.
\[
x=-6, x=1, y=e^{x} \text {, and } y=1-e^{x}
\]

Area $=\square($ Type an exact answer in terms of e.)

Step 1 ：The two functions are \(y = e^x\) and \(y = 1 - e^x\). The interval is from \(x = -6\) to \(x = 1\).

Step 2 ：The area \(A\) between the curves is given by the integral: \(A = \int_{-6}^{1} |e^x - (1 - e^x)| dx\)

Step 3 ：Simplify the absolute value: \(A = \int_{-6}^{1} |2e^x - 1| dx\)

Step 4 ：We need to split the integral at the point where \(2e^x - 1 = 0\), which is \(x = \ln(1/2)\).

Step 5 ：So, we have: \(A = \int_{-6}^{\ln(1/2)} (1 - 2e^x) dx + \int_{\ln(1/2)}^{1} (2e^x - 1) dx\)

Step 6 ：Now, we can integrate: \(A = [x - 2e^x]_{-6}^{\ln(1/2)} + [2e^x - x]_{\ln(1/2)}^{1}\)

Step 7 ：Evaluate the integrals at the limits of integration: \(A = [(\ln(1/2) - 2e^{\ln(1/2)}) - (-6 - 2e^{-6})] + [(2e - 1) - (2e^{\ln(1/2)} - \ln(1/2))]\)

Step 8 ：Simplify: \(A = [\ln(1/2) - 1 + 6 + 2/e^6] + [2e - 1 - 1 + \ln(1/2)]\)

Step 9 ：Further simplify: \(A = [\ln(1/2) + 5 + 2/e^6] + [2e + \ln(1/2)]\)

Step 10 ：Finally, \(A = 2\ln(1/2) + 5 + 2/e^6 + 2e\)

Step 11 ：This is the exact area between the curves. So, \(\boxed{A = 2\ln(1/2) + 5 + 2/e^6 + 2e}\)