Answer: \(\frac74e^3+\frac{1}{4e^5}\) square units, approximately \(35.2\).

Answer: \(\frac74e^3+\frac{1}{4e^5}\) square units, approximately \(35.2\).

At \(x=2\), the curve has value

\(y=6+e^{4(2)-5}=6+e^3.\)

So the horizontal line \(AB\) has equation

\(y=6+e^3.\)

The shaded region lies between this horizontal line and the curve \(y=6+e^{4x-5}\), from \(x=0\) to \(x=2\).

Therefore its area is

This simplifies to

Integrate:

\(\int\left(e^3-e^{4x-5}\right)\,dx =e^3x-\frac14e^{4x-5}.\)

So the area is

Substitute the limits:

\(\left(2e^3-\frac14e^3\right)-\left(0-\frac14e^{-5}\right).\)

Hence the area is

This is approximately \(35.2\).