Understand the bounded region

We are given:

• Curve: $y=e^x$
• Vertical line: $x=0$
• Horizontal line: $y=e$

First, find the intersection points:

• With $x=0$, the curve gives $y=e^0=1$, so point is $(0,1)$.
• With $y=e$, the curve gives $e^x=e \Rightarrow x=1$, so point is $(1,e)$.
• The lines $x=0$ and $y=e$ intersect at $(0,e)$.

So the bounded region is enclosed by:

• the curve from $(0,1)$ to $(1,e)$,
• the line $x=0$ from $(0,1)$ to $(0,e)$,
• the line $y=e$ from $(0,e)$ to $(1,e)$.

Compute the area

It is convenient to integrate with respect to $x$.

For $0 \le x \le 1$,

• upper boundary is $y=e$,
• lower boundary is $y=e^x$.

Hence,

So the area equals

Check each option

Option A: $e-1$

But actual area is $1$.

Therefore, A is incorrect.

Option C: $e-\int_0^1 e^x\,dx$

Evaluate:

This equals the area.

Therefore, C is correct.

Option D: $\int_1^e \ln y\,dy$

Write the curve as:

For $1\le y\le e$, the region lies between $x=0$ and $x=\ln y$.

So area is

Hence, D is correct.

We can also verify numerically:

so

Option B: $\int_1^e \ln(e+1-y)\,dy$

Let

When $y=1$, $u=e$. When $y=e$, $u=1$.

Thus,

which is exactly the same as option D, hence equals the area.

Therefore, B is correct.

Final conclusion

The correct options are:

Comparison with stored correct answer

Stored correct answer: B, C, D

This matches our derived answer exactly.