Area of Region Bounded by Exponential Curve and Lines

To find the area of the region bounded by the curve $y = e^x$, the line $x = 0$ ($y$-axis), and the line $y = e$:

Step 1: Determine the vertices and boundaries of the region

• The curve $y = e^x$ intersects the line $x = 0$ at the point $(0, 1)$.
• The curve $y = e^x$ intersects the line $y = e$ at $e^x = e \implies x = 1$, so the point is $(1, e)$.
• The line $x = 0$ intersects the line $y = e$ at $(0, e)$.

The region lies in the first quadrant, bounded by:

• $x = 0$ on the left,
• $y = e$ on the top,
• $y = e^x$ on the bottom/right.

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Step 2: Compute the area of the region

Integrating with respect to $x$ from $x = 0$ to $x = 1$:
$A = \int_0^1 (e - e^x) \, dx$

Evaluating the integral:
$A = \left[ ex - e^x \right]_0^1 = (e\cdot 1 - e^1) - (e\cdot 0 - e^0) = (e - e) - (0 - 1) = 1$

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Step 3: Evaluate each option

• Option A:
  $e - 1 \neq 1$
  Hence, Option A is incorrect.

• Option C:
  $e - \int_0^1 e^x \, dx = \int_0^1 e \, dx - \int_0^1 e^x \, dx = \int_0^1 (e - e^x) \, dx = 1$
  Hence, Option C is correct.

• Option D:
  Integrating with respect to $y$, the boundary curve is $x = \ln y$ and $y$ varies from $1$ to $e$:
  $\text{Area} = \int_1^e x \, dy = \int_1^e \ln y \, dy$
  Evaluating this:
  $\int_1^e \ln y \, dy = \left[ y \ln y - y \right]_1^e = (e\ln e - e) - (1\ln 1 - 1) = (e - e) - (-1) = 1$
  Hence, Option D is correct.

• Option B:
  Using the definite integral property $\int_a^b f(y) \, dy = \int_a^b f(a + b - y) \, dy$ with $a = 1$ and $b = e$:
  $\int_1^e \ln(e + 1 - y) \, dy = \int_1^e \ln y \, dy = 1$
  Hence, Option B is correct.

Conclusion:

The correct options are B, C, and D.