The area of the region bounded by the curve $$ y-1=\mathrm{e}^{x} $$ and $$ x $$-axis and the straight lines $$ x=0, x=\mathrm{k} $$ where $$ \mathrm{k}0 $$ equals..... square unit. $$ \begin{array}{llll} ext { (a) } \mathrm{e}^{\mathrm{k}}-\mathrm{k}+1 & ext { (b) } e^{k}+k+1 & ext { (C) } \mathrm{e}^{\mathrm{k}}+\mathrm{k}-1 & ext { (d) } e^{k}-k-1\end{array} $$

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The area of the region bounded by the curve $$ y-1=\mathrm{e}^{x} $$ and $$ x $$-axis and the straight lines $$ x=0, x=\mathrm{k} $$ where $$ \mathrm{k}>0 $$ equals..... square unit. $$ \begin{array}{llll}	ext { (a) } \mathrm{e}^{\mathrm{k}}-\mathrm{k}+1 & 	ext { (b) } e^{k}+k+1 & 	ext { (C) } \mathrm{e}^{\mathrm{k}}+\mathrm{k}-1 & 	ext { (d) } e^{k}-k-1\end{array} $$

UpStudy AI · Accepted Answer

To find the area bounded by the curve $$ y - 1 = e^{x} $$, the $$ x $$-axis, and the vertical lines $$ x = 0 $$ and $$ x = k $$ where $$ k > 0 $$, follow these steps:

1. **Understanding the Curve and Boundaries:**
   - The equation $$ y = 1 + e^x $$ represents an exponential curve shifted upwards by 1 unit.
   - The $$ x $$-axis is represented by $$ y = 0 $$.
   - The vertical lines are $$ x = 0 $$ and $$ x = k $$.

2. **Setting Up the Integral:**
   - The area between the curve and the $$ x $$-axis from $$ x = 0 $$ to $$ x = k $$ is given by the integral:
     $$
     \text{Area} = \int_{0}^{k} (1 + e^x) \, dx
     $$

3. **Evaluating the Integral:**
   $$
   \int_{0}^{k} (1 + e^x) \, dx = \left[ x + e^x \right]_{0}^{k} = (k + e^k) - (0 + e^0) = e^k + k - 1
   $$

4. **Conclusion:**
   - The area is $$ e^k + k - 1 $$ square units.

**Answer:**  
**Option (C) $$ \mathrm{e}^{\mathrm{k}} + \mathrm{k} - 1 $$**
The area is $$ e^{k} + k - 1 $$ square units.

The area of the region bounded by the curve and -axis and the
straight lines where equals… square unit.

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The area of the region bounded by the curve and -axis and the straight lines where equals… square unit.