Area Under the Curve
Area under exponential
Grade 12
Question:
<p>The area bounded by \(y=e^x\), \(y=e^{-x}\) and the line \(x=1\) (in the first quadrant) is: [MAU048]</p>
<li>\(e+\dfrac{1}{e}-2\)</li>
<li>\(e-\dfrac{1}{e}\)</li>
<li>\(e+\dfrac{1}{e}\)</li>
<li>\(2e\)</li>
Step-by-Step Solution
Key Concept: Both curves meet at x=0 (y=1). Area = \int_0^1(eˣ-e⁻ˣ)dx = [eˣ+e⁻ˣ]_0^1 = (e+1/e)-2.
<div class='solution'>
<p>At $x=0$: $e^0=e^{-0}=1$. The curves meet at $(0,1)$. On $[0,1]$: $e^x\ge e^{-x}$.</p>
<p>$$A=\int_0^1(e^x-e^{-x})dx=[e^x+e^{-x}]_0^1=(e+e^{-1})-(1+1)=e+\frac{1}{e}-2$$</p>
</div>
Correct Answer: ['A']