Definite Integration
Grade None
Question:
<p>The value of <span class="math-tex">\(\int_{-1}^{1} \frac{(1+\sqrt{|x|-x}) e^{x}+(\sqrt{|x|-x}) e^{-x}}{e^{x}+e^{-x}} d x\)</span> is equal to</p>
<p style="display:inline"><span class="math-tex">\(2+\frac{2 \sqrt{2}}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(1+\frac{2 \sqrt{2}}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(1-\frac{2 \sqrt{2}}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(3-\frac{2 \sqrt{2}}{3}\)</span></p>
Step-by-Step Solution
Key Concept: Use the definite integral property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ to eliminate the exponential denominator by summing the original and transformed integrands.
<p><span class="math-tex">$I=\int_{-1}^{1} \frac{(1+\sqrt{|x|-(-x)}) e^{-x}+(\sqrt{|x|-(-x)}) e^{x}}{e^{-x}+e^{x}} d x$</span><br />
<span class="math-tex">$=\int_{-1}^{1} \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{x}}{e^{x}+e^{-x}} d x$</span><br />
<span class="math-tex">$2 I=\int_{-1}^{1} \frac{(1+\sqrt{|x|+x}+\sqrt{|x|-x})\left(e^{x}+e^{-x}\right)}{e^{x}+e^{-x}} d x$</span><br />
<span class="math-tex">$2 I=\int_{-1}^{1}(1+\sqrt{|x|+x}+\sqrt{|x|-x}) d x$</span><br />
<span class="math-tex">$2 I=2 \int_{0}^{1}(1+\sqrt{|x|+x}+\sqrt{|x|-x}) d x$</span><br />
<span class="math-tex">$2 I=2 \int_{0}^{1}(1+\sqrt{2 x}+\sqrt{0}) d x$</span> (using symmetry)<br />
<span class="math-tex">$2 I=2\left[x+\frac{2 \sqrt{2}}{3} x^{3 / 2}\right]_{0}^{1}$</span><br />
<span class="math-tex">$=2\left(1+\frac{2 \sqrt{2}}{3}\right)=2+\frac{4 \sqrt{2}}{3}$</span><br />
Hence, <span class="math-tex">$I=1+\frac{2 \sqrt{2}}{3}$</span></p>
Correct Answer: B