Definite Integration
Grade 12
Question:
<p>The value of <span class="math-tex">\(\int_{e^{2}}^{e^{4}} \frac{1}{x}\left(\frac{e^{\left(\left(\log _{e} x\right)^{2}+1\right)^{-1}}}{e^{\left(\left(\log _{e} x\right)^{2}+1\right)^{-1}}+e^{\left(\left(6-\log _{e} x\right)^{2}+1\right)^{-1}}}\right) d x\)</span> is</p>
<p style="display:inline">2</p>
<p style="display:inline"><span class="math-tex">\(\log _{e} 2\)</span></p>
<p style="display:inline">1</p>
<p style="display:inline"><span class="math-tex">\(e^{2}\)</span></p>
Step-by-Step Solution
Key Concept: Use the substitution t = ln(x) to convert the integral, then apply the property that f(t) + f(6-t) = 1 for the integrand, which simplifies the fraction through the identity e^a/(e^a + e^b) + e^b/(e^a + e^b) = 1.
<p>Let<br />
<span class="math-tex">$I=\int_{e^{2}}^{e^{4}} \frac{1}{x}\left(\frac{e^{\left(\left(\log _{e} x\right)^{2}+1\right)^{-1}}}{e^{\left.\left(\log _{e} x\right)^{2}+1\right)^{-1}}+e^{\left.\left(6-\log _{e} x\right)^{2}+1\right)^{-1}}}\right)$</span><br />
Put <span class="math-tex">$\ln x=t \Rightarrow \frac{d x}{x}=d t$</span><br />
<span class="math-tex">$I=\int_{2}^{4} \frac{e^{\frac{1}{1+t^{2}}}}{e^{\frac{1}{1+t^{2}}}+e^{\frac{1}{1+(6-t)^{2}}}} d t$</span> ...(i)<br />
<span class="math-tex">${\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]}$</span><br />
<span class="math-tex">$I=\int_{2}^{4} \frac{e^{\frac{1}{1+(6-t)^{2}}}}{e^{\frac{1}{1+(6-t)^{2}}}+e^{\frac{1}{1+t^{2}}}} d t$</span> ...(ii)<br />
Adding eqn (i) and (ii), we get<br />
<span class="math-tex">$2 I=\int_{2}^{4} 1 d t=[t]_{2}^{4}=4-2=2$</span><br />
<span class="math-tex">$\Rightarrow I=1$</span></p>
Correct Answer: C