\(\int_{\sqrt{\log_e 2}}^{\sqrt{\log_e 4}}\displaystyle xe^{x^2}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
The factor x is half the derivative of \(x^2\), so let \(t = x^2\):
\(dt = 2x\,dx\)
Changing the limits:
\(x = \sqrt{\log_e 2} \Rightarrow t = \log_e 2\)
\(x = \sqrt{\log_e 4} \Rightarrow t = \log_e 4\)
So
\( \displaystyle \int_{\sqrt{\log_e 2}}^{\sqrt{\log_e 4}} xe^{x^2}\,dx = \frac{1}{2}\int_{\log_e 2}^{\log_e 4} e^{t}\,dt \)
\( \displaystyle = \frac{1}{2}\Big[e^{t}\Big]_{\log_e 2}^{\log_e 4} \)
Since \(e^{\log_e a} = a\):
\( \displaystyle = \frac{1}{2}(4 - 2) = 1 \)