$\int\limits_{\sqrt{\log_e2}}^{\sqrt{\log_e4}}xe^{x^2}dx$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Evaluate: $\int_{\sqrt{\ln 2}}^{\sqrt{\ln 4}} x e^{x^2} \, dx$
Let $u = x^2 \Rightarrow du = 2x \, dx \Rightarrow x \, dx = \frac{1}{2} \, du$
Change limits:
When $x = \sqrt{\ln 2} \Rightarrow u = \ln 2$
When $x = \sqrt{\ln 4} \Rightarrow u = \ln 4$
So the integral becomes:
$\int_{\ln 2}^{\ln 4} \frac{1}{2} e^u \, du = \frac{1}{2} \int_{\ln 2}^{\ln 4} e^u \, du$
$= \frac{1}{2} \left[ e^u \right]_{\ln 2}^{\ln 4}$
$= \frac{1}{2} (e^{\ln 4} - e^{\ln 2}) = \frac{1}{2} (4 - 2)$
$= \frac{1}{2} \cdot 2 = 1$