$\int\limits_{\sqrt{\log_e2}}^{\sqrt{\log_e4}}xe^{x^2}dx$ is equal to

1

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:

$\int_{\ln 2}^{\ln 4} \frac{1}{2} e^u \, du = \frac{1}{2} \int_{\ln 2}^{\ln 4} e^u \, du$

$= \frac{1}{2} \left[ e^u \right]_{\ln 2}^{\ln 4}$

$= \frac{1}{2} (e^{\ln 4} - e^{\ln 2}) = \frac{1}{2} (4 - 2)$

$= \frac{1}{2} \cdot 2 = 1$