Question

\[ \int_0^1 x e^{2x} dx \text{ is equal to} \]

• A. \(e^2 - 1\)
• B. \(\frac{1}{4}(e^2 - 1)\)
• C. \(2e^2 + 1\)
• D. \(\frac{1}{4}(e^2 + 1)\)

Hint:

Radioactive decay follows an exponential pattern where activity decreases over time.

Answer 1

The Correct Option is D

Use integration by parts: Let \(u = x\), \(dv = e^{2x} dx\) Then, \(du = dx\), \(v = \frac{1}{2}e^{2x}\) \[ \int x e^{2x} dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} \] Evaluate from 0 to 1: \[ I = \left[\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x}\right]_0^1 = \left(\frac{1}{2}e^2 - \frac{1}{4}e^2\right) - (0 - \frac{1}{4}) = \frac{1}{4}e^2 + \frac{1}{4} = \frac{1}{4}(e^2 + 1) \]