Question

\( \int \left( \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} \right) dx = \)

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

Hint:

The Taylor series for \(e^u\) is \( \sum_{r=0}^{\infty} \frac{u^r}{r!} \). Recognize common series expansions. Integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} + C \).

Answer 1

The Correct Option is D

Recall the Taylor series expansion for \( e^u \): \[ e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!} = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots \] The expression inside the integral is \( \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} = \sum_{r=0}^{\infty} \frac{(2x)^r}{r!} \).
This is the Taylor series for \( e^{2x} \).
So, we need to calculate \( \int e^{2x} dx \).
Let \( u = 2x \), then \( du = 2dx \implies dx = \frac{1}{2}du \).
\[ \int e^{2x} dx = \int e^u \left(\frac{1}{2}du\right) = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C \] Substitute back \( u=2x \): \[ = \frac{1}{2} e^{2x} + C \] This can be written as \( \frac{e^{2x}}{2} + c \).
This matches option (4).