int left(sum_{r=0}^{infty} frac{x^r 2^r}{r!}r | vedclass

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(D) We know that the Taylor series expansion for the exponential function is given by $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.Substituting $u = 2x$

Vedclass · Accepted Answer

$\frac{e^{2x}}{2} + c$

Vedclass · Answer

(D) We know that the Taylor series expansion for the exponential function is given by $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.Substituting $u = 2x$ into the series,we get $\sum_{r=0}^{\infty} \frac{(2x)^r}{r!} = \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} = e^{2x}$.Now,we need to evaluate the integral $\int e^{2x} dx$.Using the integration formula $\int e^{ax} dx = \frac{e^{ax}}{a} + c$,where $a = 2$,we get $\int e^{2x} dx = \frac{e^{2x}}{2} + c$.Thus,the correct option is $D$.

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

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