int left(sum_{r=0}^{infty} frac{x^r 3^r}{r!}r | vedclass

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Question

(B) We know that the Taylor series expansion for the exponential function is $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.Substituting $u = 3x$,we get $

Vedclass · Accepted Answer

$\frac{e^{3x}}{3} + c$

Vedclass · Answer

(B) We know that the Taylor series expansion for the exponential function is $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.Substituting $u = 3x$,we get $\sum_{r=0}^{\infty} \frac{(3x)^r}{r!} = \sum_{r=0}^{\infty} \frac{x^r 3^r}{r!} = e^{3x}$.Now,we need to evaluate the integral $\int \left(\sum_{r=0}^{\infty} \frac{x^r 3^r}{r!}\right) dx$.Substituting the series with $e^{3x}$,the integral becomes $\int e^{3x} dx$.Using the integration formula $\int e^{ax} dx = \frac{e^{ax}}{a} + c$,we get $\int e^{3x} dx = \frac{e^{3x}}{3} + c$.

$\int \left(\sum_{r=0}^{\infty} \frac{x^r 3^r}{r!}\right) dx =$

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