If I_{m, n} = int e^{mx} cdot x^n , dx,then I | vedclass

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(C) Given $I_{m, n} = \int e^{mx} \cdot x^n \, dx$. Using integration by parts,let $u = x^n$ and $dv = e^{mx} \, dx$. Then $du = nx^{n-1} \, dx$ and $

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$\frac{x^n \cdot e^{mx}}{m} + c$

Vedclass · Answer

(C) Given $I_{m, n} = \int e^{mx} \cdot x^n \, dx$. Using integration by parts,let $u = x^n$ and $dv = e^{mx} \, dx$. Then $du = nx^{n-1} \, dx$ and $v = \frac{e^{mx}}{m}$. $I_{m, n} = \int u \, dv = uv - \int v \, du$. $I_{m, n} = x^n \left( \frac{e^{mx}}{m} \right) - \int \left( \frac{e^{mx}}{m} \right) (nx^{n-1}) \, dx$. $I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} \int e^{mx} x^{n-1} \, dx$. Since $I_{m, n-1} = \int e^{mx} x^{n-1} \, dx$,we have: $I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} I_{m, n-1} + c$. Rearranging the terms,we get: $I_{m, n} + \frac{n}{m} I_{m, n-1} = \frac{x^n e^{mx}}{m} + c$. Hence,option $C$ is correct.

If $I_{m, n} = \int e^{mx} \cdot x^n \, dx$,then $I_{m, n} + \frac{n}{m} I_{m, n-1} =$

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