Given that int {e^{ax}} dx = a^m e^{ax} + C,t | vedclass

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(D) In the expression $\int {e^{ax}} dx$,the exponent $ax$ must be dimensionless. Since the dimension of $x$ is $L^1$,the dimension of $a$ must be $L^

Vedclass · Accepted Answer

None of these

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(D) In the expression $\int {e^{ax}} dx$,the exponent $ax$ must be dimensionless. Since the dimension of $x$ is $L^1$,the dimension of $a$ must be $L^{-1}$.The integral $\int {e^{ax}} dx$ results in $\frac{1}{a} e^{ax} + C$. Comparing this with the given form $a^m e^{ax} + C$,we identify that $a^m = \frac{1}{a} = a^{-1}$,which implies $m = -1$.Since the result of the integral must have the same dimensions as the term $dx$ (which is $L^1$),the constant $C$ must also have dimensions of $L^1$.Evaluating the options:$1$. $m = -1$ is correct.$2$. Dimension of $C = L^1$ is correct.$3$. Dimensions of $a = L^{-1}$ is correct.Since all statements are correct,the incorrect statement is 'None of these'.

Given that $\int {e^{ax}} dx = a^m e^{ax} + C$,then which statement is incorrect (Dimension of $x = L^1$)?

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