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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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 $x = L^1$)?

Similar Questions

If $\varepsilon_0$ and $\mu_0$ represent the permittivity and permeability of free space respectively, then the dimension of the product $\varepsilon_0 \mu_0$ is,

The dimension of $\sqrt{\frac{\mu_0}{\varepsilon_0}}$ is the same as:

Write the $CGS$ and $MKS$ units of the universal gravitational constant $G$. Also,write the dimensional formula of $G$.

Which of the following physical quantities has a unit but no dimensions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module