Question

Integrate $\int_u^v {Mvdv}$.
A) $\dfrac{M}{2}\left( {{v^2} - {u^2}} \right)$
B) $\dfrac{M}{2}\left( {{u^2} - {v^2}} \right)$
C) $\dfrac{M}{2}\left( {v - u} \right)$
D) $\dfrac{M}{2}\left( {u - v} \right)$

Answer 1

Here, we will use the concept of integration and constant multiple rule. Integration is the opposite of differentiation. It involves using a derivative of some function and ‘integrating’ it to find the original function. We know that the constant multiple rule allows you to take the constant multiples out of the integral. Here we will apply the constant multiple rule to take the constant out of the integral and integrate the rest part. Then we will apply the limits and simplify the expression to find the answer.
Formula Used: The integral of a constant multiple can be written as $\int {pxdx} = p\int {xdx}$ where $p$ is a constant multiple. The integral of a function of the form ${x^n}$ is given by $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$ where $n$ is a real number.

Complete step by step solution:
The expression $\int_u^v {Mvdv}$ is a definite integral with upper limit $v$ and lower limit $u$.
The multiple $M$ in the expression to be integrated is a constant multiple. We know that constant multiples can be taken out of the integral to simplify the calculations.
Thus, we first take out the constant multiple $M$ and get
$\int_u^v {Mvdv} = M\int_u^v {vdv}$
Now, we know that the integral of a function of the form ${x^n}$ is given by $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$ where $n$ is a real number.
We can observe that the expression $\int_u^v {vdv}$ can be written as $\int_u^v {{v^1}dv}$.
Therefore, using the formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$, we get
$\begin{array}{l} \Rightarrow \int_u^v {Mvdv} = M\left. {\left( {\dfrac{{{v^{1 + 1}}}}{{1 + 1}}} \right)} \right|_u^v\\\ \Rightarrow \int_u^v {Mvdv} = M\left. {\left( {\dfrac{{{v^2}}}{2}} \right)} \right|_u^v\end{array}$
Here, we have not used the constant $C$ because this is a definite integral.
Substituting the limits in the expression, we get
$\Rightarrow \int_u^v {Mvdv} = M\left( {\dfrac{{{v^2}}}{2} - \dfrac{{{u^2}}}{2}} \right)$
We can take out $\dfrac{1}{2}$ common from the parentheses.
Therefore, we get
$\Rightarrow \int_u^v {Mvdv} = \dfrac{M}{2}\left( {{v^2} - {u^2}} \right)$
$\therefore$ The value of the integral $\int_u^v {Mvdv}$ is $\dfrac{M}{2}\left( {{v^2} - {u^2}} \right)$.

Hence, the correct option is option (a).

Note:
A definite integral is an integral which lies within an interval. It has a lower limit and upper limit. We must know how to substitute limits to solve a definite integral. A common mistake we might make in this question is to use $u$ as the upper limit and $v$ as the lower limit. If we do this, we will get the answer $\int_u^v {Mvdv} = \dfrac{M}{2}\left( {{u^2} - {v^2}} \right)$ which is incorrect. It is also important for us to remember that we have to first integrate the integral then put the limits. If we first put the limits then integrate the integral, then we will get incorrect answers.