Question

integrate mvdv limit v to u​

Answer 1

$HEYA \: \\ \\ integration \: of \: mvdv \\ \: limits \: from \: \: v \: to \: \: u \: is \: \\ \\ m \times v {}^{2} \div 2 \: \: limits \: from \: v \: to \: u \: \\ \\ m \times (u {}^{2} - v {}^{2} ) \div 2 \: \: + c \: \\ where \: is \: called \: constant \: of \: integration \:$

Answer 2

Given that,

We need to integrate the given expression. Let it is y. So,

$y=\int\limits^v_u {Mvdv}$

To find,

$y=\int\limits^v_u {Mvdv}$

Solution,

It is an definite integral. M is constant here. It will come out of the integral. So,

$y=M\int\limits^v_u {vdv}\\\\y=M[\dfrac{v^2}{2}]_u^v$

Applying v as upper and u as lower limit such that,

$y=\dfrac{M}{2}(v^2-u^2)$

Hence, the integration of $\int\limits^v_u {Mvdv}$ is $\dfrac{M}{2}(v^2-u^2)$.

Learn more,

Integration

https://brainly.in/question/14596039