Definate intigral of Mvdv where upper limit is v and lower limit is u
To find the definite integral of Mvdv with the upper limit v and the lower limit u, you can follow these steps:
Step 1: Begin by writing down the integral expression:
∫(Mvdv)
Step 2: Apply the fundamental theorem of calculus, which states that the definite integral of a function can be found by evaluating its antiderivative at the upper and lower limits of integration.
Step 3: Determine the antiderivative of Mv with respect to v. This involves integrating M with respect to v.
Step 4: Once you have obtained the antiderivative, substitute the upper limit v into the antiderivative expression and subtract the result when the lower limit u is substituted.
Step 5: The final result will be the definite integral of Mvdv from u to v.
Note: The actual steps for finding the antiderivative of Mv with respect to v depend on the form of the function M(v). Make sure to substitute the specific form of M(v) into the integral expression during Step 3.
To find the definite integral of Mvdv with upper limit v and lower limit u, we can follow these steps:
Step 1: Determine the antiderivative of M with respect to v.
- This step involves finding the function F(v) whose derivative is equal to M. This function will be the antiderivative of M with respect to v.
- Let's say that F(v) is the antiderivative of M, so we have F'(v) = M.
Step 2: Evaluate F(v) at the upper and lower limits.
- We substitute the upper limit v into F(v) and subtract the value of F(v) at the lower limit u.
- Therefore, the definite integral of Mvdv, from u to v, is given by:
∫[u to v] Mvdv = F(v) - F(u).
So, we can find the definite integral by calculating F(v) - F(u), where F(v) is the antiderivative of M with respect to v.
Till I don't get the answer
INT M v dv over limits u,v
1/2 M v^2 over limits= 1/2 M (v^2-u^2)
This is pretty basic stuff....