First moment area theorem:

The left hand side (LHS) of the equation is the change in slope from A to B and right hand side (RHS) of the equation is the area of the M/EI   diagram between A and B. Thus, we have the first moment area theorem: Change in slope from A to B is equal to the area of the M/EI diagram among these points.

 [Here the slopes θ A , θB etc. are all measured in 'radians'.]

Next, let us multiply both sides of Eq. (2.17) by (s - x), we get

Dθ/dx (s - x) = - M/EI (s - x)

Or dθ (s - x) = - M/ EI (s - x) dx

Integrating both sides among A and B, we get

∫^B_A d θ(s - x) =-∫^B_A M/EI (s - x) dx

 LHS is integrated by parts and we get

=(∫^B_A d θ)×( s - x)- ∫^B_A(∫d θ)×d(s-x)

= [θ (s - x)] ^B_A-∫^B_A θ ( -dx) =[ θ(s - x) ] ^B_A-∫^B_A dy/dx .dx

=[ θ_B(s-a)- θ_A(s- 0)]+ ∫^B_A dy = θB(s-a)- θ_As + y_B - y_A

On substituting it back in Eq. (2.19), we get

θ_B(s-a) - θ_As + y_B - y_A= -∫^B_A (M/EI dx)(s-x)

On rearranging

[θ_A s +y_A] -[ θ_B(s-a)+y_B]= ∫^B_A(M/EI dx)(s-x)

If we look at the various intercepts on the vertical line SS′ we find that the intercept T1T2 between the tangents A′T1 and B′T2 is equal to the left hand side of the above equation, while the moment of the M/EI   diagram area between A and B about the SS′ line is the right hand side of the equation. If A is this area and  is the CG distance of this area from the line SS′, the right hand side is equal to A . Thus, we can write Eq. (2.20) as the intercept, i = A , which in words can be expressed as follows.

The moment of the M/EI   area between any two points, about a vertical line, is equal to the intercept made by the tangents at these points on the same vertical line.

This is the second moment area theorem. We shall next see in the following example how to calculate the slopes and deflections of a beam using these theorems.