Finding area given definite integral

[grace77]

Question :
https://www.physicsforums.com/attachments/71328

My question is how did the 2a and 2b come from??
Equations:
Area of trapezoid =(a+b/2)(h)
Attempt:
I know that the area of a trapezoid is (a+b/2)(h)
However why is there now a 2a and 2b in its place? Could it be related to the 2s function??

 

[FilupSmith]

Hope this helps

∫^b_a 2s ds = [s²]^b_a
this then becomes (b²-a²)

its just integral laws

 

[SteamKing]

Beats me.

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[FilupSmith]

I'm pretty sure the this is what you're looking for:
$$ \int ^{b}_{a}2xdx=\left[ x^{2}\right] ^{b}_{a}=b^{2}-a^{2} $$~| FilupSmith |~

 

[HallsofIvy]

I cannot see the attachment. If your trapezoid has parallel sides of length a and b and height h, its area is NOT (a+ b/2)h. It is (a+ b)h/2 or ((a+ b)/2)h.