Magnetar

ok thanks i understood ! I always thought why isnt the integral calculated in the first way, but now I understood

I would like to rephrase. By saying that i took the area of the rectangle as (f(x)+Δy/2)Δx in the first graph , i meant to say that i took the height of an individual rectangle as [f(x)+f(x+Δx)]/2 ( not as f(x), neither as f(x+Δx) )to get the mean height, or when Δx is big, a better approximation of the area as Σ [{f(x)+f(x+Δx)}/2]*Δx Again rephrasing, why dont we take the area under a curve as ∫ {f(x)+f(x+dx)}/2*dx? because when Δx is big, the expression for area under a curve as Σ [{f(x)+f(x+Δx)}/2]*Δx would give a more precise result. Will it give the same as Δx approaches zero?

I have attached a picture below, and the thought of this idea is confusing me too much. -> In the first graph, I have taken the area of a single rectangle (say, first rectangle) as M * Δx , where M represents the arithmetic mean of f(x) and f(x) + Δy ( Δy is f(x+Δx) - f(x) ) which i thought would givea better approximation of the area as opposed to directly taking area as f(x)Δx So the question,(please see the pic first) why dont we take the area under a curve as ∫ (f(x) + dy/2)dx? because when Δx is big, the expression of area under a curve as ∫ (f(x) + Δy/2)Δx would give a more precise result. will it give the same as Δx approaches 0? Please tell where am i going wrong. p.s. - I am a newbie at calculus, so bear with me if my question is stupid