 #1
Crystal037
 161
 5
 Homework Statement:

we know that integration is continuous summation
If we take a graph of x=y and let pts A and B on xaxis represent x=a and x=b
and try to find the area under the curve between a and b
We divide the length AB in N equal intervals.
Then the length of each interval is Δx=(ba)/N.
The height of the first shaded bar is y=x=a, of second bar is y=x=a+Δx, that of third bar is y=x=a+2Δx. The height of the Nth bar is y=x=a+(N1)Δx. The width of each bar is Δx, then total area of all bars is
I=aΔx+(a+Δx)Δx+(a+2Δx)Δx+.........+[a+(N1)Δx]Δx
={a+(a+Δx)+(a+2Δx)+.........+[a+(N1)Δx]}Δx ....(1)
=∑xi(Δx) where 1<=i<N
As Δx tends to 0 the total area of the bars becomes the area under the curve.
thus the required area is I=lim Δx tends to 0 ∑xi(Δx) where 1<=i<N
Now the terms making the series in curly bracket in eq(1) are in arithmetic progression
hence this series may be summed up using formulae S=n/2(a+l).
Thus, I=N/2{a+[a+(N1)Δx]}Δx
=NΔx/2{2a+NΔxΔx}
=ba/2{2a+baΔx}
=ba/2{a+bΔx}
Thus the area under curve is
I=lim Δx tends to 0 (ba)/2(a+bΔx)
=(ba)/2(a+b)
1/2(b^2a^2)
Hence integral from a to b = 1/2(b^2a^2)
Similarly using summation find the integral of y=x^2 from a to b
 Relevant Equations:
 y=x^2
Here, width of first bar, y=x^2=a^2
y=x^2=(a+Δx)^2
height of nth bar=y=(a+(N1)Δx)^2
Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N1)Δx]^2}Δx
I={Na^2 + 2aΔx +...}
I can't seem to get forward to get the required result which is 1/3(b^3a^3)
y=x^2=(a+Δx)^2
height of nth bar=y=(a+(N1)Δx)^2
Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N1)Δx]^2}Δx
I={Na^2 + 2aΔx +...}
I can't seem to get forward to get the required result which is 1/3(b^3a^3)