- #1

Crystal037

- 167

- 7

- Homework Statement
- we know that integration is continuous summation

If we take a graph of x=y and let pts A and B on x-axis 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=(b-a)/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+(N-1)Δx. The width of each bar is Δx, then total area of all bars is

I=aΔx+(a+Δx)Δx+(a+2Δx)Δx+.........+[a+(N-1)Δx]Δx

={a+(a+Δx)+(a+2Δx)+.........+[a+(N-1)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+(N-1)Δx]}Δx

=NΔx/2{2a+NΔx-Δx}

=b-a/2{2a+b-a-Δx}

=b-a/2{a+b-Δx}

Thus the area under curve is

I=lim Δx tends to 0 (b-a)/2(a+b-Δx)

=(b-a)/2(a+b)

1/2(b^2-a^2)

Hence integral from a to b = 1/2(b^2-a^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+(N-1)Δx)^2

Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N-1)Δx]^2}Δx

I={Na^2 + 2aΔx +...}

I can't seem to get forward to get the required result which is 1/3(b^3-a^3)

y=x^2=(a+Δx)^2

height of nth bar=y=(a+(N-1)Δx)^2

Total area,I={a^2+(a+Δx)^2+(a+2Δx^2)+...+[a+(N-1)Δx]^2}Δx

I={Na^2 + 2aΔx +...}

I can't seem to get forward to get the required result which is 1/3(b^3-a^3)