# How is the definite integral used to solve area problem?An answer with a graph will help immensely

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I'm not sure I can add a figure, but we'll see what we can do explaining the concept.

An integral is in a sense a summing over of a function, and a definite integral tells us the points to start and end with. For example, consider a simplke function:

f(x) = 2x

And say we want to integrate from x=1 to x=5. What we are doing in such an integration is finding the area under the curve of the graph. In this case, f(x)=2x is a straight line, so we would be finding the area under the line that runs from (1,2) to (5,10).

f(x)=2x after integration is x^2 (x squared), which we evaluate as 25-1=24. What we have done is added up an infinite number of infinitely narrow vertical slices of the graph of this line. Each slice is of width dx and height f(x) - making them infinitely thin rectangles.

Think of it this way: take a thick book and look at its width. Each page is the thin dx-width slice, yet these collectively make up a substantial width. The height of the book is then given by a function f(x), and so the total area of a side of the book is the sum of the height of the book times the width of each thin page.

Again, just think of integration as dividing the area under a function into a number of rectangular slices. The more slices you can make, the more precise the sum of all these rectangles' areas will match the total area under that function.

Check out some of the links below for sample graphs.

Definite integral deals with the area under a curve between two ordinates. An explanation cum example is below:

Let f(x) be the contiuous curve between x=a and x= b be the two ordinates.

Let the interval between a and b be partitioned into n partitions. (a,x), (x1,x2),(x2,x3), .....(xn-1,xn=b).

Let each of the partitons, have the length, d1,d2, d3,.....dr,....dn, where Dr is the length of the rth partition.

Now, the let f(Mr) and f(mr) be the maximum and minimum value of f(x) in the rth partition. r=1,2,3....n.

So, now the actctual area Ar, of the rth patition is between f(Mr)*Dr and f(mr)d*Dr. Therefore,

f(Mr)*Dr <Ar<f(mr)*Dr. Therefore,

Sum of (Mr).Dr < Sum of Ar < Sum of f(mr)*Dr , r=1,2,3....n.

As r ---> infinity, Dr becomes small and Limit Sum f(Mr)*Dr and Limit f(mr)*Dr exists and narrows and become equal.

So, Sum of Ar = area under the curve is Sum of f(x)dx and is called Integral f(x) dx between x=a and x=b.

Example: Find the area under the curve y = f(x) = x^2 between: x = 10 and x = 15.

Partition method: We make only one partetion (10,15)

The maximum and minimum values of f(x) are: f(115) = 225 and f(10) = 100. Dr =15-10=5 So the area A is given by the relation: f(10)*5<A<f(15)*5 or 100*5<A<225*5 or 500<A<1125.

Two equal partitions:(10,12.5) and (12.5,5). Maximumum and minimum values are f(12.5) and f(10) in 1st partition and f(15) and f(12,5) in the 2nd partition.Dr =2.5 each. So, f(10)*2.5+f(12.5)2.5 <A<f(12.5)*2.5+f(115*2.5)

250+390.625<A<390.625+562.5 or 640.25 < A < 953.125

Similarly for 5 (equal) partitions, (10,11), (11,12) ,..(14,15), Dr=1 and F(11),f(12),f(13),f(14) and f(15) are maximum and f(10),f(111), f(13) and(f(14) are minimum. Therefore, 10^2*11^2*1+12^2*1+13^2*1+14^2*1 < A < 11^2*1+12^2*1+13^2*1+14^2*1+15^2*1

or 730 < A < 855. We can see that the area lies in a narrower gap of maximum and minimum as the number of partition inceases.

So in limit, when partitions are ifinitely many both upper and lower limits come close and become equal. Then,

A = Integration x^2.dx between x=10 and x=15

= (x^3/3) at x=15 minus (x^3/3) at x = 10 or

=15^3/3-10^3/3

=1125-333.33 =791.67