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The Definite Integral
If there exists an irregularly shaped curve, y = f(x) then there is no formula to find out the area under the curve between two points x = a and x = b on the horizontal axis. If this interval [a, b] is broken into 'n' subintervals [x1, x2], [x2, x3] ... [xn-1, xn] and rectangles are constructed in such a way that the height of each rectangle is equal to the smallest value of the function in the subinterval then the sum of the areas of the rectangles i.e. will approximate the actual area under the curve, where , is the difference between any two consecutive values of x. The smaller the value of the more rectangles can be created and the closer is the sum of the areas of the rectangles so formed, i.e. , to the actual area under the curve. If the number of subintervals increases, that is 'n' approaches infinity, each subinterval becomes infinitesmally small and the area under the curve can be expressed as
Figure 1
Figure 2
The area under the graph of a continuous function between two points on the horizontal axis, x = a and
x = b, can be best described by the definite integral of f(x) over the interval x = a to x = b. This is mathematically expressed as
a and b on the left hand side of the above expression are called the upper and lower limits of the integration. Unlike the indefinite integral which represents a family of functions as it includes an arbitrary constant, the definite integral is a real number which can be found out by using the =
what is the differeance in between determinate and matrix .
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Before taking up division of polynomials, let us acquaint ourselves with some basics. Suppose we are asked to divide 16 by 2. We know that on dividing 16 by
f(x)+f(x+1/2) =1 f(x)=1-f(x+1/2) 0∫2f(x)dx=0∫21-f(x+1/2)dx 0∫2f(x)dx=2-0∫2f(x+1/2)dx take (x+1/2)=v dx=dv 0∫2f(v)dv=2-0∫2f(v)dv 2(0∫2f(v)dv)=2 0∫2f(v)dv=1 0∫2f(x)dx=1
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