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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 =
#question Show that the enveloping cylinder of the conicoid ax 2 + by 2 + cz 2 = 1 with generators perpendicular to the z-axis meets the plane z = 0 in parabolas
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A subset of the real line is called as an interval. Intervals are very significant in computing inequalities or in searching domains etc. If there are two numbers a, b € R such tha
explanation
Lines EF and GH are graphed on this coordinate plane. Which point is the intersection of lines EF and GH?
Any point on parabola, (k 2 ,k) Perpendicular distance formula: D=(k-k 2 -1)/2 1/2 Differentiating and putting =0 1-2k=0 k=1/2 Therefore the point is (1/4, 1/2) D=3/(32 1/2
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hi i would like to ask you what is the answer for [-9]=[=5] grade 7
What is Dividing Fractions? If you want to divide two fractions, you invert the second fraction (that is, turn it upside-down) and change the division sign to a multiplication
Evaluate the following integral. ∫ (x+2 / 3√(x-3)) (dx) Solution Occasionally while faced with an integral that consists of a root we can make use of the following subs
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