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Finding area of polygonal regions can be accomplished using area formulas for rectangles and triangles. Finding area bounded by a curve is more challenging. Consider that the area inside a circle is the same as the area of an inscribed n-gon where n is infinitely large.

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Summation notation simplifies representation Area under any curve can be found by summing infinitely many rectangles fitting under the curve.

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Riemann sum is the sum of the product of all function values at an arbitrary point in an interval times the length of the interval. Intervals may be of different lengths, the point of evaluation could be any point in the interval. To find an area, we must find the sum of infinitely many rectangles, each getting infinitely small.

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Let f be a function that is defined on the closed interval [a,b]. If exists, we say f is integrable on [a,b]. Moreover, called the definite integral (or Riemann integral) of f from a to be, is then given as that limit.

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The definite integral from a to b of f(x) gives the signed area of the region trapped between the curve, f(x), and the x-axis on that interval. The lower limit of integration is a and the upper limit of integration is b. If f is bounded on [a,b] and continuous except at a finite number of points, then f is integrable on [a,b]. In particular, if f is continuous on the whole interval [a,b], it is integrable on [a,b].

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Polynomial functions Sin & cosine functions Rational functions, provided that [a,b] contains no points where the denominator is 0.

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Let f be continous on the closed interval [a,b] and let x be a (variable) point in (a,b). Then

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The rate at which the area under the curve of function, f(t), is changing at a point is equal to the value of the function at that point.

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Let f be continuous (integrable) on [a,b], and let F be any antiderivative of f on [a,b]. Then the definite integral is

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Let g be a differentiable function and suppose that F is an antiderivative of f. Then

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It is the chain rule! (from differentiation) In this case, you have an integral with a function and it’s derivative both present in the integrand. This is often referred to as “u-substitution” Let u=function and du=that function’s derivative

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Let g have a continuous derivative on [a,b], and let f be continuous on the range of g. Then where u=g(x ):

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For a definite integral, when a substitution for u is made, the upper and lower limits of integration must change. They were stated in terms of x, they must be changed to be the corresponding values, in terms of u. When this change in the upper & lower limits is made, there is no need to change the function back to be in terms of x. It is evaluated in terms of the upper & lower limits in terms of u.

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Average Value of a Function: If f is integrable on the interval [a,b], then the average value of f on [a,b] is:

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If you consider the definite integral from over [a,b] to be the area between the curve f(x) and the x-axis, f-average is the height of the rectangle that would be formed over that same interval containing precisely the same area.

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If f is continuous on [a,b], then there is a number c between a and b such that

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If f is an even function then If f is an odd function, then

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If f is continuous on a closed interval [a,b], then the definite integral must exist. However, it is not always easy or possible to find the definite integral. In these cases, we use other methods to closely approximate the definite integral.

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Left (or right or midpoint) Riemann sums (estimate the area with rectangles) Trapezoidal Rule (estimate with several trapezoids) Simpson’s Rule (estimate the area with the region contained under several parabolas)

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Approximating the definite integral of f(x) over the interval from a to b.

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