FReview - Final Review Material from the previous midterms...

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Final Review Material from the previous midterms The final is cumulative but will not include prob- lems about limits, curve sketching and Riemann sums. More information about what we covered can be found in the previous midterm reviews. This review will only cover the material after the second midterm (i.e., in- tegration). Riemann sums If we want to approximate an area we can slice it into little strips each of which can be approximated by a rectangle; we then add up the individual rectan- gles. To get a better approximation we can make the slices “smaller”. This is the underlying idea of Rie- mann sums . Given a function f ( x ) and an interval [ a, b ] we start by partitioning [ a, b ] up into a partition into pieces by first choosing points x 0 = a < x 1 < x 2 < · · · < x n - 1 < x n = b. These are the bases of the rectangle and the “width” of the i th rectangle is Δ x i = x i - x i - 1 . To find the heights we choose points c i so that x i - 1 c i x i we then have that the “height” is f ( c i ). So then we have that the Riemann sum (which is an approximation of the area under the curve y = f ( x ) in the interval [ a, b ]) is n X k =1 f ( c k x k . Here “Σ” is used to indicate doing a sum. To indicate the slices getting smaller we let max { Δ x k } → 0. We are interested in functions where the limit as k P k → 0 exists, we call such functions integrable (all continuous functions are integrable) and denote the limit by lim max { Δ x k }→ 0 n X k =1 f ( c k x k = Z b a f ( x ) dx. The R sign is a stretched out “S” and indicates the idea that we are summing up little pieces. The “ x ” is a dummy variable and can be replaced by any other variable, the result will be the same. Z b a f ( x ) dx = Z b a f ( u ) du = Z b a f ( y ) dy = · · · . While our starting point is thinking of finding area, it is important to remember that the result of the in- tegration can be positive or negative, so more appro- priately it is signed area. While Riemann sums underly the principles of inte- gration, we will not test you directly on Riemann sums (including problems involving Σ). Properties of integrals Properties of integration follow from the definition of Riemann sums (as well as some geometric intuition). Z b a f ( x ) dx = area above x -axis - area below x -axis . We can find the values of some integrals by finding the area is composed of combinations of triangles, rectan- gles and circles (right now this is the only way we can handle integrals of the form r 2 - x 2 ). If our upper and lower bound match then there is no “area” and so the integral is 0, i.e., Z a a f ( x ) dx = 0 .
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