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CALCULUS CONCEPTS: Integrals Lecture 20 (11/6/08): Integrals Riemann Sum y=f(x) find area 1. Partition Interval into ‘n’ parts Widest part = w 2. Choose a point in each sub-interval Sample point 3. Draw rectangles Total Area = sum of all the rectangles Should be the approximate area under a curve in the interval [a,b] to the x-axis Sigma Notation R = ∑ f(xi)∆Xi (i=1;3) (n=0;b) ∑2n+1 = (b+1) (i=1;n) ∑f(xi)∆Xi = ∑(i/n)(i/n) (i=1,n) i/n ∑I (1/n)(n(n+1)/2) (n+1)/(2n) Only approximations We want WIDTH to go to 0 and n to ∞ We want lots of skinny rectangles for better approximation (n∞) lim (n+1)/2n = ½ We expect the area of the curve to be ½ Lecture 21 (11/11/08): Fundamental Theorum of Calculus Definite Integrals f(x)=F’(x) (F is the antiderivative of f) (ab) ∫ f(x)dx = F(b)-F(a) This finds the area under a curve to the x-axis Lecture 22 (11/13/08): Finding the area between two curves Functions can be in terms of x or y Whichever variable they are in terms of, use those bounds If in terms of x use bounds from x=a to x=b If in terms of y use bounds from y=f(a) to y=f(b) (ab) ∫(f(x)-g(x))dx f(x) = top most curve in the interval g(x) = bottom most curve in the interval If symmetrical= can double it If not symmetrical (usually because of two curves) then the function needs to be put in terms of the other variable Bounds need to be found (setting both equations equal two each other then finding the value of x) f(x) and g(x) then usually have to be switched If doing U-Substitution for definite integrals (integrals with bounds/limits), new limits need to be calculated by plugging the x-value into the u-equation to get the value for u. Lecture 23 (11/18/08): Shell, Washer, & Disk Method All use variations of the VOLUME equation (need π) Disk Method y=x want to rotate around y-axis from 0 to 1 Rotation about y-axis = horizontal slice (needs to be in terms of y) This is a bowl shaped (if in terms of x) it would be the excess space) Volume = πr radius = x (which also equals y) Since it needs to be in terms of y (rotation around y-axis) a new function is needed: x=√y Integration (01) ∫π(√y)dy = π∫ydy = (01) ½π[y] = π/2 y=x want to rotate around x-axis from 0 to 1 Rotation of the area underneath this curve around x-axis using horizontal slices Needs to be in terms of ‘y’ since using horizontal slices x=√y But the radius = (1-√y) Integration (01) ∫π(1-√y)dy = π∫(1-2√y+y)dy = π/6 Washer Method Find the area of the region bounded by the curves y=1 and y=x as it rotates about the x-axis Slices at x look like washers Area = πr(outer) - πr(inner) dV = π((1)-(x))dx Volume = (01) π∫(1-x⁴)dx (01) π[x-(1/5)x⁵] = 4π/5 Shell Method V=(ab)∫2π(shell radius)(shell height)dx Find the area bounded between the functions y=2-x & y=x as it rotates about the y-axis In terms of x (shell being rotated around y-axis has vertical stripsin terms of x) Set equations equal to find upper limit x=1,-1 (use 01) Still need to take in consideration top and bottom equation dV=2π(x)(2-x-x)dx V=(01) 2π∫(2x-x-x)dx = (01) 2π[x-(1/3)x-¼x⁴] = 5π/6

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