16.2 Notes Average Value of a Function

UM Name Period Honors Brief Calc. 16.2 Notes Average Value of a Function The average value of a function is a function f on the interval [a, b] is that y-value which specifies the height of the rectangle which has an area exactly equal to the define integral . 4th tvi K 1 rb The average value (AV) of a continuous function f over the interval [a,b] is AV = Geometric Interpretation: If we rearrange the formula for AV, then ( A V ) ( b >' ftYtngt YtiM = i b -a * —w a) b = fa f ( x ) d x - y =/ : : : :: — j/ W d x Va leu 44VM ZfvtXfS " :: i * rhe rectangle has tLe same area as :he shaded region under the curve . f Q j f t K o- / Tie LeigUt of this >, horizontal line is the average value of lie \ function . b .r M EX 1: Find the average value of f { x ) = x 3 over the interval [0,2] -a L- D EX 2: Find the average value of f { x ) = 16 ttv= - 1 f tfw - if ) -7- ^ — x 2 over the interval [-4,4] ^ 4 EX 3: Find the average value of f i x ) = e * over the interval [0,1] t s 0 do ( A - ' ^ 6 1 - he " £ r I I 0' EX 4: Suppose the current world population is 6 • 109 and the population in t years is assumed to grow exponentially at a 2% growth rate according to the law P (t) = (6 • 109 )e 02 t . What will be the average world population during the next 20 years? > , * I b* 10 e ^A = / ( 70 o 6 ,0 } V. 1- $ x EX 5: A car starting from rest accelerates at the rate of 3 meters per second per second . Find its average speed over the first 8 seconds. [ /t l o a t -V A A )^ 3 0 mi t~ ~) - Car i s nb i -cj f - y/ o ) - 0 Vtf ' 3( 0 ) f C 6» - 0 ± C \J / o ) -fc fifld C AP EXTENSION: The Second Fundamental Theorem of Calculus . If / is continuous on an open interval I containing n, then, for every x in the interval, = ~ CL\ / ( -v ) and a What this says that the constant on the bottom doesn' t matter when we take the derivative of an integral. The derivative of the integral is the original integrand (but with the variable changed). r EX 6: d a. Integral and derivatives are inverses and cancel out . r f 5 -1 ; - 1 dt .1 b. dv [ ' rsinr dr c. cos t — dv f UA dt n ., * , /d5 X •WlVW*''*'***' 9 X -| " ' X^ ~ \ AP KIDS: Work on pages 165- 166 in AP book . Also, read about Mean Value Theorem for Integrals on page 163 Name Period Honors Brief Calc. Volumes of Solids of Revolution / Method of Rings Solid of revolution is obtained when a plane region is revolved about a fixed line, called the axis of revolution. To get a solid of revolution we start out with a function, y = / ( x) , on an interval [ a, bl We then rotate this curve about a given axis to get the surface of the solid of revolution . For purposes of this discussion let’s rotate the curve about the jc-axis, although it could be any vertical or horizontal axis. Doing this for the curve above gives the following three dimensional region . I 'M How can we find the volume of this object? V = * A( x )tfx f V = * A{ y )dy \ where, A( JC) and A( y ) is the cross-sectional area of the solid. There are many ways to get the crosssectional area. Whether we will use A( x ) or A( y ) will depend upon the method and the axis of rotation used for each problem . One of the easier methods for getting the cross-sectional area is to cut the object perpendicular to the axis of rotation . Doing this, the cross section will be a solid disk if the object is solid . The disk is a cylinder whose radius r, is the height of the rectangular strip, and whose height is the thickness of the strip, Ax . Thus EX: Find the volume of the solid generated when the region is bounded by y 2 y= Jx 'f A i 0 I 3 2 AJ 'd# DB fl ' : fry x x r V y ' 1iv wmr 4 V^TTr’- ^ X £? - = yfx and the x-axis. / fell A' X ^ - v* TTf . ^ ^">0 -i j a r 0 EX: Find the volume of the solid generated when the region bounded by y 0 is rotated about the x-axis = x 2 , x = 2 and y = 2 ns = ir EX: Find the volume of the solid generated when the region bounded by y 0 is rotated around the x-axis. 'A 3x - y* - exn vfi - x)- <3 < 0 = 3x IVOS L —x 2 and y ) V 1 ' = ° puteiw* Os / ' x o : y -5 V s Tfr 2 A X 4 T O *-* • » * A ^ X 2. Si - £I ( < t J What if we rotate the object and get a ring? Then the area can be expressed as, A=n outer radius inner , radius , 7> , and creates a washer (a disk with a hole in it). The volume can be regarded as the difference in S the volumes of two disks. ( ) )\u V Wb 1C -^ J 2 0 4 Find volume of the solid 0 EXis: rotated yabout the the 0 axis Vs H J.. t 3 i Ai t' f 8 6 ^ X generated when the region bounded by y 7 (j S {' C In wf - f -\ — x ,x = 2 , /V/ f £ r ! ' / _— ^ Wjgrri ~ • J Ml) EX: Find the volume obtained when the region bounded by y x-axis. C t* = y ) J = TT / VM d < j - ir) 0 2 , and y ^ = x 2 , y = 2 x is rotafed Sbout the ' IAM U * \ jz Jf IfrUMi V " ( After J AX ' / if j o s Am 5 - 7) 5 7? /5 ) 2. I o Straight Line Motion - Revisited - Classwork When we looked at straight line motion in our derivative section , we generated a relationship between position, velocity, and acceleration. Given position function $( ? ) , the velocity function v ( f ) = sY /) and the acceleration We can now move in the opposite direction as well. If you have an object traveling in function a( ? ) = v '( ? ) = a straight line, its velocity can be written as v ( f ) = Ja( t ) d t + C .. Examples) Find v ( / ) given «( / ) and v (0) 1) a[ t ) = At - 6 and v (0 ) = 3 2) a( t ) = sin / + 21 and v (0) = 5 v ( t ) = -cos t + t 2 + C ( /) = 2/ 2 - 6/ + C v (0) = C = 3 v ( f ) = 2 r 2 6/ + 3 V v (0) = - l + C = 5 => C = 6 — v [ t ) = - cos t + t 2 + 6 — — We now define two terms: distance: how far you travel between time t a and time t b. displacement: the difference in distance from where you start and where you stop. Start then turn Travel 80 miles to the right 0 to Travel 60 miles to the left Finish Displacement = 20 miles ( 80 - 60) Distance = 140 miles O -v ifAvU K \ >T> /Y\ Can displacement equal distance? How? When motion is always forward So if we are given a velocity function which is always positive, we know that distance = displacement. But if the velocity function is not always positive , we must find when the time interval when the velocity is negative (moving backwards) so we can find the distance traveled in that interval and subtract it (which will be subtracting a negative number or adding a positive one) . This can be summarized as: Given a particle moving on a straight line with velocity Displacement — a and time t — b. Then Distance = JEW I * between time t = JvM d‘ =t Example 3) A particle is moving along a straight line with velocity displacement and distance traveled on the time interval [ l , 7 j 2 - I t + 10 c, (ft / sec) Find the ( Cjr \ \ J +- vI 1 •v* Displacement (easy) Distance (more difficult) -/ 2 Distance 7 Disp - f ( t - I t + 10) d t = 6 2 Distance = (5DbXI da W 1 f ( t 2 - l t + 10) d t - f ( t 2 - I t + 10) d t + f ( t 1 \$i r 2 - 7r + 1 0 d t 2 5 2 - l t +\0 ) dt =\5 ![ 2 + V) jh ^ ' Pi This can be confirmed using the calculator. Pv 15 \ V \ rv^r\ V / v° \ \ \ yYA- a \H2 1 4 3 / / / 5. / 6 * 7 8 pd $ S> D CA^ This is the graph of v = / 2 - 7 f + 10 , notice that v is negative (going backward) between t 2 and / = 5. When you take the integral This is the graph of V = t 2 - I t + 1 0 . Note that between t - 2 and / = 5, the function is now above the axis and distance is positive. So the integral is = 1 f (t 1 2 - 7 t + w d t , this will be taken into account. ytok-V ) —^ u T - —^ <: 7 SY -7 / + 10 dt= 1 2 f (t * 7 5 2 - 7 t + 10 ) d t - 1 If you are allowed to use your calculator, you use FnInt (X2- 7 X + 10,X,l ,7) SrVflJrvCfi. f (t 2 - 7 t + 10 ) d t + f (t 2 - l t + \0 ) d t 5 2 If you are allowed to use your calculator, you use FnInt (abs (X2- 7 X +10) ,X, l ,7) Example 4) Given an object moving in a straight line with sec 2 , v (0 ) = -18, / = 0 to f = 16 , find v ( / ) and the displacement and distance of the object . v ( i ) = - ? 3/ 2 + C => v (0 ) = C = - 18 3 Dist = - 3 2 0 2 v() A subway train accelerates as it leaves one station, then decelerates as it comes into the next station following chart measures the velocity v given in miles per hour . time t (sec) velocity Distance 0 0 0 5 15 . 010 10 33 . 033 15 42 . 052 20 44 . 060 25 44 . 061 30 44 . 061 35 44 . 061 40 43 . 060 45 38 . 056 . 50 24 . 043 every 5 second interval. Use the trapezoid rule for each entry . For instance , the first entry is a) find the distance the train travels 1 hr —1 5 sec 3600 sec 2 miles hr \ .010 \ b) find the total distance between subway stops . Either add up all the results or use : 1 — 5 sec 2 1 hr (0 + 2(15 ) + 2( 33) + . . . + 2( 24 ) + 2( 8 ) + 0) = . 526 miles 3600 sec 55 8 .022 , 6 0 0 . 006