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Calculus 1 Chapter 4 notes

by: Jack Magann

Calculus 1 Chapter 4 notes MTH 161

Marketplace > University of Miami > Mathematics (M) > MTH 161 > Calculus 1 Chapter 4 notes
Jack Magann
GPA 3.865
161 Calculus 1
Dr. Beverly

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These notes cover chapters 4.1, 4.2, 4.3, 4.4, and 4.5 Topics include: Basics of integration, Definition of integration, properties of integrals, fundamental theorem of calc, and Mean value theo...
161 Calculus 1
Dr. Beverly
Class Notes
Calculus 1 University of Miami
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This 12 page Class Notes was uploaded by Jack Magann on Sunday February 15, 2015. The Class Notes belongs to MTH 161 at University of Miami taught by Dr. Beverly in Fall2014. Since its upload, it has received 82 views. For similar materials see 161 Calculus 1 in Mathematics (M) at University of Miami.


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Date Created: 02/15/15
Chapter 41 Start by looking at an area problem M We are looking for the area between the curve 7 and the Xaxis as shown in the graph Here we see area S is bound by the curve and j the lines xa and xb o Notice that this is very different from such regular shapes as a triangle or rectangle One way to approach this problem is to use rectangles to approximate the area under a curve Start with a more simple curve like y x2 EX Use rectangles to estimate the area under the parabola y x2 from 01 Here is the graph y x2 m If we draw a triangle by making the hypotenuse the straight line from the origin to 11 we can see that the area will be less then 5 This is because the area under the curve takes up less space then the area under the triangle Now break it up into four equally spaced sections and make rectangles that are anchored on the right A 77 II I l X 39 L Now by taking the sum of the area of these rectangles we will get an approximation of the area under the curve S 1 1 2 1 1 2 1 3 2 1 2 5 2 2 z a 2 2 z 1 4687 The proper notation for this is R4 4687 As shown in the graph this is an over estimation so we can also anchor the rectangles on the left This then gives us the underestimation L4 21875 The true area lies between these two values If we break the section of the graph into even more rectangles an even closer estimation can be found as shown ll As infinitely more rectangles are used the estimation gets closer and limits can be used to describe this behavior A limnoo R limnoo L Where n is the A similar method can be used where the rectangles are anchored in the middle of the intervals The rectangles may not always all be over or under the curve depending on the increasing and decreasing intervals 1 A i I f quot1 I Yet this graph also illustrates how Rn f x1Ax f x2Ax f x3Ax f ngx This can be written as 12m f xiMac where 2 means the sum of what comes after 11 is the value the interval ends on m is the value the interval starts on Chapter 42 Many graphs will have sections below the XaXis These sections then have negative areas and so the areas of the rectangles we use to estimate the area would also be negative The area below the Xaxis is then subtracted from the area above the XaXis as shown in the below picture y A I l a This is called the net area Definition of Definite Integral If f is a function defined on 6119 the definite integral of f from a to b is the number If fxdx mm gt0 211 folMac Fill the area under the graph with an infinite number of rectangles Ax is the width of the rectangle Doesn t matter where rectangles are anchored since they are so small Can also be written as If fxdx limnsw 221 folMac n is number of rectangles Ax is length of interval over number of rectangles In this section this formula will only be applied to graphs that have areas under them that are of shapes we know how to find the area of This includes rectangles triangles and circles ex 1 Evaluate the integral by interpreting it in terms of areas f x 1dx If we draw the graph of X1 out we see that the area under the graph on 03 makes two triangles One under the Xaxis with a negative area One above with a positive area The problem then comes down to just adding the areas of these two triangles A1 11001 lt2lt2 2 A2 lt1lt 1gt f x 1dx A1 A2 15 This answer is the net area under the curve eX 2 f 01 V 1 x2 dx This equation makes a semicircle with a radius of 1 A However the interval means we are only looking at a quarter m circle as seen in the drawing of it 1 l The area of a circle is A m 2 so 1V1 x2 dx l7T12 E This is the total net area 39 o 4 4 ex 3 L65le dx If we draw the graph there are two triangles under the curve One from 06 A1 and another from 50 A2 Both are above the Xaxis and so they are added together A 66 18 Aztsxs so ff5x dx 18 305 eX 4 f044 V16 x2dx Remembering the second example the equation makes a semi circle but this time it is moved up by 4 units This makes the area into two distinct shapes whose areas can be added to find the net area The first is the semicircles that has a radius of 4 but the interval makes it a quarter circle 1 The area 1s A Zn rz The second is a 4 by 4 square that is created under the quarter circle by moving it up 4 units The area is A bh L040 V16 x2dx in42 44 16 4n Chapter 43 L12 f x dx Is asking to find the area between the xaxis and the curve If curve makes recognizable shapes the area can be found by adding those respective areas as seen in the previous chapter If the curve makes an irregular shape we only know how to estimate the area using the rectangle method discussed in chapter 1 There is a way to find the area underneath an irregular curve exactly If f is a continuous function on the interval a b then f f x dx F b F a where Fx is the antiderivative of fx ex I x2 dx Previously predicted to be WP 50 Fltxgt x3c folxzdx m3 cm3 This is the exact answer Notice how the C was taken out since it would just cancel itself out A definite integral is one where the interval is defined meaning a and b are given values it will produce an exact answer as above An indefinite integral does not have a defined interval It will produce another function where the C is included ex 1 L12 x3 dx This is read as evaluate the integral of x3 on the interval 21 Find the antiderivative f x x3 F x 2 ix C 1 3 l 4l 4l 15 f2x dye 41 4 2 4 4 4 Rules for integration f Cfx dx cf fo dx mm gxdx I fxdx I 900 dx f k dx kx c f xquot dx antiderivative power rule f sec2xdx tanx C f csc2xdx cotx C fsinx dx cosx c fcosx dx sinx c fsecx tanx dx secx C fcscx cotx dx cscx C ex 2 f10x4 2 sec2xdx fx 10x4 2 sec2x Fx 2 2x5 2 tanx C This uses the sum rule and the trig identity If the two functions are broken up then their C values are combined into the above function 10364 2 sec2xdx 2x5 2 tanx C 2 2 ex 3 fig dt for this it is best to try and simplify 9 2t2t2t 1 9 212 15 1 9 t2 dt2j t 2 t2 t 2 dtj2f 1t2dt 1 1 1 1 2 E fx2tz t 2 Fx2t t2t391 9 2 E 1 2 292 112 E1t 2dt29923 2 1T Integrals can then be applied to familiar concepts like displacement velocity and acceleration Since vt s t f vtdt 3t2 3t1 This is the net area under the velocity curve It can also be referred to as the displacement of the object This is different for the total distance since this means the negative areas need to be added rather then subtracted So Total distance f 12 vt dt Ex4 Forvt t2 t 6 a find the displacement of the object on the interval 14 1 3 1 2 St Vt Et Et 6t 4 4 64 16 1 1 9 Jvtdt 2 t2 t 6dt 24 6 1 1 3 2 3 2 This means that the object ends up 3 units behind from where it started b Find the total distance the object travels on the interval 14 First find the sections of vt that are negative since they need to be made positive which means finding the critical numbers of vt vt t 3 t 2 so the critical numbers are 3 and 2 but only 3 fits in the interval By plugging in numbers into the intervals created vt is negative 13 vt is positive 34 so total distance f 120 dt f34vt dt f23t2 t 6dt ft2 t 6dt E z 1017 6 The object traveled a total distance of about 1017 units Chapter 44 The fundamental Theorem of Calculus Bridges the gap between the area problem integrals and derivatives The idea is that antiderivatives are used to nd the area under a curve gx fffoc dx then we fx Differentiation and integration are inverse processes We can then say that b fa fx dx Frb Fltagt Mean Value Theorem of integrals If f is continuous on ab then there eXists a number c in ab such that 1 b fC fave fa dx so If fx dx fcb a Unlike MVT of derivatives c can be on the endpoints with integrals eX 1a Find fave offx 1 x2 on 12 This function would be a parabola but with the vertex moved up one All the area under the curve is above the XaXis so the answer will be positive It is possible to have a negative average but not here Keep in mind thatF x x x3 1 2 2 l E E l fave 2 1f 11 x dx 3 2 3 1 3 3 6 2 Iffave 2 then the total are is 22 1 6 There will be a time when the area reaches the average area or fc 1b Find the values of c 1 b fC fave fa dx 1C2 ci1 Chapter 45 Substitution Rule Substitution Rule Used to do antiderivative of a derivative taken using the chain rule Previously only found antiderivatives of derivatives found with power rule U substitution foVI xzdx Pick one equation between summation and dX to be u Equation picked needs to also have it s antiderivative present in the original equation So Let u 1 x2 2 2x and du Zxdx Now substitute in so 1 foVI x2dx fuidu This is much easier to integrate 1 3 u du 2 Eu C now plug 1 x2 back in 3 l 2 E 2 3 fuzdu u2C 251x22C This answer can be checked by taking its derivative to see if it matches the original equation 1 EX 1 f2x 1 dx u2x1 du2dx Here we have to make the equation balanced to match the u values 122 13d 1 lat 123 C 12 1E C 5f x 2 x Efuz u u2 x 2 3 Answer 2x 1 C 1 EX 2 dx fx1 4x2 dx I V1 4x2 u 1 4x2 du 8xdx Balance equation 1 1 1 1 1 l x x 2x uz u uz 8f 8142 d 8f d 82 C 1 1 answer Z 1 4x22 C These two examples were for inde nite integrals and so produced equations De nite integrals will produce number values Can also have multiple substitutions in one problem EX 3 EB 5x 2dx u3 5x du 5dx fu 2du Can t use the same interval since u is a new function so the interval has to be plugged into u u1 2 u2 7 f 27u 2du U f 1 72 2 1 1i 5 f Z u du 5 7 5 2 14 1 Answer 14 EX 4 f xVx 5 dx ux 5 dudx x5u0 x6u1 However here we have an extra X that can t be taken out like 1 l 1 1 f0 xu2du or xfo uzdu Do Not Do This rather solve for X in terms of u So x u 5 5 f xVx 5 dx 10101 fol du U Eu 1 E l 2 10 56 f01135uzdug 0E Practice Problems Evaluate the following de nite integrals 21139 lf3n3coszxsmxdx ucosx du 51nxdx Uu3 21139 1 7 3C052xsinxdx j3u3du 1cos3x 11 317quot 8 8 1 2 f018y2 y 1T5 32y 2dy u 8y2 y 1 du 32y 2 U 31123 f018y2 y 1quotg 32y 2dy 2 f01 u du 381 1 1 31 2 1 x 1 1 3f0mdx f0xx12 u x 1 du dx x u 1 fol x dx 2 KW 1u du ff 11 du U 211 x1 1 f12u5 11 du 3905 4 9 xE 4 fldeZJA 1 l 19x 2 1dx Fx 2 18x2 x 149x 1 dx 36 3 17 15 All these notes are from a Calculus 1 class run by Dr on Beverly at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print


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