Calculus 1120, Review for Prelim 1 Dan Barbasch September 25, 2012 - PowerPoint PPT Presentation

Calculus 1120, Review for Prelim 1 Dan Barbasch September 25, 2012 Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 1 / 8

Fundamental Theorem of Calculus Practice Exercises chapter 5: any 43-72, 73-112, 121-128 Techniques of Integration ◮ Substitution ◮ Integration by parts ◮ Trigonometric Integrals Practice Exercises chapter 5: Section 5.6: any 43-112 Practice Exercises chapter 8: any 1-8, 37-44 Applications ◮ Areas of regions between curves ◮ Volumes by slicing, disks/washers, shells ◮ Arclength ◮ Surface area ◮ Distance traveled, displacement, average value of a function Section 5.4: 73-75 Section 5.5: 77, 78 Section 5.6: any 93 -110 Practice Exercises chapter 6: any 3, 4, 7-10, 13, 14, 19, 20, 22, 23 Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 2 / 8

Fundamental Theorem of Calculus � x F ( x ) = f ( t ) dt a is an antiderivative of y = f ( x ); the unique one which satisfies F ( a ) = 0 . The function f ( x ) must be continuous throughout an interval [ a , b ] containing x . � b f ( x ) dx = F ( b ) − F ( a ) . a A general formula (which you should not memorize but understand) �� h ( x ) � d f ( t ) dt = f ( h ( x )) · h ′ ( x ) − f ( g ( x )) · g ′ ( x ) . dx g ( x ) � x 3 � 2 + sin 2 t dt . Compute F ′ ( x ) for F ( x ) = x 2 Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 3 / 8

Techniques of Integration Substitution. If you make a substitution u = h ( x ) you must be able to solve uniquely x = g ( u ), and then you substitute x = g ( u ) in the function, and dx = g ′ ( u ) du . Usually one takes short cuts. For definite integrals you must either carry the x and substitute at the end, or else change the limits of integration. � e x sec 2 ( e x − 7) dx . Compute � � Integration by Parts. udv = uv − vdu . Trigonometric Integrals. You need to know which trigonometric formulas to use. � Compute sin 3 θ cos 2 θ d θ. Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 4 / 8

Area � b A = | f ( x ) − g ( x ) | dx . a You need to divide up the interval [ a , b ] into pieces where f or g is the larger of the two functions. Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 5 / 8

Volumes � b Slices. V = A ( x ) dx . a You need to choose an axis, say x , and then compute the area A ( x ) of the cross section at x . Disks/Washers. A special case of the method of slices; the cross sections are washers A ( x ) = π [ r 2 ( x ) 2 − r 1 ( x ) 2 ] . � b π [ r 2 ( x ) 2 − r 1 ( x ) 2 ] dx . V = a r 2 , r 1 need to be positive. The integrals may be in y depending on the axis of rotation. � b Shells. V = 2 π r ( x ) h ( x ) dx . a h , r need to be positive. Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 6 / 8

Volumes 1 Sketch the region. 2 Choose a variable for integration. 3 Draw a slice perpendicular to the variable, and rotate to decide washers/shells. 4 Write the general formula with the endpoints for the variable. 5 Compute r 1 , r 2 or r , h in terms of the chosen variable. Label them. 6 Compute the integral. 6 (page 414). Find the volume of the solid with base the region bounded by y 2 = 4 x , and x = 1 with coross section perpendicular to the x − axis equilateral triangles with one edge in the xy − plane. 7 (page 414). Find the volume of the solid generated by revolving the region bounded by the x − axis, y = 3 x 4 and the lines x = 1 and x = − 1 revolved by the (a) x − axis (b) y − axis (c) x = 1 (d) y = 3 Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 6 / 8

Arclength/Surface Area � b � b L = ds , A = 2 π rds . a a � � 1 + f ′ ( x ) 2 dx 1 + g ′ ( y ) 2 dy ds = ds = � x ′ ( t ) 2 + y ′ ( t ) 2 dt ds = The first two are special cases of the third. Using parametric forms for curves are very convenient. a 2 + y 2 Example: length of the ellipse x 2 b 2 = 1 . Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 7 / 8

Other Applications Motion: On a line: ds dv dt = v ( t ) , dt = a ( t ) , F = ma . � b Displacement is s ( b ) − s ( a ) = v ( t ) dt . a � b Distance travelled is | v ( t ) | dt . a Along a curve: � b � x ′ ( t ) 2 + y ′ ( t ) 2 dt . If the motion does not go Distance travelled is a back on itself this is the same as arclength. Average Value of a Function: � b 1 f ( x ) dx . b − a a Dan Barbasch Calculus 1120, Review for Prelim 1 September 25, 2012 8 / 8