Math 1120 Class 2 Dan Barbasch Aug. 28, 2012 Dan Barbasch () Math - PowerPoint PPT Presentation

Math 1120 Class 2 Dan Barbasch Aug. 28, 2012 Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 1 / 19

Fundamental Theorem of Calculus The Fundamental Theorem of Calculus (FTC) says the following: Theorem: Let y = f ( x ) be continuous on the interval [ a , b ] . Then f ( x ) has an antiderivative F ( x ), and � b f ( x ) dx = F ( b ) − F ( a ) . a What does this mean?! The LHS(left hand side) is the definite integral. The RHS(right hand side) involves evaluating the antiderivative F of f . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 2 / 19

Fundamental Theorem of Calculus � b The definition of the definite integral (LHS) a f ( x ) dx is very deep. You approximate y = f ( x ) by step functions over the interval, and take a limit. The definition is in section 5.3. You may need 5.2 for the notation and other technicalities. The definite integral is what arises in applications. You can see them in section 5.1; area, distance travelled, displacement, average of a function. When you compute a definite integral by finding an antiderivative and evaluating, you are invoking FTC (mostly without explicit mention). How many antiderivatives can a function y = f ( x ) have? Question: The mean value theorem implies that if F ′ ( x ) = G ′ ( x ) over some interval, then there is a constant C such that F ( x ) = G ( x ) + C . Question: How does this fit with FTC? Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 2 / 19

Fundamental Theorem of Calculus Example: Compute the area under the graph of y = 1 − x 2 and above the x − axis. The notion of the area of a region is by definition a definite integral; the limit of Riemann sums. � b A = | f ( x ) | dx a is the area bounded by x = a , x = b y = f ( x ) and the x − axis. Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 2 / 19

Fundamental Theorem of Calculus Solution to Exercise: The graph of y = 1 − x 2 in the region − 1 ≤ x ≤ 1 is FTC reduces the problem of computing the area to finding the antiderivative of f ( x ) = 1 − x 2 . We know that F ( x ) = x − x 3 / 3 satisfies F ′ ( x ) = 1 − x 2 . So by FTC, � 1 1 − x 2 � x − x 3 / 3 | 1 � � � A = dx = − 1 = − 1 = (1 − 1 / 3) − ( − 1 + 1 / 3) = 4 / 3 . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 2 / 19

FTC The Fundamental Theorem of Calculus (FTC) also says the following: Theorem: Assume that y = f ( x ) is continuous in the interval [ a , b ] . Then � x a f ( t ) dt (as a limit of Riemann sums) exists for any a ≤ x ≤ b . The � x function F ( x ) := a f ( t ) dt therefore exists for any a ≤ x ≤ b . Furthermore F ( x ) is differentiable in the interval, and F ′ ( x ) = f ( x ) . I say appropriate because the way one proves the theorem is by applying properties of the definite integral proved from the definition in terms of Riemann sums and their limit. Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 3 / 19

Differentiating Integrals Exercises: Compute the following. �� x � 1. d � 1 + t 2 dt dx 0 �� x 2 � 2. d � 1 + y 2 dy dx 0 �� tan x � 3. G ( x ) := d csc 2 t dt dx sin x Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 4 / 19

Differentiating Integrals � b a f ( t ) dt = F ( b ) − F ( a ) where F ′ ( x ) = csc 2 ( x ) . So The Key: � tan x csc 2 t dt = F (tan x ) − F (sin x ) sin x The chain rule implies G ( x ) = F ′ (tan x ) (tan x ) ′ − F ′ (sin x ) (sin x ) ′ . Exercise: Finish the calculation. Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 4 / 19

Answer 1 (tan x ) ′ = (sin x ) ′ = cos x , cos 2 x , So 1 G ( x ) = csc 2 (tan x ) · cos 2 x − csc 2 (sin x ) · cos x Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 5 / 19

A Few Other Notions in 5.4 Area between the graph of a function y = f ( x ) and the x -axis Signed Area. Exercise: Find the area of the region between the graph of y = ( x − 2) 2 , the x -axis and the lines x = 1 and x = 3 . Exercise: Find the area of the region between the graph of ln x x , the x -axis, and the lines x = 1 / e and x = e . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 6 / 19

Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 7 / 19

Area, Marginal Cost, Distance, Displacement � b If f ( x ) ≥ 0 for a ≤ x ≤ b , then the area of the region is A = a f ( x ) dx . � b If f ( x ) �≥ 0 throughout the interval, then the integral a f ( x ) dx represents the signed area of the region. � b The area is A = a | f ( x ) | dx . Marginal cost is the derivative of total cost C ( x ) with respect to quantity x . Similar for Marginal Revenue . Velocity is the derivative of position with respect to time v ( t ) = s ′ ( t ) . So position (displacement) is the integral of velocity � b s ( b ) − s ( a ) = a v ( t ) dt . � b Total distance traveled is D = a | v ( t ) | dt . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 8 / 19

(Hints to) Answers The function y = ( x − 2) 2 is positive in the interval [1 , 3] , so � 3 1 ( x − 2) 2 dx = ( x − 2) 3 1 = (3 − 2) 3 − (1 − 2) 2 | 3 A = 3 3 3 The function ln x is nonpositive in the interval 1 / e ≤ x ≤ 1 and x nonnegative in the interval 1 ≤ x ≤ e . Thus � 1 � e � e | ln x ln x ln x A = | dx = − dx + dx . x x x 1 / e 1 / e 1 IS (ln x ) 2 The antiderivative of ln x because of the chain rule! x 2 ln x = (ln x ) (1 / x ) = (ln x ) (ln x ) ′ . x Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 9 / 19

Examples 1. 5.4, problem 74. Suppose a company’s marginal revenue r ( x ) from selling x eggbeaters is 2 r ( x ) = 2 − ( x + 1) 2 r ( x ) in thousands of $, x in thousands of units. How much money does the company earn from 3000 eggbeaters? � x 1 f ( t ) dt = x 2 − 2 x + 1 . Find f (4) and 2. 5.4, problem 77. Suppose f ( x ). 3. 5.4 problem 75b. The temperature T ( t ) (in ◦ F ) of a room at time t minutes is given by T ( t ) = 85 − 3 √ 25 − t for 0 ≤ t ≤ 25. Find the room’s average temperature for 0 ≤ t ≤ 25. Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 10 / 19

Hints to the answers � x 1. R ( x ) = 0 r ( u ) du . 2. The function f ( x ) is the derivative of x 2 − 2 x + 1 . 3. The average of a function y = f ( x ) over an interval a ≤ x ≤ b is � b 1 Avg ( f ) := a f ( x ) dx . b − a Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 11 / 19

Some Integrals � 3 � 3 ( x − 2) 2 dx ( x − 2) 2 dx (1) (2) 0 1 x + √ x � 2 � 1 √ x dx (3) (4) dx x 2 3 1 � e ln x (5) dx x 1 � � cos 2 x sin x dx (6) (7) csc x cot x dx Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 12 / 19

(Hints to) Answers x n dx = x n +1 � In the first three problems we use n +1 + C for n � = − 1 . In problem (4) we need to use the algrbraic identity x + √ x = x − 1 + x − 3 / 2 . x 2 For (6), the antiderivative is − cos 3 x + C . Check the answer by 3 differentiating; the chain rule is crucial. For (7), the expression is (sin x ) − 2 cos x = (sin x ) − 2 (sin x ) ′ . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 13 / 19

Substitution In (5)-(7) we had to rely on the chain rule to obtain the correct answer. The method of substitution makes this procedure systematic. � Substitute u = h ( x ) in f ( x ) dx . The substitution u = h ( x ) must be 1-1; compute/solve x = h − 1 ( u ) . h − 1 � ′ ( u ) du . � Compute dx = � � h − 1 � ′ ( u ) du . h − 1 ( u ) � � � Substitute f ( x ) dx = f � � If already x = r ( u ) , f ( x ) dx = f ( r ( u ))) r ′ ( u ) du . For definite integrals, � b � h − 1 ( b ) h − 1 � ′ ( u ) du , h − 1 ( u ) � � � f ( x ) dx = f h − 1 ( a ) a � b � r ( b ) f ( x ) dx = f ( r ( u ))) r ′ ( u ) du . r ( a ) a Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 14 / 19

Computing a Definite Integral � b f ( x ) dx a Check that the function is continuous in the interval [ a , b ], or at least piecewise continuous. NO ASYMPTOTES! Find an antiderivative F ( x ) of f ( x ) F ′ ( x ) = f ( x ) . Evaluate F ( b ) − F ( a ) . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 15 / 19

Examples of Substitutions Exercise: Compute the area of the circle of radius r > 0 . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 16 / 19

We use r = 2 , the general case is not that different. � 2 � 4 − x 2 dx . A = 2 − 2 Make the substitution x = 2 sin θ with − π/ 2 ≤ θ ≤ π/ 2 . Then dx = 2 cos xdx . If x ranges between − 2 ≤ x ≤ 2 , then − π/ 2 ≤ θ ≤ π/ 2. So we change the endpoints of the integral: � π/ 2 4 − 4 sin 2 θ · 2 cos θ d θ. � A = 2 − π/ 2 √ Because 4 − 4 sin 2 θ = 4(1 − sin 2 θ ) = 4 cos 2 θ, 4 − 4 x 2 = 2 | cos θ | ; the square root must be nonnegative, but cos θ may be negative. !Again, note that − π/ 2 ≤ θ ≤ π/ 2 , cos θ ≥ 0! Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 17 / 19

The area is � π/ 2 cos 2 θ d θ. A = 8 − π/ 2 Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 18 / 19

The area is � π/ 2 cos 2 θ d θ. A = 8 − π/ 2 Important Formulas for Integration sin( a ± b ) = sin a cos b ± sin b cos a cos( a ± b ) = cos a cos b ∓ sin a sin b cos 2 x = cos 2 x − sin 2 x sin 2 x = 2 sin x cos x cos 2 x = 2 cos 2 x − 1 cos 2 x = 1 − 2 sin 2 x . Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 18 / 19