LU10 Integration_student.ppt - LU10 Integration Antiderivatives and the Rules of Integration Integration by Substitution Area and the Definite Integral

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Unformatted text preview: LU10 Integration Antiderivatives and the Rules of Integration Integration by Substitution Area and the Definite Integral The Fundamental Theorem of Calculus Evaluating Definite Integrals Area Between Two Curves Applications of the Definite Integral to Business and Economics 10.1 Antiderivatives and the Rules of Integration Antiderivatives The concept of the antiderivative of a function. ✦ A function F is an antiderivative of f on an interval I if F ′(t) = f(t) for all of t in I. Theorem 1 Let G be an antiderivative of a function f. Then, every antiderivative F of f must be of the form F(x) = G(x) + C where C is a constant. Example 1 Prove that the function G(x) = x2 is an antiderivative of the function f(x) = 2x. Write a general expression for the antiderivatives of f. Solution Since G′(x) = 2x = f(x), we have shown that G(x) = x2 is an antiderivative of f(x) = 2x. By Theorem 1, every antiderivative of the function f(x) = 2x has the form F(x) = x2 + C, where C is a constant. The Indefinite Integral The process of finding all the antiderivatives of a function is called antidifferentiation or integration. We use the symbol ∫, called an integral sign, to indicate that the operation of integration is to be performed on some function f. Thus, 1 dx x C and 2 2 x dx x K where C and K are arbitrary constants. Basic Integration Rules Rule 1: The Indefinite Integral of a Constant kdx kx C ( k , a constant) Example 2 Find each of the following indefinite integrals: a. 2dx b. 2 dx Solution Each of the integrals have the form f(x) = k, where k is a constant. Applying Rule 1 in each case yields: a. 2dx 2 x C b. 2 2 dx x C Basic Integration Rules From the rule of differentiation, d n x nx n 1 dx we obtain the following rule of integration: ✦ Rule 2: The Power Rule 1 n 1 x dx n 1 x C n (n 1) Examples 3 Find the indefinite integral: 3 x dx 3/2 x dx 1 x 3/2 dx Solution – Examples 3a Find the indefinite integral: 1 4 x dx 4 x C 3 Basic Integration Rules ✦ Rule 3: The Indefinite Integral of a Constant Multiple of a Function cf ( x)dx c f ( x)dx where c is a constant. Examples 4 Find the indefinite integral: 3 2t dt 2 3x dx Solution – Examples 4a Find the indefinite integral: 3 3 2t dt 2 t dt 1 4 2 t K 4 1 4 t 2K 2 1 4 t C 2 Basic Integration Rules ✦ Rule 4: The Sum Rule f ( x) g ( x) dx f ( x )dx g ( x)dx f ( x) g ( x ) dx f ( x )dx g ( x)dx Examples 5 Find the indefinite integral: 3x 5 4 x 3/2 2 x 1/2 dx 3x 5dx 4 x 3/2dx 2 x 1/2dx 3x 5dx 4 x 3/2dx 2 x 1/2dx 1 6 2 5/2 3 x 4 x 2 2 x1/2 C 6 5 1 6 8 5/2 x x 4 x1/2 C 2 5 Basic Integration Rules ✦ Rule 5: The Indefinite Integral of the Exponential Function x x e dx e C Examples 6 Find the indefinite integral: x x 3 2 e (2 e x ) dx dx 3 x dx 2 e x dx 3 x dx 1 4 2e x C 4 x Basic Integration Rules ✦ Rule 6: The Indefinite Integral of the Function f(x) = x–1 1 x dx x dx ln x C 1 ( x 0) Examples 7 Find the indefinite integral: 3 4 3 4 2x x x 2 dx 2 xdx x dx x 2 dx 1 2 xdx 3 dx 4 x 2dx x 1 2 2 x 3ln x 4( 1) x 1 C 2 4 2 x 3ln x C x Differential Equations Consider the function f ′(x) = 2x – 1 from which we want to find f(x). We can find f by integrating the equation: 2 f ( x ) dx (2 x 1) dx x x C where C is an arbitrary constant. Thus, infinitely many functions have the derivative f ′, each differing from the other by a constant. Differential Equations Equation f ′(x) = 2x – 1 is called a differential equation. In general, a differential equation involves the derivative of an unknown function. A solution of a differential equation is any function that satisfies the differential equation. For the case of f ′(x) = 2x – 1, we find that f(x) = x2 – x + C gives all the solutions of the differential equation, and it is therefore called the general solution of the differential equation. Differential Equations f(x) = x2 – x + 3 y Different values of C yield different functions f(x). But all these functions have the same slope for any given value of x. For example, for any value of C, we always find that f ′(1) = 1. f(x) = x2 – x + 2 5 4 f(x) = x2 – x + 1 3 f ′(1) =f(x) 1 = x2 – x + 0 2 f ′(1) =f(x) 1 = x2 – x – 1 1 f ′(1) = 1 f ′(1) = 1 –1 –1 1 2 f ′(1) = 1 3 x Differential Equations It is possible to obtain a particular solution by specifying the value the function must assume for a given value of x. For example, suppose we know the function f must pass through the point (1, 2), which means f(1) = 2. Using this condition on the general solution we can find the value of C: f(1) = 12 – 1 + C = 2 C =2 Thus, the particular solution is f(x) = x2 – x + 2 Differential Equations y Here is the graph of the f(x) = x2 – x + 2 particular solution of f when C = 2. Note that this graph does go through the point (1, 2). 5 4 3 2 (1, 2) 1 –1 –1 1 2 3 x Initial Value Problems The problem we just discussed is of a type called initial value problem. In this type of problem we are required to find a function satisfying 1. A differential equation. 2. One or more initial conditions. Example 8 Find the function f if it is known that f ( x ) 3x 2 4 x 8 and f (1) 9 Solution Integrating the function f ′, we find f ( x ) f ( x )dx (3x 2 4 x 8)dx x3 2 x2 8x C Example 9 Find the function f if it is known that f ( x ) 3x 2 4 x 8 and f (1) 9 Solution Using the condition f(1) = 9, we have f ( x) x3 2 x2 8x C 9 f (1) (1)3 2(1) 2 8(1) C 9 7 C C 2 Therefore, the required function f is f ( x) x3 2 x2 8x 2 10.2 Integration by Substitution Integration by Substitution The method of substitution is related to the chain rule for differentiating functions. It is a powerful tool for integrating a large class of functions. The Method of Integration by Substitution Step 1 Let u = g(x), where g(x) is part of the integrand, usually the “inside function” of the composite function f(g(x)). Step 2 Find du = g′(x)dx. Step 3 Use the substitution u = g(x) and du = g′(x)dx to convert the entire integral into one involving only u. Step 4 Evaluate the resulting integral. Step 5 Replace u by g(x) to obtain the final solution as a function of x. How the Method of Substitution Works Consider the indefinite integral 5 2(2 x 4) dx One way to solve this integral is to expand the expression and integrate the resulting integrand term by term. An alternative approach simplifies the integral by making a change of variable. Write u = 2x + 4 with differential du = 2dx How the Method of Substitution Works Substitute u = 2x + 4 and du = 2dx in the original expression: 5 5 5 2(2 x 4) dx (2 x 4) (2 dx ) u du Now it’s easy to integrate: 1 6 u du 6 u C 5 Replacing u by u = 2x + 4, we obtain: 1 6 2(2 x 4) dx 6 u C 1 (2 x 4)6 C 6 5 How the Method of Substitution Works We can verify the result by finding its derivative: d 1 1 6 5 (2 x 4) C 6 (2 x 4) (2) dx 6 6 2(2 x 4)5 The derivative is indeed the original integrand expression. Examples 10 Find 2 4 2 x ( x 3) dx Examples 11 Find 3x e dx Solution Let u = –3x, so that du = –3dx, or dx = – ⅓ du. Substitute to express the integrand in terms of u: 1 1 u 3x u e dx e 3 du 3 e du Evaluate the integral: 1 u 1 u e du e C 3 3 Replace u by –3x to find the solution: 1 3x e dx 3 e C 3x Examples 12 Find x 3x 2 1 dx Solution 1 Let u = 3x2 + 1, so that du = 6xdx, or xdx = 6 du. Substitute to express the integrand in terms of u: x 1 1 1 1 1 3x 2 1 dx 3x 2 1 ( xdx ) u 6 du 6 u du Evaluate the integral: 1 1 1 du ln u C 6 u 6 Replace u by 3x2 + 1 to find the solution: x 1 2 dx ln 3 x 1 C 3x 2 1 6 Examples 12 Find (ln x ) 2 2 x dx Solution Let u = ln x, so that du = 1/x dx, or dx/x = du. Substitute to express the integrand in terms of u: (ln x ) 2 1 1 2 2 dx dx (ln x ) ( ) u du 2x 2 x 2 Evaluate the integral: 1 2 1 1 3 1 3 u du u C u C 2 2 3 6 Replace u by ln x to find the solution: (ln x ) 2 1 3 dx (ln x ) C 2x 6 10.3 Area and the Definite Integral Integrability of a Function Let f be continuous on [a, b]. Then, f is integrable on [a, b]; that is, the definite integral b f ( x)dx a exists. Geometric Interpretation of the Definite Integral If f is nonnegative and integrable on [a, b], then b f ( x)dx a is equal to the area of the region under the graph of f on [a, b]. Geometric Interpretation of the Definite Integral The definite integral is equal to the area of the region under the graph of f on [a, b]: y y f ( x) b A f ( x )dx a a b x Geometric Interpretation of the Definite Integral If f is continuous on [a, b], then b f ( x )dx a is equal to the area of the region above [a, b] minus the area of the region below [a, b]. Geometric Interpretation of the Definite Integral The definite integral is equal to the area of the region above [a, b] minus the area of the region below [a, b]: y b A f ( x )dx R1 R2 R3 a R3 R1 a y f ( x) R2 b x 10.4 The Fundamental Theorem of Calculus Theorem 2 The Fundamental Theorem of Calculus Let f be continuous on [a, b]. Then, b f ( x)dx F (b) F (a) a where F is any antiderivative of f; that is, F ′(x) = f(x). Example 13 Let R be the region under the graph of f(x) = x on the interval [1, 3]. Use the fundamental theorem of calculus to find the area A of R and verify your result by elementary means. Example 13 Solution The graph shows the region to be evaluated. Since f is nonnegative on [1, 3], the area of R is given by the definite integral of f from 1 to 3. 4 y y f ( x) x 3 2 x=3 1 R x=1 1 2 3 4 x Example 13 Solution By the fundamental theorem of calculus, we have 1 2 3 A xdx x C 1 1 2 9 1 C C 2 2 4 3 Thus, the area A of region R is 4 square units. Note that the constant of integration C dropped out. This is true in general. Example 13 Solution Using elementary means, note that area A is equal 1 to R1 (base ☓ height) plus R2 ( 2☓ base ☓ height). 1 Thus, A = R1 + R2 = 2(1) + (2)(2) = 4 2 4 y y f ( x) x 3 2 2 R2 1 R1 1 2 2 3 4 x Example 14 Find the area of the region under the graph of y = x2 + 1 from x = –1 to x = 2. Net Change Formula The net change in a function f over an interval [a, b] is given by b f (b) f (a ) f ( x )dx a provided f ′ is continuous on [a, b]. Applied Example 15: Assembly Time of Workers An efficiency study conducted for Elektra Electronics showed that the rate at which Space Commander walkietalkies are assembled by the average worker t hours after starting work at 8:00 a.m. is given by the function f (t ) 3t 2 12t 15 (0 t 4) Determine how many walkie-talkies can be assembled by the average worker in the first hour of the morning shift. Applied Example 15: Assembly Time of Workers Solution Let N(t) denote the number of walkie-talkies assembled by the average worker t hours after starting work in the morning shift. Then, we have N (t ) f (t ) 3t 2 12t 15 Therefore, the number of units assembled by the average worker in the first hour of the morning shift is 1 1 0 0 N (1) N (0) N (t )dt ( 3t 2 12t 15)dt 3 2 1 ( t 6t 15t ) 1 6 15 0 0 20 or 20 units. 10.5 Evaluating Definite Integrals Properties of the Definite Integral Let f and g be integrable functions, then a 1. 2. 3. 4. 5. f ( x)dx 0 a b a a b f ( x )dx f ( x )dx b b a a cf ( x )dx c f ( x )dx (c, a constant ) b b b a a a f ( x) g ( x) dx f ( x )dx g ( x )dx b c b a a c f ( x)dx f ( x )dx f ( x)dx ( a c b) Examples 16: Using the Method of Substitution Evaluate 4 2 x 9 x dx 0 Solution First, find the indefinite integral: I x 9 x 2 dx ✦ Let u = 9 + x2 so that d du (9 x 2 )dx dx 2 xdx 1 xdx du 2 Examples 16: Using the Method of Substitution Evaluate 4 2 x 9 x dx 0 Solution First, find the indefinite integral: I x 9 x 2 dx 1 ✦ Then, integrate by substitution using xdx = du: 2 I x 9 x 2 dx 1 1 1/2 udu u du 2 2 1 3/2 u C 3 1 (9 x 2 )3/2 C 3 Examples 16: Using the Method of Substitution Evaluate 4 2 x 9 x dx 0 Solution Using the results, we evaluate the definite integral: 4 1 2 2 3/2 x 9 x dx (9 x ) 0 3 0 4 1 [(9 (4) 2 )3/2 (9 (0) 2 )3/2 ] 3 1 (125 27) 3 98 3 32 23 Examples 17: Using the Method of Substitution Evaluate x2 0 x 3 1 dx 1 Solution Let u = x3 + 1 so that d 3 ( x 1)dx dx 3x 2dx du 1 du x 2dx 3 Examples 17: Using the Method of Substitution Evaluate x2 0 x 3 1 dx 1 Solution Find the lower and upper limits of integration with respect to u: ✦ When x = 0, the lower limit is u = (0)3 + 1 = 1. ✦ When x = 1, the upper limit is u = (1)3 + 1 = 2. Substitute x 2 dx 13 du , along with the limits of integration: 1 21 1 x2 1 1 21 2 0 x 3 1 dx 0 x 3 1 x dx 1 u 3 du 3 1 u du 1 2 1 1 1 ln u (ln 2 ln1) ln 2 3 3 3 1 Average Value of a Function Suppose f is integrable on [a, b]. Then, the average value of f over [a, b] is 1 b f ( x )dx a b a Applied Example 18: Automobile Financing The interest rates changed by Madison Finance on auto loans for used cars over a certain 6-month period in 2008 are approximated by the function 1 3 7 2 r (t ) t t 3t 12 12 8 (0 t 6) where t is measured in months and r(t) is the annual percentage rate. What is the average rate on auto loans extended by Madison over the 6-month period? Applied Example 18: Automobile Financing Solution The average rate over the 6-month period is given by 1 6 1 3 7 2 t t 3 t 12 dx 0 6 0 12 8 6 1 1 4 7 3 3 2 t t t 12t 6 48 24 2 0 1 1 7 3 2 4 3 (6) (6) (6) 12(6) 6 48 24 2 9 or 9% per year. 10.6 Area Between Two Curves The Area Between Two Curves Let f and g be continuous functions such that f(x) g(x) on the interval [a, b]. Then, the area of the region bounded above by y = f(x) and below by y = g(x) on [a, b] is given by b f ( x) g ( x) dx a Examples 19 Find the area of the region bounded by the x-axis, the graph of y = –x2 + 4x – 8, and the lines x = –1 and x = 4. Solution The region R is being bounded above by the graph f(x) = 0 and below by the graph of g(x) = y = –x2 + 4x – 8 on [–1, 4]: y –2 2 4 6 x R x = –1 –4 x=4 –8 y = –x2 + 4x – 8 – 12 Examples 19 Find the area of the region bounded by the x-axis, the graph of y = –x2 + 4x – 8, and the lines x = –1 and x = 4. Solution Therefore, the area of R is given by b 4 2 f ( x ) g ( x ) dx 0 ( x 4 x 8) dx a 1 4 ( x 2 4 x 8)dx 1 4 1 3 x 2 x2 8x 3 1 1 3 (4) 2(4)2 8(4) 3 31 23 1 3 2 ( 1) 2( 1) 8( 1) 3 Examples 20 Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x2 – 4. Solution First, find the points of intersection of the two curves. To do this, you can set g(x) = f(x) and solve for x: x 2 4 2 x 1 x 2 2 x 3 0 ( x 1)( x 3) 0 so, the graphs intersect at x = – 1 and at x = 3. Examples 20 Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x2 – 4. Solution The graph of f always lies above that of g for all x in the interval [– 1, 3] between the two intersection points: y (3 , 5) y = 2x – 1 4 y = x2 – 4 2 R –4 –2 2 (– 1 , – 3) –4 4 x Examples 20 Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x2 – 4. Solution Since the graph of f always lies above that of g for all x in the interval [– 1, 3], the required area is given by b 3 2 f ( x ) g ( x ) dx (2 x 1) ( x 4) dx a 1 3 1 3 2 2 ( x 2 x 3)dx x x 3x 1 3 1 3 1 3 (3) (3) 2 3(3) 3 2 10 3 1 ( 1)3 ( 1) 2 3( 1) 3 Examples 21 Find the area of the region bounded by f(x) = x3 – 3x + 3 and g(x) = x + 3. Solution The region R being considered is composed of two subregions R1 and R2: y y=x+3 5 y = x3 – 3x + 3 R1 3 R2 1 –3 –1 1 2 3 x Examples 21 Find the area of the region bounded by f(x) = x3 – 3x + 3 and g(x) = x + 3. Solution To find the points of intersection, we solve simultaneously the equations y = x3 – 3x + 3 and y = x + 3. x 3 3x 3 x 3 y x 3 4 x 0 x ( x 2)( x 2) 0 5 (2 , 5) y = x3 – 3x + 3 R1 So, x = 0, x = – 2, and x = 2. The points of intersection of the two curves are (– 2, 1), (0, 3), and (2, 5). y=x+3 3 (0 , 3) R2 (– 2, 1) –3 1 –1 1 2 3 x Examples 21 Find the area of the region bounded by f(x) = x3 – 3x + 3 and g(x) = x + 3. Solution Note that f(x) g(x) for [– 2, 0], so the area of region R1 is given by b R1 f ( x ) g ( x ) dx y a 0 3 ( x 3 x 3) ( x 3) dx 2 0 3 ( x 4 x )dx 5 y=x+3 (2 , 5) y = x3 – 3x + 3 R1 3 (0 , 3) 2 0 1 4 x 2 x2 4 2 (4 8) 4 (– 2, 1) –3 1 –1 1 2 3 x Examples 21 Find the area of the region bounded by f(x) = x3 – 3x + 3 and g(x) = x + 3. Solution Note that g(x) f(x) for [0, 2], so the area of region R2 is given by b R2 g ( x ) f ( x ) dx y a 2 3 ( x 3) ( x 3x 3) dx 5 0 y=x+3 (2 , 5) y = x3 – 3x + 3 2 ( x 3 4 x )dx 3 (0 , 3) 0 2 1 4 x 2 x 2 4 0 4 8 4 R2 (– 2, 1) –3 1 –1 1 2 3 x Examples 21 Find the area of the region bounded by f(x) = x3 – 3x + 3 and g(x) = x + 3. Solution Therefore, the required area R is R R1 R2 4 4 8 y square units. 5 y=x+3 (2 , 5) y = x3 – 3x + 3 R1 3 (0 , 3) R2 (– 2, 1) –3 1 –1 1 2 3 x 10.7 Applications of the Definite Integral to Business and Economics Consumers’ and Producers’ Surplus Suppose p = D(x) is the demand function that relates the price p of a commodity to the quantity x demanded of it. Now suppose a unit market price p has been established, along with a corresponding quantity demanded x. Those consumers who would be willing to pay a unit price higher thanp for the commodity would in effect experience a savings. This difference between what the consumer would be willing to pay and what they actually have to pay is called the consumers’ surplus. Consumers’ and Producers’ Surplus We obtain the consumers’ surplus CS formula: x CS D ( x )dx p x 0 where D(x) is the demand function, p is the unit market price, andx is the quantity demanded. p CS p = D(x) p x x Consumers’ and Producers’ Surplus Suppose p = S(x) is the supply function that relates the price p of a commodity to the quantity x supplied of it. Again, suppose a unit market price p has been established, along with a corresponding quantity supplied x. Those sellers who would be willing to sell at unit price lower thanp for the commodity would in effect experience a gain or profit. This difference between what the seller would be willing to sell for and what they actually can sell for is called the producers’ surplus. Consumers’ and Producers’ Surplus Geometrically, the producers’ surplus is given by the area of the region bounded above the straight line p =p and below the supply curve p = S(x) from x = 0 to x =x: p p = S (x ) p PS x x Consumers’ and Producers’ Surplus The producers’ surplus PS is given by PS p x x S ( x )dx 0 where S(x) is the supply function,p is the unit market price, andx is the quantity supplied. p p = S (x ) p PS x x Example 22 The demand function for a certain make of 10-speed bicycle is given by p D( x ) 0.001x 2 250 where p is the unit price in dollars and x is the quantity demanded in units of a thousand. The supply function for these bicycles is given by p S ( x ) 0.0006 x 2 .02 x 100 where p stands for the price in dollars and x stands for the number of bicycles that the supplier will want to sell. Determine the consumers’ surplus and the producers’ surplus if the market price of a bicycle is set at the equilibrium price. Example 22 Solution To find the equilibrium point, equate S(x) and D(x) to solve the system of equations and find the point of intersection of the demand and supply curves: 0.0006 x 2 .02 x 100 0.001x 2 250 0.0016 x 2 .02 x 150 0 16 x 2 200 x 1,500,000 0 2 x 2 25 x 187,500 0 (2 x 625)( x 300) 0 Thus, x = – 625/2 or x = 300. The first number is discarded for being negative, so the solution is x = 300. Example 22 Solution Substitute x = 300 to find the equilibrium value of p: p 0.001(300)2 250 160 Thus, the equilibrium point is (300, 160). That is, the equilibrium quantity is 300,000 bicycles, and the equilibrium price is \$160 per bicycle. Example 22 Solution To find the consumers’ surplus, we set x = 300 and p = 160 in the consumers’ surplus formula: x CS D ( x )dx p x 0 300 ( 0.001x 2 250)dx (160)(300) 0 300 1 x 3 250 x 48,000 3000 0 3003 250(300) 48,000 18,000 3000 or \$18,000,000. Example 22 Solution To find the producers’ surplus, we setx = 300 andp = 160 in the producers’ surplus formula: PS p x x S ( x )dx 0 (160)(300) 300 0 (0.0006 x 2 0.02 x 100)dx 3 2 300 48,000 (0.0002 x 0.01x 100 x ) 0 0 48,000 [0.0002(300)3 0.01(300) 2 100(300)] 11,700 or \$11,700,000. Example 22 Solution Consumers’ surplus and producers’ surplus when th...
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