The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a, b], then the function g defined by Z x g (x) = f (t) dt a ≤ x ≤ b a is continuous on [a, b] and differentiable on (a, b), and g 0(x) = f (x). The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then Z b f (x) dx = F (b) − F (a), a where F (x) is any antiderivative of f (x), that is, a function such that F 0(x) = f (x). Application of FTC Example Evaluate R6 1 3 x dx. Solution An antiderivative for x1 is F (x) = ln x. So, by FTC, Z 6 1 dx = F (6) − F (3) = ln 6 − ln 3. x 3 Example Evaluate R3 1 e x dx. Solution Note that an antiderivative for e x is F (x) = e x . So, by FTC, Z 3 e x dx = F (3) − F (1) = e 3 − e. 1 Example Find area A under the cosine curve from 0 to b, where 0 ≤ b ≤ π2 . Solution Since an antiderivative of f (x) = cos(x) is F (x) = sin(x), we have b Z b A= cos(x) dx = sin(x) = sin(b)−sin(0) = sin(b). 0 0 The Fundamental Theorem of Calculus Suppose f is continuous on [a, b]. Rx 1. If g (x) = a f (t) dt, then g 0(x) = f (x). Rb 2. a f (x) dx = F (b) − F (a), where F (x) is any antiderivative of f (x), that is, F 0(x) = f (x). Notation: Indefinite integral R f (x) dx = F (x) means F 0(x) = f (x). R We use the notation f (x) dx to denote an antiderivative for f (x) and it is called an indefinite integral. A definite integral has the form: b Z b Z f (x) dx = f (x) dx = F (b) − F (a) a a Table of Indefinite Integrals R R R c · f (x) dx = R c · f (x) dx R [f (x) + g (x)] dx = f (x) dx + g (x) dx R k dx = kx + C R n n+1 x dxR= xn+1 + C (n 6= −1) 1 = ln |x| + C R x dx e x dx = e x + C Table of Indefinite Integrals R Rsin x dx = − cos x + C R cos2 x dx = sin x + C R sec2 x dx = tan x + C R csc x dx = − cot x + C R sec x tan x dx = sec x + C csc csc x + C R x1cot x dx = − −1 x +C 2 +1 dx = tan x R 1 −1 √ dx = sin x +C 2 1−x R2 3 3 2x − 6x + 0 x 2 +1 dx and interpret the result in terms of areas. Solution FTC gives Z 2 3 3 2x − 6x + 2 dx x +1 0 2 x2 x4 −1 = 2 − 6 + 3 tan x . 4 2 0 2 1 = x 4 − 3x 2 + 3 tan−1 x 2 0 1 = (24 ) − 3(22 ) + 3 tan−1 2 − 0 2 = −4 + 3 tan−1 2 Example Find The Net Change Theorem The Net Change Theorem The integral of a rate of change is the net change: Z b F 0(x) dx = F (b) − F (a). a Example A particle moves along a line so that its velocity at time t is v (t) = t 2 − t − 6 (measured in meters per second). Find the displacement of the particle during the time period 1 ≤ t ≤ 4. Solution By the Net Change Theorem, the displacement is Z 4 Z 4 s(4) − s(1) = v (t)dt = (t 2 − t − 6) dt 1 t3 t2 = − − 6t 3 2 1 4 1 9 =− . 2 This means that the particle moved 4.5 m toward the left. Example A particle moves along a line so that its velocity at time t is v (t) = t 2 − t − 6 (measured in meters per second). Find the distance traveled during the time period 1 ≤ t ≤ 4. Solution Note that v (t) = t 2 − t − 6 = (t − 3)(t + 2) and so v (t) ≤ 0 on the interval [1, 3] and v (t) ≥ 0 on [3, 4]. From the Net Change Theorem, the distance traveled is Z 4 Z 3 Z 4 |v (t)| dt = [−v (t)] dt + v (t) dt 1 1 Z = 1 3 (−t 2 + t + 6) dt + 3 Z 4 (t 2 − t − 6) dt 3 3 3 3 4 t t2 t t2 − − 6t = − + + 6t + 3 2 3 2 1 3 61 = ≈ 10.17m 6 The Substitution Rule The Substitution Rule is one of the main tools used in this class for finding antiderivatives. It comes from the Chain Rule: [F (g (x))]0 = F 0(g (x))g 0(x). So, Z F 0(g (x))g 0(x) dx = F (g (x)). The Substitution Rule The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Find x 3 cos(x 4 + 2) dx. The Substitution Rule The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Find x 3 cos(x 4 + 2) dx. Solution 1. Make the substitution: u = x 4 + 2. 3 2. Get Z du = 4x dx. Z Z 1 1 3 4 cos u du x cos(x + 2) dx = cos u · du = 4 4 1 1 = sin u + C = sin(x 4 + 2) + C . 4 4 Note at the final stage we return to the original variable x. The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. Example Evaluate R√ 2x + 1 dx. The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R√ Example Evaluate 2x + 1 dx. Solution Let u = 2x + 1. Then du = 2 dx, so dx = the Substitution Rule gives Z Z Z √ √ du 1 1 2x + 1 dx = u = u 2 du 2 2 3 du . 2 3 1 u2 1 3 1 = · 3 + C = u 2 + C = (2x + 1) 2 + C . 2 2 3 3 Thus, The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Calculate e 5x dx. The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Calculate e 5x dx. Solution If we let u = 5x, then du = 5 dx, so dx = 15 du. Therefore Z Z 1 1 1 5x e dx = e u du = e u + C = e 5x + C . 5 5 5 The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Calculate tan x dx. The Substitution Rule If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then Z Z 0 f (g (x))g (x) dx = f (u) du. R Example Calculate tan x dx. Solution Z Z sin x tan x dx = dx. cos x This suggests substitution u = cos x, since then du = − sin x dx and so, sin x dx = −du: Z Z Z sin x du dx = − tan x dx = cos x u = − ln |u| + C = − ln | cos x| + C . The Substitution Rule for Definite Integrals Substitution Rule for Definite Integrals If g 0 is continuous on [a, b] and f is continuous on the range of u = g (x), then Z b Z g (b) f (g (x))g 0(x) dx = f (u) du. a g (a) The Substitution Rule for Definite Integrals If g 0 is continuous on [a, b] and f is continuous on the range of u = g (x), then Z b Z g (b) 0 f (g (x))g (x) dx = f (u) du. a Example Calculate g (a) Re 1 ln x x dx. The Substitution Rule for Definite Integrals If g 0 is continuous on [a, b] and f is continuous on the range of u = g (x), then Z b Z g (b) 0 f (g (x))g (x) dx = f (u) du. a g (a) Re Example Calculate 1 lnxx dx. Solution We let u = ln x because its differential du = dx x occurs in the integral. When x = 1, u = ln 1 = 0; when x = e, u = ln e = 1. Thus 1 Z 1 Z e ln x u2 1 u du = dx = = . x 2 0 2 0 1 Areas between curves Area between curves The area A of the region bounded by the curves y = f (x), y = g (x), and the lines x = a, x = b, where f and g continuous and f (x) ≥ g (x) for all x in [a, b], is Z b A= [f (x) − g (x)] dx a Example Find the area of the region bounded above by y = e x , bounded below by y = x, and bounded on the sides by x = 0 and x = 1. Example Find the area of the region bounded above by y = e , bounded below by y = x, and bounded on the sides x by x = 0 and x = 1. Solution The region is shown in the figure. The upper boundary curve is y = e x and the lower boundary curve is y = x. So we use the area formula with f (x) = e x , g (x) = x, a = 0, and b = 1: 1 Z 1 1 2 x x A= (e − x) dx = e − x 2 0 0 1 = e − − 1 = e − 1.5 2 Example Find the area of the region enclosed by the parabolas y = x 2 and y = 2x − x 2 . Example Find the area of the region enclosed by the parabolas y = x 2 and y = 2x − x 2 . Solution We first find the points of intersection of the parabolas by solving their equations simultaneously. This gives x 2 = 2x − x 2 , or 2x 2 − 2x = 0. Thus 2x(x − 1) = 0, so x = 0 or 1. The points of intersection are (0, 0) and (1, 1). So the total area is: h 2 i1 R1 R1 x x3 2 2 A = 0 (2x − 2x ) dx = 2 0 (x − x ) dx = 2 2 − 3 0 = 2 12 − 31 = 13 Areas between curves The area between the curves y = f (x) and y = g (x) and between x = a and x = b is Z b A= |f (x) − g (x)| dx a . Example Find the area enclosed by the line y = x − 1 and the parabola y 2 = 2x + 6. Example Find the area enclosed by the line y = x − 1 and the parabola y 2 = 2x + 6. Volumes Volumes V = volume ≈ Pn ∗ i=1 A(xi )∆x Definition of volume Definition of Volume Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane Px , through x and perpendicular to the x−axis, is A(x), where A is a continuous function, then the volume of S is Z b n X V = lim A(xi∗ )∆x = A(x) dx n→∞ i=1 a Computing volume of a sphere Computing volume of a sphere Example Show the volume of a sphere of radius r is V = 43 πr 3 . Computing volume of a sphere Example Show the volume of a sphere of radius r is V = 43 πr 3 . Solution x 2 + y 2 = r 2 p y = r 2 − y 2. 2 So the is A(x) = πy 2 = π(r x 2 ). So, R rcrossectionalRarea Rr − r 2 2 2 V = −r A(x) dx = −r π(r − x ) dx = 2π 0 (r − x 2 ) dx h ir 3 3 = 2π r 2 x − x3 = 2π r 3 − r3 = 34 πr 3 0 Computing a volume of revolution Example Find the volume of the solid obtained by rotating about the x-axis the region under the curve y = from 0 to 1. √ x Computing a volume of revolution Example Find the volume of the solid obtained by rotating about the x-axis the region under the curve y = from 0 to 1. Solution√The area of the crossection is: A(x) = π( x)2 = πx. i1 R1 R1 2 So, V = 0 A(x) dx = 0 πx dx = π x2 = 0 π 2 √ x Volume of a solid paraboloid Example Find the volume of the solid obtained by rotating the region bounded by y = x 3 , y = 8 and x = 0 about the y -axis. Volume of a solid paraboloid Example Find the volume of the solid obtained by rotating the region bounded by y = x 3 , y = 8 and x = 0 about the y -axis. √ Solution Note that x = 3 y . The area of crossection is: 2 √ A(y ) = πx 2 = π( 3 y )2 = πy 3 . So, h 5 i8 R∞ R ∞ 2 dy 3 V = 0 A(y ) dy = 0 πy = π 53 y 3 = 0 96π . 5 Other volumes Example 2 The region R enclosed by the curves y = x and y = x is rotated about the x-axis. Find the volume of the solid region. Other volumes Example The region R enclosed by the curves y = x and 2 y = x is rotated about the x-axis. Find the volume of the solid region. Solution The curves y = x and y = x 2 intersect at the points (0, 0) and (1, 1). Crossection of rotated surface has the shape of a washer (annular ring). So the crossectional area is: A(x) = πx 2 − π(x 2 )2 = π(x 2 − x 4 ). h 3 i R1 R1 5 So, V = 0 A(x) dx = 0 π(x 2 − x 2 ) dx = π x3 − x5 = 2π . 15 Example The region R enclosed by the curves y = x and y = x 2 is rotated about the line y = 2. Find the volume. Example The region R enclosed by the curves y = x and y = x 2 is rotated about the line y = 2. Find the volume. Solution The crossection is again a washer. The crossectional area is: A(x) = π(2 − x 2 )2 − π(2 − x)2 = π(x 4 − 5x 2 + 4x). R1 R1 So, V = 0 A(x) dx = π 0 (x 4 − 5x 2 + 4x) dx = h 5 i1 3 2 π x5 − 5 x3 + 4 x2 = 8π . 15 0 Integration by parts We now return to integration methods. Recall our first method was substraction which came from the chain rule. Integration by parts comes from the product rule. d [f (x)g (x)] = f (x)g 0 (x) + g (x)f 0 (x). dx So, f (x)g 0 (x) = (f (x)g (x))0 − g (x)f 0 (x) Thus, after taking antiderivatives, we get Z Z f (x)g 0 (x) = f (x)g (x) − g (x)f 0 (x) dx. Integration by parts Z f (x)g 0 (x) dx = f (x)g (x) − Z g (x)f 0 (x) dx The above is called the formula for integration by parts. Integration by parts Z f (x)g 0 (x) dx = f (x)g (x) − Z g (x)f 0 (x) dx The above is called the formula for integration by parts. If we let u = f (x) and v = g (x), then du = f 0 (x) dx and dv = g 0 (x) dx, so the formula becomes: Z Z u dv = uv − v du. Strategy for using integration by parts Recall the integration by parts formula: Z Z u dv = uv − v du. To apply this formula we must choose dv so that we can integrate it! Frequently, we choose u so that the derivative of u is simpler than u. If both properties hold, then you have made the correct choice. Examples using strategy Z 1. x sin x dx : Choose u = x and dv = sin x dx Examples using strategy Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. xe x dx : Choose u = x and dv = e x dx Examples using strategy Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. Z 3. xe x dx : Choose u = x and dv = e x dx t 2 e t dt : Choose u = t 2 and dv = e t dt Examples using strategy Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. Z 3. 4. Z xe x dx : Choose u = x and dv = e x dx t 2 e t dt : Choose u = t 2 and dv = e t dt x 2 sin 2x dx : Choose u = x 2 and v = sin 2x dx Examples using strategy Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. Z 3. 4. Z xe x dx : Choose u = x and dv = e x dx t 2 e t dt : Choose u = t 2 and dv = e t dt x 2 sin 2x dx : Choose u = x 2 and v = sin 2x dx Z 5. ln(x) dx : Choose u = ln x and v = dx