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 R 6 1 Example Evaluate 3 x dx. 1 Solution An antiderivative for x is F (x) = ln x. So, by FTC, Z 6 1 dx = F (6) − F (3) = ln 6 − ln 3. 3 x R 3 x Example Evaluate 1 e dx. Solution Note that an antiderivative for ex is F (x) = ex . So, by FTC, Z 3 ex dx = F (3) − F (1) = e3 − e. 1 Example Find area A under the cosine curve π from 0 to b, where 0 ≤ b ≤ 2 . Solution Since an antiderivative of f (x) = cos(x) is F (x) = sin(x), we have Z b 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]. R x 1. If g(x) = a f (t) dt, then g 0(x) = f (x). R b 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). We use the notation R f (x) dx to denote an antiderivative for f (x) and it is called an indefinite integral. A definite integral has the form: Z b Z b f (x) dx = f (x) dx = F (b) − F (a) a a Table of Indefinite Integrals R c · f (x) dx = c · R f (x) dx R [f (x) + g(x)] dx = R f (x) dx + R g(x) dx R k dx = kx + C R n xn+1 x dx = n+1 + C (n =6 −1) R 1 x dx = ln |x| + C R ex dx = ex + C Table of Indefinite Integrals R sin x dx = − cos x + C R cos x dx = sin x + C R sec2 x dx = tan x + C R csc2 x dx = − cot x + C R sec x tan x dx = sec x + C R csc x cot x dx = − csc x + C R 1 −1 x2+1 dx = tan x + C R √ 1 dx = sin−1 x + C 1−x2 R 2 3 3 Example Find 0 2x − 6x + x 2+1 dx and interpret the result in terms of areas. Solution FTC gives Z 2 3 3 2x − 6x + 2 dx 0 x + 1 x 4 x 2 2 = 2 − 6 + 3 tan−1 x . 4 2 0 1 2 = 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 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) = t2 − 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 = (t2 − t − 6) dt 1 1 t3 t2 4 9 = − − 6t = − . 3 2 1 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) = t2 − t − 6 (measured in meters per second). Find the distance traveled during the time period 1 ≤ t ≤ 4. Solution Note that v(t) = t2 − 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 3 Z 3 Z 4 = (−t2 + t + 6) dt + (t2 − t − 6) dt 1 3 t3 t2 3 t3 t2 4 = − + + 6t + − − 6t 3 2 1 3 2 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)). Solution 1. Make the substitution: u = x 4 + 2. 2. Get du = 4x 3 dx. Z Z 1 1 Z x 3 cos(x 4 + 2) dx = cos u · du = 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 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 f (g(x))g 0(x) dx = f (u) du. Example Find R 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 f (g(x))g 0(x) dx = f (u) du. Example Find R x 3 cos(x 4 + 2) dx. Solution 1. Make the substitution: u = x 4 + 2. 2. Get du = 4x 3 dx. Z Z 1 1 Z x 3 cos(x 4 + 2) dx = cos u · du = 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. du Solution Let u = 2x + 1. Then du = 2 dx, so dx = 2 . Thus, the Substitution Rule gives Z √ Z √ du 1 Z 1 2x + 1 dx = u = u 2 du 2 2 3 1 u 2 1 3 1 3 2 2 = · 3 + C = u + C = (2x + 1) + C. 2 2 3 3 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 f (g(x))g 0(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 f (g(x))g 0(x) dx = f (u) du. √ Example Evaluate R 2x + 1 dx. du Solution Let u = 2x + 1. Then du = 2 dx, so dx = 2 . Thus, the Substitution Rule gives Z √ Z √ du 1 Z 1 2x + 1 dx = u = u 2 du 2 2 3 1 u 2 1 3 1 3 2 2 = · 3 + C = u + C = (2x + 1) + C. 2 2 3 3 1 Solution If we let u = 5x, then du = 5 dx, so dx = 5 du. Therefore Z 1 Z 1 1 e5x dx = eu du = eu + C = e5x + 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 f (g(x))g 0(x) dx = f (u) du. Example Calculate R e5x 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 f (g(x))g 0(x) dx = f (u) du. Example Calculate R e5x dx. 1 Solution If we let u = 5x, then du = 5 dx, so dx = 5 du. Therefore Z 1 Z 1 1 e5x dx = eu du = eu + C = e5x + C. 5 5 5 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 sin x Z du tan x dx = dx = − cos x u = − ln |u| + C = − ln | cos x| + C. 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 f (g(x))g 0(x) dx = f (u) du. Example Calculate R 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 f (g(x))g 0(x) dx = f (u) du. Example Calculate R 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 sin x Z du tan x dx = 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) dx Solution We let u = ln x because its differential du = x occurs in the integral.