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Home , Antiderivative, Continuous function

Antiderivatives

Definition A F is called an of f on an I if F 0(x) = f (x) for all x in I . Example Let f (x) = x 2. Then an antiderivative 2 x 3 F (x) for x is F (x) = 3 . If F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is F (x) + C where C is an arbitrary . Table of Antidifferentiation Formulas

Function Particular antiderivative c · f (x) c · F (x) f (x) + g(x) F (x) + G(x) n xn+1 x (n =6 −1) n+1 1 x ln |x| ex ex cos x sin x Table of Antidifferentiation Formulas

Function Particular antiderivative sin x − cos x sec2 x tan x sec x tan x sec x √ 1 sin−1 x 1−x2 1 −1 1+x2 tan x Example A is thrown upward with a speed of 48 ft/s from the edge of a cliff 432 ft above the ground. Find its height above the ground t seconds later. When does it reach its maximum height? When does it hit the ground? Solution The motion is vertical and we choose the positive direction to be upward. At time t the distance above the ground is s(t) and the v(t) is decreasing. Therefore, the must be negative and dv a(t) = = −32 dt Taking v(t) = −32t + C To determine C we use that v(0) = 48. This gives 48 = 0 + C, so v(t) = −32t + 48 The maximum height is reached when v(t) = 0, that is after 1.5 s. Since s0(t) = v(t), we antidifferentiate again and obtain s(t) = −16t2 + 48t + D Using the fact that s(0) = 432, we have 432 = 0 + D and so s(t) = −16t2 + 48t + 432 The expression for s(t) is valid until the ball hits the ground. This happens when s(t) = 0, that is when −16t2 + 48t + 432 = 0 or, equivalently, t2 − 3t − 27 = 0 √ 3 ± 3 13 t = 2 We reject solution with the minus since it gives a negative value for√t. Therefore, the ball hits the ground after 3(1 + 13)/2 ≈ 6.9 s. under a graph Area under y = x 2 from 0 to 1.

Example Use rectangles to estimate the area under the parabola y = x 2 from 0 to 1. Area estimate using right and points

1 12 1 12 1 32 1 15 R = · + · + · + · 12 = 4 4 4 4 2 4 4 4 32 15 Note area A < 32 = .46875 Area estimate using left end points

1 1 1 1 12 1 32 7 L = ·02+ · + · + · = = .21875 4 4 4 4 4 2 4 4 32 Note area satisfies .21875 ≤ A ≤ .46875 General calculation using right end points Area definition using right end points

Definition The area A of the region S that lies under the graph of the f is the of the sum of the of approximating rectangles: A = lim Rn n→∞

= lim [f (x1)∆x + f (x2)∆x + ··· + f (xn)∆x] n→∞ Note in the above definition that if I = [a, b], then b − a ∆x = , n where n is the number of rectangles or divisions Sigma notation and A = Area

n X f (xi )∆x = f (x1)∆x + f (x2)∆x + ··· + f (xn)∆x i=1 n X A = lim f (xi )∆x n→∞ i=1 n X = lim f (xi−1)∆x n→∞ i=1 n X ∗ ∗ = lim f (xi )∆x, xi ∈ [xi−1, xi ]. n→∞ i=1 Definition of a Definite

Definition If f is a continuous function defined for a ≤ x ≤ b, we divide the interval [a, b] into n subintervals of equal width ∆x = (b − a)/n. We let x0(= a), x1, x2,..., xn(= b) be the endpoints of ∗ ∗ ∗ these subintervals and we let x1 , x2 ,..., xn be any ∗ sample points in these subintervals, so xi lies in

the i-th subinterval [xi−1,xi ]. Then the definite integral of f from a to b is Z b X ∗ f (x) dx = lim f (xi )∆x n→∞ a i=1 Midpoint Rule

n Z b X f (x) dx ≈ f (xi )∆x = ∆x[f (x1)+···+f (xn)] a i=1 where b − a ∆x = n and 1 x = (x + x ) = midpoint of [x , x ] i 2 i−1 i i−1 i Properties of the Integral

R b 1. a c dx = c(b − a), where c is any constant R b R b R b 2. a [f (x) + g(x)] = a f (x) dx + a g(x) dx R b R b 3. a c · f (x) dx = c a f (x) dx, where c is any constant

R b R b R b 4. a [f (x) − g(x)] dx = a f (x) dx − a g(x) dx The Fundamental Theorem of , 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 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

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 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 for finding antiderivatives. It comes from the : [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 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 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 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 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 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) 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) f (g(x))g 0(x) dx = f (u) du. a g(a)

R e ln x Example Calculate 1 x dx. dx Solution We let u = ln x because its differential du = x occurs in the integral. When x = 1, u = ln 1 = 0; when x = e, u = ln e = 1. Thus Z e ln x Z 1 u2 1 1 dx = u du = = . 1 x 0 2 0 2