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. 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
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. 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 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 = ex , 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 = ex , bounded below by y = x, and bounded on the sides by x = 0 and x = 1.
Solution The region is shown in the figure. The upper boundary curve is y = ex and the lower boundary curve is y = x. So we use the area formula with f (x) = ex , g(x) = x, a = 0, and b = 1: Z 1 1 1 A = (ex − x) dx = ex − x 2 0 2 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: 1 R 1 2 R 1 2 h x2 x3 i A = 0 (2x − 2x ) dx = 2 0 (x − x ) dx = 2 2 − 3 0 1 1 1 = 2 2 − 3 = 3 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
Pn ∗ V = volume ≈ 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 n Z b X ∗ V = lim A(xi )∆x = A(x) dx n→∞ i=1 a Example Show the volume of a sphere of radius r 4 3 is V = 3πr . Solution x 2 + y 2 = r 2 y = pr 2 − y 2. So the crossectional area is A(x) = πy 2 = π(r 2 − x 2). So, R r R r 2 2 R r 2 2 V = −r A(x) dx = −r π(r − x ) dx = 2π 0 (r − x ) dx r h 2 x3 i 3 r 3 4 3 = 2π r x − 3 = 2π r − 3 = 3 πr 0
Computing volume of a sphere Solution x 2 + y 2 = r 2 y = pr 2 − y 2. So the crossectional area is A(x) = πy 2 = π(r 2 − x 2). So, R r R r 2 2 R r 2 2 V = −r A(x) dx = −r π(r − x ) dx = 2π 0 (r − x ) dx r h 2 x3 i 3 r 3 4 3 = 2π r x − 3 = 2π r − 3 = 3 πr 0
Computing volume of a sphere
Example Show the volume of a sphere of radius r 4 3 is V = 3πr . Computing volume of a sphere
Example Show the volume of a sphere of radius r 4 3 is V = 3πr . Solution x 2 + y 2 = r 2 y = pr 2 − y 2. So the crossectional area is A(x) = πy 2 = π(r 2 − x 2). So, R r R r 2 2 R r 2 2 V = −r A(x) dx = −r π(r − x ) dx = 2π 0 (r − x ) dx r h 2 x3 i 3 r 3 4 3 = 2π r x − 3 = 2π r − 3 = 3 πr 0 Solution√The area of the crossection is: A(x) = π( x)2 = πx. 1 R 1 R 1 x2 i π So, V = 0 A(x) dx = 0 πx dx = π 2 = 2 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 = x from 0 to 1. 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 = x from 0 to 1.
Solution√The area of the crossection is: A(x) = π( x)2 = πx. 1 R 1 R 1 x2 i π So, V = 0 A(x) dx = 0 πx dx = π 2 = 2 0 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 √ 2 2 A(y) = πx = π( 3 y) = πy 3 . So, 8 ∞ ∞ 2 h 3 5 i 96π R R 3 dy 3 V = 0 A(y) dy = 0 πy = π 5 y = 5 . 0 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). R 1 R 1 2 2 h x3 x5 i 2π So, V = 0 A(x) dx = 0 π(x − x ) dx = π 3 − 5 = 15 .
Other volumes
Example The region R enclosed by the curves y = x and y = x 2 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 y = x 2 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). R 1 R 1 2 2 h x3 x5 i 2π So, V = 0 A(x) dx = 0 π(x − x ) dx = π 3 − 5 = 15 . 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). R 1 R 1 4 2 So, V = 0 A(x) dx = π 0 (x − 5x + 4x) dx = 1 h x5 x3 x2 i 8π π 5 − 5 3 + 4 2 = 15 . 0
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). R 1 R 1 4 2 So, V = 0 A(x) dx = π 0 (x − 5x + 4x) dx = 1 h x5 x3 x2 i 8π π 5 − 5 3 + 4 2 = 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. 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.
Integration by parts
Z Z f (x)g 0(x) dx = f (x)g(x) − g(x)f 0(x) dx
The above is called the formula for integration by parts. Integration by parts
Z Z f (x)g 0(x) dx = f (x)g(x) − 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. Z 2. xex dx : Choose u = x and dv = ex dx Z 3. t2et dt : Choose u = t2 and dv = et dt Z 4. 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
Examples using strategy
Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 3. t2et dt : Choose u = t2 and dv = et dt Z 4. 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
Examples using strategy
Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. xex dx : Choose u = x and dv = ex dx Z 4. 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
Examples using strategy
Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. xex dx : Choose u = x and dv = ex dx Z 3. t2et dt : Choose u = t2 and dv = et dt Z 5. ln(x) dx : Choose u = ln x and v = dx
Examples using strategy
Z 1. x sin x dx : Choose u = x and dv = sin x dx Z 2. xex dx : Choose u = x and dv = ex dx Z 3. t2et dt : Choose u = t2 and dv = et dt Z 4. 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. xex dx : Choose u = x and dv = ex dx Z 3. t2et dt : Choose u = t2 and dv = et dt Z 4. 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