Lecture 5: Integrals and Applications Lejla Batina Institute for Computing and Information Sciences Digital Security Version: spring 2012 Lejla Batina Version: spring 2012 Wiskunde 1 1 / 21
Outline The Definite Integral The Definite Integral Lejla Batina Version: spring 2012 Wiskunde 1 2 / 21
Intro The Definite Integral So far: from f to f : tangent line, monotonicity, extrema,... From f to f : If F (x) is a function such that F (x) = f (x), which information we get about f (from F )? f (x) = F (x) = lim h 0 F (x+h) F (x) h f (x) h F (x + h) F (x). So, F (x) gives some information about the surface under the graph of f. Lejla Batina Version: spring 2012 Wiskunde 1 3 / 21
The area problem and the definite integral Let y = f (x) be a continuous function defined on [a, b]. In order to estimate the area under y = f (x) from a to b we divide [a, b] into n subintervals: [x 0, x 1 ], [x 1, x 2 ], [x 2, x 3 ],..., [x n 1, x n ], where a = x 0, b = x n, each of length x = b a n (x i = a + i x, i = 0,, n). The area S i of the strip between x i 1 and x i can be approximated as the area of the rectangle of width x and height f (x i ), where x i [x i, x i+1 ], i = 0,, n. So, the total area under the curve is close to the following sum: A n i=1 f (x i ) x = f (x 1 ) x + f (x 2 ) x +... + f (x n ) x. A = lim n n i=1 f (x i ) x = b a f (x)dx. Lejla Batina Version: spring 2012 Wiskunde 1 4 / 21
The evaluation theorem Theorem If f is a continuous function and F (x) = f (x) (F is an antiderivative of f ), then: b a f (x)dx = F (b) F (a). This value gives the area below the graph of f on [a, b]. Example Compute the following definite integrals using the evaluation theorem: 1 0 x 2 dx π 2 0 sinxdx Lejla Batina Version: spring 2012 Wiskunde 1 5 / 21
Properties of the definite integrals Due to linearity and interval additivity we get: b a f (x)dx = c a f (x)dx + b c f (x)dx, where a < c < b a a f (x)dx = 0 b a f (x)dx = a b f (x)dx b a [f (x) ± g(x)]dx = b a f (x)dx ± b a g(x)dx c b a dx = c(b a) c b a f (x)dx = b a c f (x)dx Comparison: f (x) 0 b a f (x)dx 0 f (x) g(x) b a f (x)dx b a g(x)dx m f (x) M m(b a) b a f (x)dx M(b a) Lejla Batina Version: spring 2012 Wiskunde 1 6 / 21
Indefinite integrals Definition A function F such that F (x) = f (x) is called an antiderivative (or a primitive) function of f. Then for any constant C, F (x) + C is another antiderivative of f (x). The family of all antiderivatives of f is called indefinite integral of f and denoted as: f (x)dx = F (x) + C. Lejla Batina Version: spring 2012 Wiskunde 1 7 / 21
Table of indefinite integrals 0 dx = C 1 dx = x + C, so dx = x x n dx = xn+1 n+1 + C, n 1 1 x dx = ln x + C e x dx = e x + C a x dx = ax lna + C sinxdx = cos x + C cosxdx = sin x + C 1 dx = tan x + C cos 2 x dx = arctan x + C 1+x 2 = arcsin x + C dx 1 x 2 Lejla Batina Version: spring 2012 Wiskunde 1 8 / 21
Examples The Definite Integral Example (3x 5 2x 2 + 1)dx = 3x 5 dx 2x 2 dx + dx = = 3 x 5 dx 2 x 2 dx + dx = 3 x6 6 2 x3 3 + x + C. ( 3 x 2 1 )dx = x 2 x 2 3 dx x 2 dx = x 1 1 + C = 3 5 x 3 x 2 + 1 x + C. = x5/3 5 3 Lejla Batina Version: spring 2012 Wiskunde 1 9 / 21
The fundamental theorem of calculus Theorem Let f be a continuous function on [a, b]. Then: 1 The function g(x) = x a f (t)dt is an antiderivative of f, i.e., g (x) = f (x). 2 (Evaluation theorem) If F is an antiderivative of f, i.e. F (x) = f (x), then b a f (x)dx = F (b) F (a). We can rewrite it as follows: 1 2 d dx x a f (t)dt = f (x). b a F (x)dx = F (b) F (a). Lejla Batina Version: spring 2012 Wiskunde 1 10 / 21
Example The Definite Integral Example Find d x 2 dx 0 t 3 dt. We can solve this in two ways. Let g(x) = x 2 0 t 3 dt, then using the theorem: for h(u) = u 0 t3 dt h (u) = u 3. We have g(x) = h(x 2 ), end from the chain rule: g (x) = h (x 2 ) 2x = (x 2 ) 3 2x = 2x 7. Or directly: g(x) = x 2 0 t 3 dt = [ t4 4 ]x2 g (x) = 8x7 4 = 2x 7. 0 = x8 4, end then Lejla Batina Version: spring 2012 Wiskunde 1 11 / 21
The substitution rule As it holds: du = u dx, we can write: f (u)u dx = f (u)du = f (g(x))g (x)dx, where u = g(x). Example cos x 3 dx = [u = x 3, du = 1 3 dx dx = 3du] = cos u 3du = = 3 cos udu = 3 sin u + C = 3 sin x 3 + C. x sin(x 2 )dx = [u = x 2, du = 2xdx xdx = 1 2 du] = sin u 1 2 du = 1 2 sin udu = 1 2 cos x 2 + C. x 2 x + 1dx = [x + 1 = u, dx = du x = u 1] = = (u 1) 2 udu = u 5 2 du 2 u 3 2 du + u 1 2 du =... cos x x dx = [u = x, du = dx 2 x dx x = 2du] = = 2 cos udu = 2 sin x + C. Lejla Batina Version: spring 2012 Wiskunde 1 12 / 21
Integration by parts Recollect the product rule for differentiation: [f (x)g(x)] = f (x)g(x) + f (x)g (x), or f (x)g (x) = [f (x)g(x)] f (x)g(x). After integration we get: f (x)g (x)dx = [f (x)g(x)] f (x)g(x)dx. For u = f (x), v = g(x), we get du = f (x)dx and dv = g (x)dx, hence: udv = uv vdu. For definite integrals we have: b a udv = uv b a b a vdu Lejla Batina Version: spring 2012 Wiskunde 1 13 / 21
Examples The Definite Integral Example xe x dx = [u = x du = dx, dv = e x dx v = e x dx = e x ] = xe x e x dx = xe x e x + C. x ln xdx = [u = ln x du = dx x = x2 2 ln x x 2 2 1 x dx = x2 2 ln x 1 2 x2, dv = xdx v = 2 ] = xdx = x 2 2 ln x x2 4 + C. Lejla Batina Version: spring 2012 Wiskunde 1 14 / 21
Applications The Definite Integral The area below y = f (x) between x = a and x = b is A = b a f (x)dx. Let f (x) and g(x) are continuous functions such that g(x) f (x) when a x b. The area between f and g and x = a and x = b is A = b a [f (x) g(x)]dx. Let f be differentiable function on [a, b]. The arc length of f (between a and b) is: L = b a 1 + (f (x) 2 )dx. Lejla Batina Version: spring 2012 Wiskunde 1 15 / 21
Examples The Definite Integral Example Compute the area below y = sin 2 xcosx between x 1 = 0 and x 2 = π 2. π 2 0 sin2 xcosxdx = [u = sin x du = cos xdx] = 1 0 u2 du = = u3 3 1 0 = 1 3. Compute the area bounded by y 2 = 4x and 4x 5y + 4 = 0. Solution: A = 9 8. Find the length of the following curve x = 1 4 y 2 1 2 ln y from y = 1 to y = e. x (y) = y 2 1 2 1 y = y 2 1 y. s = e 1 1 + (y 2 1) 2 = e y 4 +2y 2 +1 4y 2 1 2y = 1 e 2 1 (y + 1 y )dy = 1 2 ( y 2 2 + ln y) e 1 = 1 4 (e2 + 1). Lejla Batina Version: spring 2012 Wiskunde 1 16 / 21
More applications Applications of the indefinite integral Displacement and velocity formulas As v = ds dt so ds = vdt s = vdt. a = dv dt = d 2 s dt v = adt. 2 Voltage across a capacitor i = dq dt q = idt, i-current, q-charge. Applications of the definite integral Computing volumes (derived by rotation) Average value of a function For y = f (x) from x = a to x = b: y ave = b a f (x)dx b a. Lejla Batina Version: spring 2012 Wiskunde 1 17 / 21
Examples: from Interactive mathematics (www.inmath.com) Example A proton moves in an electric field such that its acceleration (in cm 2 /s) is: a(t) = 20. Find the velocity as a (1+2t) 2 function of time if v = 30 cm s when t = 0. v = adt v = 20dt (1+2t) 2 = [u = 1+2t, du = 2dt] = 10 du u 2 = 10 u +C. For t = 0 v(0) = 30 cm s so 30 = 10 + C C = 20 cm s and then v(t) = 10 1+2t + 20 cm s. The temperature T (in C) recorded during a day followed the curve T = 0.001t 4 0.280t 2 + 25, where t are hours from noon 12 t 12. Find the average temperature during a day. y ave = 12 12 T (t)dt 24 = = 15.7C. Lejla Batina Version: spring 2012 Wiskunde 1 18 / 21
Here, we make two extensions to the definition of the definite integral. The first covers integrals of functions over intervals of the form [a, ) and (, b], where a, b R (or generally over (, )). The second covers integrals of functions which have infinite discontinuities. An integral of either one of these two types is called an improper integral. Definition If f is defined on [a, ) and integrable on [a, B] for all a < B <, then we define: B a f (x)dx = lim B a f (x)dx, provided that this limit exists. Similarly, if f is defined on (, b] and integrable on [A, b] for all < A < b, then we define: b f (x)dx = lim b A A f (x)dx, provided that this limit exists. Lejla Batina Version: spring 2012 Wiskunde 1 19 / 21
Improper integrals due to discontinuities Definition If f is defined on the interval (a, b], with lim x a +f (x) =, and is integrable on every interval [C, b], where a < C < b, then we define: b a f (x)dx = lim b C a+ C f (x)dx, provided that this limit exists. Example 3 1 B dx dx = lim x 3 B 3 x 3 = lim B [ 1 2x 2 ] B 3 = 1 2 18. 1 B x dx = lim B 2 = lim B [2 B 2 2] =. = lim B x 2 2 B 3 = dx x = lim x B B 2 = 1 2 Lejla Batina Version: spring 2012 Wiskunde 1 20 / 21
Example: improper integral due discontinuity Example Find out whether or not the following improper integral exist. 1 0 1 (x 1) 2/3 dx lim α 0 1 α 0 dx (x 1) 2 3 = lim α 0 α 1 t 2/3 dt = lim α 0 3t 1/3 α 1 = = lim α 0 3 3 t α 1 = lim α 03( 3 α + 1) = 3. Lejla Batina Version: spring 2012 Wiskunde 1 21 / 21