Lecture 4: Integrals and applications

The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Lecture 4: Integrals and applications Lejla Batina Institute for Computing and Information Sciences – Digital Security Radboud University Nijmegen

Version: autumn 2013

Lejla Batina

Version: autumn 2013

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Outline

The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Lejla Batina

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Intro So far: from f to f 0 : tangent line, monotonicity, extrema, ... From f 0 to f : If F (x) is a function such that F 0 (x) = f (x), which information we get about f (from F )? (x) f (x) = F 0 (x) = limh→0 F (x+h)−F h ⇒ f (x) · h ≈ F (x + h) − F (x).

So, F (x) gives some information about the surface under the graph of f .

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

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: [x0 , x1 ], [x1 , x2 ], [x2 , x3 ], . . . , [xn−1 , xn ], where a = x0 , b = xn , each of length ∆x = b−a n (xi = a + i∆x, i = 0, · · · , n). The area Si of the strip between xi−1 and xi can be approximated as the area of the rectangle of width ∆x and height f (xi ∗ ), where xi ∗ ∈ [xi , xi+1 ], i = 0, · · · , n. So, the total area under the curve is closeP to the following sum: A ≈ ni=1 f (xi ∗ )∆x = f (x1 ∗ )∆x + f (x2 ∗ )∆x + . . . + f (xn ∗ )∆x. Rb P A = limn→∞ ni=1 f (xi ∗ )∆x = a f (x)dx. Lejla Batina

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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The evaluation theorem Theorem If f is a continuous function and F 0 (x) = f (x) (F is an Rb antiderivative of f ), then: 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: R1 • 0 x 2 dx Rπ • 02 sinxdx

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Properties of the definite integrals Due to linearity and interval additivity we get: Rb Rc Rb • a f (x)dx = a f (x)dx + c f (x)dx, where a < c < b Ra • a f (x)dx = 0 Rb Ra • a f (x)dx = − b f (x)dx Rb Rb Rb • a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx Rb • c a dx = c(b − a) Rb Rb • c a f (x)dx = a c · f (x)dx Comparison: Rb

• f (x) ≥ 0 ⇒ a f (x)dx ≥ 0 Rb Rb • f (x) ≥ g (x) ⇒ a f (x)dx ≥ a g (x)dx

Rb

• m ≤ f (x) ≤ M ⇒ m(b − a) ≤ a f (x)dx ≤ M(b − a) Lejla Batina

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Indefinite integrals

Definition A function F such that F 0 (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 Rf is called indefinite integral of f and denoted as: f (x)dx = F (x) + C .

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Table of indefinite integrals • • • • • • • • • • •

R 0 · dx = C R R 1 · dx = x + C , so dx = x R n n+1 x dx = xn+1 + C , n 6= −1 R 1 dx = ln |x| + C R xx e dx = e x + C R x ax a dx = lna +C R sinxdx = − cos x + C R cosxdx = sin x + C R 1 2 x dx = tan x + C R cos dx = arctan x + C 1+x 2 R dx √ = arcsin x + C 1−x 2

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Examples

Example R R R (3x 5 − 2x 2 + 1)dx = 3x 5 dx − 2x 2 dx + dx = R R R 6 3 = 3 x 5 dx − 2 x 2 dx + dx = 3 x6 − 2 x3 + x + C . R 2 R R √ 3 • ( x 2 − x12 )dx = x 3 dx − x −2 dx = √ 5/3 −1 3 = x 5 − x−1 + C = 53 x x 2 + x1 + C .

•

R

3

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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The fundamental theorem of calculus Theorem Let f be a continuous function on [a, b]. Then: Rx 1 The function g (x) = a f (t)dt is an antiderivative of f , i.e., g 0 (x) = f (x). 2

(Evaluation theorem) If F is an antiderivative of f , i.e. Rb F 0 (x) = f (x), then a f (x)dx = F (b) − F (a).

We can rewrite it as follows: Rx d 1 dx a f (t)dt = f (x). Rb 0 2 a F (x)dx = F (b) − F (a).

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Example

Example Find

d dx

R x2 0

t 3 dt.

R x2 We can solve this in two Let g (x) = 0 t 3 dt, then using the R u ways. theorem: for h(u) = 0 t 3 dt ⇒ h0 (u) = u 3 . We have g (x) = h(x 2 ), and from the chain rule: g 0 (x) = h0 (x 2 ) · 2x = (x 2 )3 · 2x = 2x 7 . R x2 8 4 2 Or directly: g (x) = 0 t 3 dt = [ t4 ]x0 = x4 , and then 7 g 0 (x) = 8x4 = 2x 7 .

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

The substitution rule As it holds: 0 du weRcan write: R R = u dx, f (u)u 0 dx = f (u)du = f (g (x))g 0 (x)dx, where u = g (x).

Example R cosR x3 dx = [u = x3 , du = 13 dx ⇒ dx = 3du] = cos u · 3du = = 3 cos udu = 3 sin u + C = 3 sin x3 + C . R • x sin(x 2 )dx = [u = x 2 , du = 2xdx ⇒ xdx = 12 du] R R = sin u · 21 du = 12 sin udu = − 21 cos x 2 + C . R √ • x 2 x + 1dx = [x + 1 = u, dx = du ⇒ x = u − 1] = R R 5 R 3 R 1 √ = (u − 1)2 udu = u 2 du − 2 u 2 du + u 2 du = . . . R cos √x √ dx √ √ ⇒ √ • dx = [u = x, du = 2dx = 2du] = x x x R √ = 2 cos udu = 2 sin x + C . •

R

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Integration by parts Recollect the product rule for differentiation: [f (x)g (x)]0 = f 0 (x)g (x) + f (x)g 0 (x), or f (x)g 0 (x) = [f (x)g (x)]0 − f 0 (x)g (x). After integration we get: R R f (x)g 0 (x)dx = [f (x)g (x)] − f 0 (x)g (x)dx. For u = f (x), v = g (x), we get du = f 0 (x)dx and dv = g 0 (x)dx, hence: R R udv = uv − vdu. For definite integrals we have:

Lejla Batina

Version: autumn 2013

Rb a

udv = uv |ba −

Rb a

vdu

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Examples

Example R xe x dx = [uR = x ⇒ du = dx, dv = e x dx ⇒ v = e x dx = e x ] = xe x − e x dx = xe x − e x + C . R 2 • x ln xdx = [u = ln x ⇒ du = dx , dv = xdx ⇒ v = x2 ] = x R 2 R 2 2 2 2 = x2 ln x − x2 · x1 dx = x2 ln x − 12 xdx = x2 ln x − x4 + C . •

R

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Radboud University Nijmegen

Applications

• The area below y = f (x) between x = a and x = b is

A=

Rb 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 Rb x = a and x = b is A = 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 =

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Rbp 1 + (f 0 (x)2 )dx. a

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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Examples Example • Compute the area below y = sin2 xcosx between x1 = 0 and

x2 = π2 . R π2 R1 2 2 0 sin xcosxdx = [u = sin x ⇒ du = cos xdx] = 0 u du = 3 = u3 |10 = 13 . • Compute the area bounded by y 2 = 4x and 4x − 5y + 4 = 0.

Solution: A = 89 . • Find the length of the following curve x = 41 y 2 − 12 ln y from 2

y = 1 to y = e. x 0 (y ) = y2 − 12 · y1 = y 2y−1 . Re q R e √y 4 +2y 2 +1 R (y 2 −1)2 1 e 1 s = 1 1 + 4y 2 = 1 = 2y 2 1 (y + y )dy = 1 y2 2( 2

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+ ln y )|e1 = 14 (e 2 + 1). Version: autumn 2013

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

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More applications • Applications of the indefinite integral • Displacement and velocity formulas R As v = ds ⇒ s = vdt. dt so ds = vdt R d 2s a = dv adt. dt = dt 2 ⇒ v = • Voltage across Ra capacitor i = dq idt, i-current, q-charge. dt ⇒ q = • Applications of the definite integral • Computing volumes (derived by rotation) • Average value of a function

For y =R f (x) from x = a to x = b: yave = • ···

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b a

f (x)dx b−a .

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The Definite Integral The Indefinite Integral The Fundamental Theorem of Calculus The Substitution Rule Integration by Parts

Examples:

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from “Interactive mathematics” (www.inmath.com)

Example • A proton moves in an electric field such that its acceleration 20 (in cm2 /s) is: a(t) = − (1+2t) 2 . Find the velocity as a cm function R of time if v = 30 s when t = 0. v = R adt R du 20dt v = − (1+2t) = 10 2 = [u = 1+2t, du = 2dt] = −10 u +C . u2 cm cm For t = 0 ⇒ v (0) = 30 s so 30 = 10 + C ⇒ C = 20 s and 10 then v (t) = 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. R yave = Lejla Batina

12 −12

T (t)dt 24

= · · · = 15.7C .

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