# Integration. Antiderivatives and Indefinite Integration.

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Integration

Antiderivatives and Indefinite Integration

A function F is an antiderivative of f on an interval I if F’(x) = f (x) for all x in I.

If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval I if and only if G is on the form: G(x) = F(x) + C, for all x in I where C is a constant

C - is called the constant of integration G(x) = x 2 + c is the general solution of the differential equation G’(x) = 2x Integration is the “inverse” of Differentiation

Is an equation that involves x, y and derivatives of y. EXAMPLES y’ = 3x and y’ = x 2 + 1

When solving a differential equation of the form dy/dx = f(x) you can write dy = f(x)dx and is called antidifferentiation and is denoted by an integral sign ∫

y = ∫ f(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

y = ∫ F’(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

DIFFERENTIATION FORMULA INTEGRATION FORMULA  d/dx [c] = 0  d/dx [kx] = k  d/dx [kf(x)] = kf’(x)  d/dx [f(x) ± g(x)] = f’(x) ± g’(x)  d/dx [x n ] = nx n-1  ∫ 0 dx = c  ∫ kdx = kx + c  ∫ kf(x)dx = k ∫ f(x) dx  ∫ f(x) ± g(x)]dx = ∫ f(x) dx ± ∫ g(x) dx  ∫ x n dx = (x n+1 )/(n+1) + c, n≠ - 1  (Power Rule)

1. ∫ 3x dx 2. ∫ 1/x 3 dx 3. ∫ (x + 2) dx 4. ∫ (x + 1)/√x dx

EXAMPLE F’(x) = 1/x 2, x > 0 and find the particular solution that satisfies the initial condition F(1) = 0

1. Find the general solution by integrating, - 1/x + c. x > 0 2. Use initial condition F(1) = 0 and solve for c, F(1) = -1/1 + c, so c = 1 3. Write the particular solution F(x) = - 1/x + 1, x > 0

A ball is thrown upward with an initial velocity of 64 ft/sec from an initial height of 80 ft. a) Find the position function giving the height s as a function of the time t b) When does the ball hit the ground?

a) Let t = 0 represent the initial time; s(0) = 80 and s’(0) = 64 b) Use -32 ft/sec as the acceleration due to gravity, then s”(t) = - 32 c) ∫ s”(t) dt = ∫ -32 dt = -32 t + c d) s’(0) =64 = -32(0) = c, so c = 64 e) s(t) = ∫ s’(t) = ∫ (-32t + 64) dt = -16t 2 + 64t + C f) s(0) = 80 = -16(0 2 ) + 64(0) + C, hence C = 80 g) s(t) = -16t 2 + 64t + 80 = 0, solve and t = 5

Area

The sum of n terms a 1, a 2, a 3 …,a n is written as n ∑ a i = a 1 + a 2 + a 3 + …+ a n i = 1 Where i the index of summation, a 1 is the i th term of the sum, and the upper and lower bounds of summation are n and 1, respectively.

6 ∑ i= 1 + 2 + 3 + 4 + 5 + 6 i = 1 5 ∑ ( i + 1)= 1 + 2 + 3 + 4 + 5 + 6 i = 0 7 ∑ j 2 = 9 + 16 + 25 + 36 + 49 j = 3

n 1. ∑ c = cn i = 1 n 2. ∑ i = n(n + 1)/2 i = 1 n 3. ∑ i 2 = n(n + 1)(n + 2)(2n + 1)/6 i= 1 n 4. ∑ i 3 = n 2 (n + 1) 2 /4 i = 1

n n 1. ∑ ka i = k ∑ a i i = 1 n n n 2. ∑ (a i ± b i ) = ∑ a i ± ∑ b i i = 1 i = 1 i = 1

Find the sum for n = 10 and n = 100 n 1. ∑ ( i + 1)/n 2 i = 1

Find the area of the region lying between the graph of f(x) = - x 2 + 5 and the x-axis between x = 0 and x = 2 using five rectangles to find an approximation of the area. You should use both inscribed rectangles and circumscribed rectangles. In doing so you will be able to find a lower and upper sum.

Let f be continuous and nonnegative on the interval [a,b]. The limits as n→∞ of both the lower and upper sums exist and are equal to each other. That is, n lim s( n ) = lim ∑ f (m i )  x and n→∞ n→∞ i = 1

n lim s( n ) = lim ∑ f (M i )  x n→∞ n→∞ i = 1 n lim S( n ) = lim ∑ f (Mi)  x n→∞ n→∞ i = 1

Let f be continuous and nonnegative on the interval [ a,b ]. The area of the region bounded by the graph of f, the x-axis, and the vertical lines x= a and x = b is n Area = lim ∑ f (c i )  x, x i -1  c i  x i n→∞ i = 1 Where  x = (b-a)/n

Fine the area of the region bounded by the graph of f(x) =x 3, the x-axis, and the vertical lines x=0 and x =1. 1. Partition the interval [0,1] into n subintervals each of width 1/n =  x 2. Simplify using the formula below and A = 1/4 n Area = lim ∑ f (c i )  x, x i -1  c i  x i n→∞ i = 1 Where  x = (b-a)/n

Riemann Sums and Definite Integrals

Let f be defined on the closed interval [ a, b ] and let  be a partition of [ a,b ] given by a = x o < x 1 < x 2 < …<x n-1 <x n =b Where  x i is the width of the i th subinterval. If c i is any point in the i th subinterval, then the sum n  f ( c i )  x i, x i-1  c i  x i is called a Riemann i = 1 sum of f for the partition 

The width of the largest subinterval of a partition  is the norm of the partition and is denoted by   . If every subinterval is of equal width, the partition is regular and the norm is denoted by    =  x = (b – a)/n

If f is defined on the closed interval [ a, b ] and the limit n lim ∑ f (c i )  x    → 0 i = 1 exists, then f is integrable on [ a,b ] and the limit is b ∫ f(x) dx a The number a is the lower limit of integration, and the number b is the upper limit of integration

If a function f is continuous on the closed interval [a,b], then f is integrable on ]a,b]

Evaluate the definite integral 1 ∫ 2xdx -2

If f is continuous and nonnegative closed interval [ a, b ] the the area of the region bounded by the graph of f, the x-axis, and the vertical lines x =a and x = b is given by b ∫ f(x) dx a

Evaluate the Definite Integral 0 ∫ ( x + 2)dx 3 Sketch the region and use formula for trapezoid

If f is integrable on the three closed intervals determined by a, b and c, then, b c b ∫ f(x) dx = ∫ f(x) dx = ∫ f(x) dx a a c

If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and b b 1. ∫ kf dx = k ∫ f(x) dx a a

If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and b b b 1. ∫ [ f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ f(x) dx a a a

THE FUNDAMENTAL THEOREM OF CALCULUS

If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then b ∫ f(x) dx = F(b) – F(a) a

1. Find the antiderivative of f if possible 2. Evaluate the definite integral Example: ∫ x 3 dx on the interval [1,3]

Find the area of the region bounded by the graph of y = 2x 2 – 3x +2, the x-axis, and the vertical lines x=0 and x= 2. 1. Graph 2. Find the antiderivative 3. Evaluate on your interval

If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that ∫ f(x)dx = f(c)(b-a)

If f is integrable on the closed interval [a,b], then the average value of f on the interval is b 1/(b-a) ∫ f(x)dx a

If f is continuous on an open interval I containing a, then, for every x in the interval x d/dx [ ∫ f(t) dt ] = f(x) a

INTEGRATION BY SUBSTITUTION

Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then ∫ f(g(x))g’(x)dx = F(g(x)) + C If u = g(x), then du = g’(x)dx and ∫ f(u)du = F(u) + C

You completely rewrite the integral in terms of u and du. This is useful technique for complicated intergrands. ∫ f(g(x))g’(x)dx = ∫ f(u) du = F(u)+ C

Find ∫ (2x – 1).5 dx Let u = 2x - 1, then du/dx = 2dx/dx Solve for dx and substitute back to obtain the antiderivative. Check your answer.

If g is a differentiable function of x, then, ∫ ((g(x)) n g’(x)dx = ∫ (g(x)) n+1 /(n+1) + C

If the function u = g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then, b g(b) ∫ (g(x)g’(x)dx = ∫ f(u)du a g(a)

Let f be integrable on the closed interval [ -a,a]. If f is an even function, then a a ∫ f(x) dx=2 ∫ f(x) dx -a 0

Let f be integrable on the closed interval [ -a,a]. If f is an odd function, then a ∫ f(x) dx= ∫ f(x) dx = 0 -a

NUMERICAL INTEGRATION

Let f be continuous on [a,b]. The trapezoidal Rule for approximating ∫ f(x) dx  (b-a)/2n [f(x 0 ) = 2(f(x 1 ) +…..+2f(x n-1 ) + f(x n )]

If p(x) = Ax2 + Bx + c, then b ∫ p(x) dx = a (b-a)/6 [p(a) + 4p[(a+b)]/2) + p(b)]

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