Unit 6: Antidifferentiation (6.1-6.8)
Approximate 4 ∫ (x^2 + 3) using a right reimann sum 0
2 rectangles x^2 + 3 = y (2)^2 + 3 = 7 (4)^2 + 3 = 19 A ≈ (2)(7) + (2)(19) ≈ 52 4 rectangles x^2 + 3 = y (1)^2 + 3 = 4 (2)^2 + 3 = 7 (3)^2 + 3 = 12 (4)^2 + 3 = 19 A ≈ (4) + (7) + (12) + (19) ≈ 42
Property 1 of series
3 ∑ 5i = 5(1) + 5(2) + 5(3) = 30 i = 1 OR 3 5 ∑ i = 5(1 + 2 + 3) = 30 i = 1
∫ 1/x+1 dx
ln |x + 1| + C
How do you find the midpoint Riemann sum?
you take your table of values (this one with 2 rectangles) ex: (1, 8), (3,7), (5,10), (7,2), (9, 4) there are 5 spots and two rectangles, we will split into 1-5 and 5-9 since those are both 4 unit intervals the 1-5 interval has y values in order of 8, 7, 10 corresponding to the 1, 3, 5 x values -- the midpoint sum says we should pick the middle one which is 7 the 5-9 interval has y values in order of 10, 2, 4 corresponding to the 5, 7, 9 x values -- the midpoint sum says we should pick the middle one which is 2 (interval amount)(first midpoint value) + (interval amount)(second midpoint value) = the answer (4)(7) + (4)(2) = 36 which is the answer
4 ∫ (x^2 + 3) dx is what? 0
Δx = b - a / n = 4 - 0/n = 4/n REP = Δxi + a = 4/n i + 0 = 4/n i A ≈ n ∑ bh i = 1 n = ∑ 4/n ( f (REP) i = 1 n = ∑ 4/n ( f (4/n i) i = 1 n = 4/n ∑ [(4/n i)^2 + 3] i = 1 n = 4/n ∑ [(16/n^2/n^2 i^2) + 3] i = 1 n n = 4/n ( ∑ (16/n^2/n^2 i^2) + ∑ 3 ) i = 1 i = 1 n = 4/n ( 16/n^2/n^2 ∑ (i^2) + 3n) i = 1 = 4/n (16/n^2 * n(n+1)(2n + 1) / 6 + 3n) AFTER TONS OF SIMPLIFYING A ≈ 100n^2 + 96n + 32 / 3n^2 NOW WE USE FORMULA TO SEE APPROX A = lim (100n^2 + 96n + 32 / 3n^2) n -> infinity YOU ONLY LOOK AT THE n^2 terms = 100/3 upper and lower sum will produce same END RESULT (100/3)
∫ 1/x^2 dx
-1/x + C or -x^-1 + C
What are the three types of algebraic function solving with u-substitution?
1) Straightforward 2) finagling constant 3) finessing variable
4 ∫ |x - 3| dx 0
1/2 (3)(3) + 1/2 (1)(1) = 9/2 + 1/2 = 10/2 = 5
∫ (x^2 + x)/x dx
1/2 x^2 + x + C
3 ∫ 2xdx -2
1/2(3)(6) - 1/2(2)(4) its minus bc the second part is below the x axis 9-4 = 5 for TOTAL area it would be |2x|dx or 9+4 = 13
What is the formula for area of a trapezoid?
1/2(b1+b2)h
Property 2 of series
3 ∑ (i^2 + 2i) = plugging in all of the x values into this equation=26 i = 1 OR 3 3 ∑ i^2 + 2 ∑ i = 1 + 4 + 9 + 2(1 + 2 + 3) = 26 i = 1 i = 1
∫ (5x^2 + 7) dx
5/3 * x^3 + 7x + C
What would you do with a curve where 8 un were above x axis, 2 un below, and 16 units above?
8 + 16 - 2 = 22
Rule for n ∑ c i = 1
= c*n
∫ (5/2√x^3) dx
= ∫ (5/x^-3/2) dx = 5/2 ∫ (x^-3) dx = 5/2 (-2)(x^-1/2) + C = -5x^-1/2 + C
Definition of Area
A = lim n sum i = 1 delta x f(LEP or REP) n-> infin
Approximate 14 ∫ f(x) dx using 4 trapezoids 2
set: (2, 1), (5, 4), (8, 2), (11, 3), (14, 4) A ≈ b - a/2n (f(2) + 2f(5) + 2f(8) + 2f(11) + f(14) A ≈ 14 - 2 / 2(4) ( 1 + 2(4) + 2(2) + 2(3) + 4) A ≈ 12/8 (1 + 8 + 4 + 6 + 4) A ≈ 12/8 (23) ≈ 3/2 (23) ≈ 69/2
∫ (sec^2x) dx
tanx + C
5 ∫ √(25 - x^2) dx -5
this is a semicircle y = √25 - x^2 y^2 = 25 - x^2 x^2 + y^2 = 25 circle but its positive so it is the top half of the circle A = 1/2 pi r^2 = 25pi/2
What is an integral?
anti-differentiate
What is an indefinite integral?
antiderivative ∫ 2xdx = x^2 + C
What is a definite integral?
area under the curve 3 ∫ 2xdx --- this says what is the area under the function 1 y = 2x on [1,3] use area of a triangle to do this -- 1/2(2)(4) = 4 area of square under it = 2x2 = 4 answer = 8
What is h, or Δx equal to for trapezoids?
b - a / n = interval length/number of trapezoids
∫ 1/x dx = ln |x| + C
d/dx [ln x] = 1/x
What is the left endpoint with left Riemann sum?
for 4 ∫ (x^2 + 1)dx with 4 subintervals 0 we use the leftmost point on the rectangle, which will produce an underapproximation (1)(10) + (1)(5) + (1)(2) + (1)(1) = 18 each rectangle has a base of 1 and then you plug x in to equation to find y to multiply by this is the left Riemann sum
What is the right endpoint with left Riemann sum?
for 4 ∫ (x^2 + 1)dx with 4 subintervals 0 we use the rightmost point on the rectangle, which will produce an overapproximation (1)(17) + (1)(10) + (1)(5) + (1)(2) = 37 each rectangle has a base of 1 and then you plug x in to equation to find y to multiply by this is the right Riemann sum
What is the substitution rule?
if u = g(x) is a differentiable function and if f is continuous over the range of g(x), then ∫ f(g(x)) * g'(x) dx = ∫ f(u) du
What is u-substitution?
integration by substitution-- you let u represent the function in the parentheses and try to work backwards in an integral
∫ 3x^2 + 5/ x^2 + 5x -1 dx
its deriv of denominator / denominator = ln let u = x^3 + 5x - 1 (denom) du = (3x^2 + 5) dx = ln |u| + C = ln |x^3 + 5x - 1| + C written as ∫ du/u or ∫ 1/u du
How does the reimann sum compare to the actual area?
left reimann sum </= area </= right reimann sum
∫ (2x^2 + 1)^4 * 4xdx
let u = 2x^2 + 1 du/dx = 4x du = 4xdx = ∫ u^4 * du = 1/5 u^5 + C = 1/5 (2x^2 + 1)^5 + C
∫ 6x(3x^2 + 5)^-5 dx
let u = 3x^2 + 5 du/dx = 6x du = 6xdx = ∫ u^-5 * du = -1/4 u^-4 + C = -1/4 (3x^2 + 5)^-4 + C
∫ 15x^4 (3x^5 - 1 )^2/3 dx
let u = 3x^5 - 1 du = 15x^4 * dx = ∫ u^2/3 * du = 3/5 u^5/3 + C = 3/5 (3x^5 - 1)^5/3 + C
∫ (4x^3 + 3)^-3 * 48x^2 dx
let u = 4x^3 + 3 du = 12x^2 dx = 4 ∫ du * u^-3 = 4 * -1/2 u^-2 + C = -2(4x^3 + 3)^-2 + C
∫ 5cos(5x) dx
let u = 5x du/dx = 5 du = 5dx ∫ 5cos(5x) dx = ∫ cos(u) * du = sin(u) + C = sin(5x) + C
∫ x(x^2 + 1)^2 dx
let u = x^2 + 1 du/dx = 2x du = 2xdx now we must rewrite the problem to make x -> 2x like in the du expression = 1/2 ∫ 2x (x^2 + 1)^2 dx = 1/2 ∫ u^2 du = 1/2 * (1/3)*(u^3) + C = 1/6 u^3 + C = 1/6 (x^2 + 1)^3 + C
∫ 2x (x^2 + 1)^2 dx
let u = x^2 + 1 du/dx = 2x du = 2xdx ∫ 2x (x^2 + 1)^2 dx = ∫ u^2 * du now we integrate using power rule = 1/3 u^3 + C = 1/3 (x^2 + 1)^3 + C
∫ 3x^2 (x^3 + 1)^4 dx -- straightforward
let u = x^3 + 1 du/dx = 3x^2 du = 3x^2 dx = ∫ u^4 du = 1/5 u^5 + C = 1/5 (x^3 + 1)^5 + C
∫ 15x^2 (x^3 + 2)^1/2 dx
let u = x^3 + 2 du = 3x^2 dx = 5 ∫ 3x^2dx (x^3 + 2)^1/2 = 5 ∫ du * u^1/2 = 5 * (2/3 u^3/2) + C = 10/3 (x^3 + 2)^3/2 + C
∫ 5x^4 √(x^5 + 3) dx
let u = x^5 + 3 du = 5x^4 dx = ∫ u^1/2 du = 2/3 u^3/2 + C = 2/3 (x^5 + 3) ^3/2 + C
∫ x^2 sin(5x^3 - 7) dx
let u == 5x^3 - 7 du/dx = 15x^2 du = 15x^2 dx = 1/15 ∫ 15x^2 sin (u) dx = 1/15 ∫ sin u * du = 1/15 * -cos u + C = -1/15 cos u + C = -1/15 cos (5x^3 + 7) + C
∫ 1/x dx
ln |x| + C
Rule for (i is subscript in equation) n ∑ ai + bi (also works with - bi) i = 1
n n ∑ ai + ∑ bi also works with - bi i = 1 i = 1
What is the trapezoidal rule?
A ≈ (b - a)/2n ( f(x0) + 2f(x1) + 2f(x2) + .... + 2f(x n-1) + f(xn) ) (the numbers are subscripts for the f(x) section) ONLY WORKS WHEN ALL TRAPEZOIDS HAVE THE SAME HEIGHT
n ∑ i i = 1
[ n (n + 1) ] / 2
n ∑ i^2 i = 1
[ n(n + 1)(2n + 1) ] / 6
