Calculus study realm

1) Approximate the area under the curve
a) y = 2 - x^2, between x=0 and x=1, using 4 rectangles and left endpoint.

rectangle base: 1/4
height: 2 - x^2, where x = (i-1)/4, i from 1 to 4

1/4 Σ(2 - ((i-1)/4)^2) = 1/4(8 - 1/16 Σ(i^2 - 2i + 1)) = 2 - 1/64 ((4(4+1)(2(4)+1)/6 - 2 (4(4+1)/2) + 4) = 2 - 1/64 (30 - 20 + 4) = 57/32

It takes you so long to do one problem. -.-

That's because I'm watching Yuri no Boke at the same time.
Who the hell study so early before the exam?

You're using sigma for this? In the teacher's correction, it just adds the 4 areas.

See, Spirit always to the long way. It's why I want to see he's work.

Why do you need to know the long way when you can use the short way?

Cool! A study realm XD
I should have come here for the linear algebra exam. =.=

b) y = 3x - 7 between x=3 and x=6, using 6 rectangles and right endpoint.

rectangle base: 1/2
height: 3x - 7, where x = (i/2)+3, i from 1 to 6

1/2 Σ(3((i/2)+3) - 7) = 1/2 3Σ(i/2 + 3) - 7(6)/2 = 1/2 (3/2 6(6+1)/2 + 3(3)(6)) - 21 = 87/4

c) y = 4x^2 + 1, between x=-1 and x=2, using 3 rectangles and the midpoint

rectangle base: 1
height: 4x^2 + 1, where x = ((i-2) + (i-1))/2 = i - 3/2, 1 from 1 to 3

Σ(4(i - 3/2)^2 + 1) = 4Σ(i^2 - 3i + 9/4) + 3 = 4(3(3+1)(2(3)+1)/6 - 3(3)(3+1)/2 + (3)9/4) + 3 = 14

2) Using Riemann Sums, find the area under the curve.
a) y = 1 - x, between x=0 and x=1

rectangle base: 1/n
height: 1 - x, where x = i/n, i from 1 to n

lim(n-->infinity) 1/n Σ(1 - i/n) = lim(n-->infinity) 1/n (n - 1/n n(n+1)/2) = lim(n-->infinity) (1 - (n+1)/2n) = 1 - 1/2 = 1/2

b) y = 4x + 1, between x=1 and x=3

i from 1 to n

lim(n-->infinity) 2/n Σ(4(1 + 2i/n) + 1) = lim(n-->infinity) 2/n (+ 8/n n(n+1)/2 + 5n) = lim(n-->infinity) (8(n+1)/n + 10) = 8 +10 = 18

c) y = 2x + 3, between x = -1 and x = 0

i from 1 to n

lim(n-->infinity) 1/n Σ(2(-1 + i/n) +3) = lim(n-->infinity) 1/n (2/n n(n+1)/2 +n) = lim(n-->infinity) ((n+1)/n + 1) = 1 + 1 = 2

d) y = x^2 + 1, between x=0 and x=2

i from 1 to n

lim(n-->infinity) 2/n Σ(4i^2 /n^2 + 1) = lim(n-->infinity) 2/n (4/n^2 n(n+1)(2n+1)/6 + n) = lim(n-->infinity) 4(2n^2 + 3n +1)/3n^2 +2 = 8/3 + 2 = 14/3

e) y = x^2 + 5x + 1, between x=0 and x=3

i from 1 to n

lim(n-->infinity) 3/n Σ(9i^2 /n^2 + 15i/n + 1) = lim(n-->infinity) 3/n (9/n^2 n(n+1)(2n+1)/6 + 15/n n(n+1)/2 +n) = lim(n-->infinity) 9(2n^2 + 3n + 1)/2n^2 + 45(n+1)/2n + 3 = 9 + 45/2 + 3 = 69/2

f) y = x^3, between x=1 and x=2

i from 1 to n

lim(n-->infinity) 1/n Σ(1 + i/n)^3 = lim(n-->infinity) 1/n Σ(1 + 3i/n + 3i^2 /n^2 + i^3 /n^3) = lim(n-->infinity) 1/n (n + 3/n n(n+1)/2 + 3/n^2 n(n+1)(2n+1)/6 + 1/n^3 n^2 (n+1)^2 /4) = lim(n-->infinity) (1 + 3(n^2 +n)/3n^2 + (2n^3 + 3n^2 + n)/2n^3 + (n^4 + 2n^3 + n^2)/4n^4) = 1 + 3/2 + 1 + 1/4 = 15/4

g) y = 3x^3 + 2x^2 + x, between x=0 and x=1

i from 1 to n

lim(n-->infinity) 1/n Σ(3i^3 /n^3 + 2i^2 /n^2 + i/n) = lim(n-->infinity) (3n^2 (n+1)^2 /4n^4 + 2n(n+1)(2n+1)/6n^3 + n(n+1)/2n^2) = 3/4 + 2/3 + 1/2 = 23/12

h) y = 4x^4, between x=0 and x=1

i from 1 to n

lim(n-->infinity) 1/n 4/n^4 Σi^4 = lim(n-->infinity) 4/n^4 (n+1)(6n^3 + 9n^2 +n -1)/30 = 4/5

y = x^2, between x=a to x=b

i from 1 to n

lim(n-->infinity) (b-a)/n Σ(a+ (b-a)i/n)^2 = lim(n-->infinity) (b-a)/n Σ(a^2 +2a(b-a)i/n + (b-a)^2 i^2 /n^2) = lim(n-->infinity) (b-a)/n (na^2 + 2a(b-a)/n n(n+1)/2 + (b-a)^2 /n^2 n(n+1)(2n+1)/6) = lim(n-->infinity) ((b-a)a^2 + a(b-a)^2 /n (n+1) + (b-a)^3 /n^2 (n+1)(2n+1)/6) = (b-a)a^2 + a(b-a)^2 + (b-a)^3 /3 = ba^2 - a^3 + ab^2 -2ba^2 +a^3 + b^3 /3 - ab^2 + ba^2 - a^3 /3 = (b^3 - a^3)/3

3) Interpret as an area and evaluate.
a) ∫(2 to 2) x tanx dx = 0

b) ∫(-1 to 2) (3 - |x|)dx = ∫(-1 to 0) (3 + x)dx + ∫(0 to 2) (3 - x)dx = (3x + x^2 /2)|(-1 to 0) + (3x - x^2 /2)|(0 to 2) = 3 - 1/2 + 6 - 2 = 6.5

Yo, why did you stop without permission?

c) ∫(0 to 2) (4-y^2)^(1/2) dy --> from the function x = (4-y^2)^(1/2) ~> x^2 + y^2 = 4 --> circle

radius = 4/2 = 2

Area (first quadrant) = pi2^2 /4 = pi

d) ∫(0 to 6) (x - 2) dx

Spoiler:

Between x=0 and x=2, the function y=x-2 is negative, thus area = -(2)(2)/2 = -2
Between x=2 and x=6, area = (4)(4)/2 = 8
Total area = 8-2=6

e) ∫(0 to 6) |x - 2| dx
Between x=0 to x=2, area = 2
Between x=2 to x=6, area = 8
Total area = 8+2=10

Huh? I didn't your help.

c) Let y = 2sinθ; dy = 2cosθ dθ
∫(0 to 2) (4-y^2)^(1/2) dy = ∫(0 to π/2) (4-4(sinθ)^2)^(1/2) 2cosθ dθ = ∫(0 to π/2) 2cosθ 2cosθ dθ = ∫(0 to π/2) 4(cosθ)^2 dθ = ∫(0 to π/2) 4(1/2 + 1/2 cos2θ) dθ = (2θ - sin2θ) |(0 to π/2) = π - 0 = π

d) ∫(0 to 6) (x - 2) dx = (x^2 / 2 - 2x)|(0 to 6) = 18 - 12 = 6