#Day 2

Area under a curve could be anything, negative or positive Area between two curves will always be positive

Even function has symmetry across y-axis Odd function has symmetry across origin

even \( f(-x) = f(x) \)

odd \( f(-x) = -f(x) \)

6.1 problem 32 sketch region closed by \( y = \frac{x}{x^2 + 1} \quad and \quad y = \frac{x}{5} \)

We we integrate w/ respect to x axis because one function is always top and another is always bottom we shouldn’t with y because towards the top of the function, the outer function is it’s own top and bottom

\(top \quad - \quad bottom \quad Area = 2 \int_0^2 \frac{x}{x^2 + 1} - \frac{x}{5} dx \
u = x^2 + 1 \quad du = 2xdx \quad make sure to switch boundries u(x) \
A = \int_1^5 1/u du - 2/5 \int_0^2 xdx \
ln u |_1^5 - (\frac{x^2}{5}) |_0^2 = ln5-4/5 \)

Volume y __|__b__ | | | | | | |————|—————–| <— slice |————|—————–| delta y | | | | | a | —————————————-x

\(V_slice = A_cross-section * A_y \
V_solid = \int_a^b A(y) dy \
Pyramid \quad A_pyr = 1/3 A_base height \)

Sample Problem Volume of pyramid with height 12 and square base of length 4

12 / | \ a / |
/|\ b / |
/__|__\ c 0 x

a -> b = 12 - y b -> c = y similar triangles 12x = 24 - 2y x = 2 - 1/6y

Area of slice \( = (2x)^2 = (2(2 = 1/6 y ))^2 = (4 - 1/3 y)^2 \
V = \int_0^12 (4 - 1/3 y)^2 dy \
u = 4 - 1/3 y \quad du = 1/3 dy \
-3 \int_{u(0) = 4}^{u(12) = 0} u^2 du = -3 \frac{u^3}{3} |_4^0 = 64 \)

An integral is an application of a reimann sum

Neighbourhoods are all equal density circles, find population

population density function

\( population \quad density \quad function \quad p(r) circle O \
|—-|-\Delta r-|——-R \
take tiny ring, with tiny width, find numb of people living in ring\
width is \Delta r \
Area = \pi r^2 \
Area = 2 \pi r \Delta r * p(r) \
\int_0^R 2 \pi r p(r) dr \)

Problem number 29, 6.2 Find population, if population density is given as \( p(r) = 4 (1+r^2) ^ {1/3} \\ R = 10 \quad pop =? \
Population = 2 \pi \int_0^10 4 (1+r^2) ^ {1/3} dr \
u = 1 + r^2 \quad du = 2 r dr \
4 \pi \int_1^101 u^{1/3} du \)

Average Value theorem \(A = \int_a^b f(x) dx = f(c) * (b-a) \
f(c) = \frac{1}{b - a} \int_a^b f(x) dx \) Assuming function is continuous, gives approximate value, as if it was rectangular

#Day 1

Two midterms and one final first is integration second is series final is cumulative quizes and workshps

missed quiz = 0

Review

Indefinite integrals \( \int f(x) dx = F(x) + c \) no limits

Definite integral \( \int_a^b f(x) dx \quad = A_1 - A_2 \quad (signed) \; area \; under \; graph \
\int_a^a f(x) dx = 0 \quad \quad \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx \\ \int_a^b f(x) dx = - \int_b^a f(x) dx \)

A1 is area above x-axis (positive), A2 is area below x-axis (negative)

Fudamental Theorm of Calc

I. \( \int_a^b f(x) dx = F(x) |_a^b = F(b) - F(a) \)

II. \( \frac{d}{dx} \int_a^x f(t) dt = f(x) \)

piece wise

2	| y1/\   y2
	|   /   \
	|_/___\_______
	| /1 3   4
-1	|/
	

geometry calc

(3,2) , (0,-1)

m= y2 - y1 = 2 - -1 —————— = 1 x2 - x1 = 3 - 0

y - 2 = x - 3 y = x - 1

so x intercept is 1

(1,0) triangle

\( \frac{b h}{2} = \frac{3 * 2}{ 2} = 3 = positive \quad Area \
\frac{b h}{2} = (1 * 1) / 2 \
3 - .5 = 2.5 \)

\(y_1 = x - 1 \quad y_2 = -2x+8\)

integration, need to split, negative doesn’t matter \(\int_0^3 y_1(x) dx + \int_3^4y_2(x) dx = 1.5 + 1 = 2.5 \)

diff int

\( (e^x)’ = e^x \quad \quad \int e^x dx = e^x +c \
sinx’ = cosx \int cosx = sinx + c \
cosx = -sinx sinx = -cosx +c \
tanx = sec^2 x sec^2 x = tanx + c\
x^n ‘ = nx^{n-1} x^n = x^n+1/n+1 + c \
lnx’ = 1/x 1/x = ln|x| + c\
b^x ‘ = b^x ln(b) \
arctan x’ = 1 / 1 + x^2 \
chain \quad rule \
\frac{d}{dx} [f(u)] = f’(u) * u’ \
subsitution \int f’(u) * u’ du = f(u) + c\
\)

\(

1. \int x sqrt{x^2 +9} dx \\
2. \int_2^7 \)