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Hi students, Please use the opportunity
of being here in my Padlet. Take everything as part of your reviewer for your upcoming final assessment next week. I hope you learn a lot with my Padlet. Till next time, See yahhh! Please comment below your comments or suggestion with my Padlet. Thank You! 💛💙

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Fundamental Theorem of Calculus
Let f be continuous on [a, b] and F be an antiderivative of f in the interval. then
∫b a f(x)dx = F (x)l b a = F(b) - F(a)
On the left side of the equation, an and be are called the limits of integration. and the numbers a and b are the lower and upper limits, respectively.

for example:
∫⁻¹ 3 (x²/2 - 5x + 1)dx                         Find the antiderivative
(x³/6 - 5x²/2 + x ) l⁻¹ 3
Note that in evaluating a definite integral , the arbitrary constant C does not have to be included. Just take a particular antiderivatives. this because the constant will just zero out if we evaluate F(-1) - F(3).
(x³/6 - 5x²/2 + x ) l⁻¹ 3
Apply the limits of integration then simplify (Upper limits - Lower limits)
= [(-1)³/6 - 5(-1)²/2 + (-1)] - [(3)³/6 - 5(3)²/2 + (3)]
= -11/3 - (-15)
= 34 / 3

Try Me!
1. ∫⁵ 3 (3x² - 2x + 1)dx

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Sum and Difference Rule
∫ [f(x) + g(x)]dx = ∫ [f(x)dx + ∫g(x)]dx      ADDITION
∫ [f(x) -  g(x)]dx = ∫ [f(x)dx - ∫g(x)]dx       SUBTRACTION

Example:
∫ (x + 7) dx                               Applying sum rule to split the given integral into two
= ∫ xdx + ∫ 7dx.                        Perform the integration term by term
= (x¹⁺¹ / 1+1 +c₁ ) + (x + c₂)            Simplify
= (x² / 2 + c₁ ) + (x + c₂)
= x ² / 2 + x + c
What happened to c1 and c₂? We recall that those c's stand for arbitrary constant that is, meaning they can be any real number. When we add them, what we get is another arbitrary constant that is c 1 + c 2 can be also be any real number. Thus we denote c 1 + c 2 = c

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Integrals of Trigonometric Functions
Let x be in the respective domains of the trigonometric functions. Then we have the following formulas:

1. ∫ sinx dx = -cosx + C
2. ∫ cosx dx = sinx + C
3. ∫ sec²x dx = tanx + C
4. ∫ csc²x dx = -cotx + C
5. ∫ secx tanx dx = secx + C
6. ∫ cscx cotx dx = -cscx + C

Examples:
1. ∫ sec²x / 5 dx
=  ∫ sec²x / 5 dx     Rewrite, and it gives as 1/5 as constant multiplier
= 1/5 ∫ sec²x dx     Integrate using #3 ∫ sec²x dx = tanx + C
=1/5 tanx + C        Final answer

2. ∫ (secx tanx + 4) dx
= ∫ (secx tanx + 4) dx         Apply Sum and Difference rule
= ∫ (secx tanx dx + ∫ 4 dx   Use #5 ∫ secx tanx dx = secx + C and Constant rule
= sec x + 4x + C                  Answer

3. ∫ (secx) (secx + tanx) dx
Can we perform the integration ? We notice that the given is a product of two functions, meaning we need to apply the distribution law first before we integrate. Alway note that before doing the integration, make sure to observe the given first ☺️

it gives us,
∫ (secx) (secx + tanx) dx       Use Distribution Law
= ∫ (sec²x + secxtanx) dx     Use Sum and Difference Rule
=∫ sec²x dx + ∫secxtanx dx   Use #3 and #5
= tanx + secx + C                  Answer

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Happy Valentine's Day Everyone! 💙💛

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A. Constant Rule - Suppose k is constant. Then the following is true: ∫ kdx = kx + C.

examples:
1.  ∫ 4 dx
∫ 4 dx = 4x + C
Since 4 is a constant, to find the integral using constant rule just add x + C.

2. ∫ ½ dx
∫ ½ dx = ½ x + C

B. Power Rule - ∫ x n dx = (x n+1) / (n+1) + C.
The formula is exclusive for "any variable" raised to "any constant"

examples:
1. ∫ x⁴ dx
∫ x⁴ dx = (x⁴⁺¹) / (4+1) + C
=  x ⁵ / 5 + C

2. ∫ √θ dθ
∫ √θ dθ = (θ½+1) / (½+1) + C       using law of exponent n√am = am/n
=  θ3/2 / 3/2 + C           simplify: a/b / c/d = a/b⋅d/c
= 2/3 θ 3/2 + C              answer

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Differentiation and Integration are two building blocks of calculus. Differential calculus and Integral calculus are just the opposite of each other. Differential calculus is basically dealing with the process of dividing something to get track of the changes. On the other hand, Integral calculus adds all the pieces together.
In general, the antiderivative is expressed mathematically as:
∫ f(x)dx = F(x) + C, where;

∫                - the integral sign
f(x)dx       - the integrand
x               - the variable of integration
F(x) + C    - the indefinite integral (Unknown function)
C               - the arbitrary constant

Note:
- An arbitrary constant is a constant whose value has yet to be determined.
- ∫ f(x)dx is read as "the indefinite integral of f(x) with respect to x".

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Your answer for Exercises 2 and 2.1 will serve as your attendance for this week.

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Solve the given problem
1. The air is being pumped into a spherical balloon so that its volume increases at a rate of 85 cm³/s. How fast is the radius of the balloon increasing when the radius is 25cm?

G =
A =
F =
S =
A =

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Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable x.

To do implicit differentiation, follow the steps below.
1. Differentiate both sides of the equation with respect to x.
2. Collect dy/dx on one side of the equation.
3. Solve for dy/dx.

As easy as 1,2,3. So are you ready now for the example?

EXAMPLE
1.  2x² + 3y² - 5xy = 25

Differentiate with respect to x.
2x² + 3y² - 5xy = 25

4x + 6y(dy/dx) - [5y + 5x (dy/dx)] = 0
Distribute the (-) sign

4x + 6y(dy/dx) - 5y - 5x (dy/dx) = 0
Isolate dy/dx on one side

6y(dy/dx) - 5x (dy/dx) = -4x + 5y
Factor out dy/dx

dy/dx (-5x + 6y) = -4x + 5y
Divide (-5x + 6y) on both side

dy/dx = -4x + 5y / -5x + 6y
Simplify

dy/dx = 4x -  5y / 5x - 6y
Answer

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If two related quantities are changing over time, the rates at which the quantities change are related.
For example, if a balloon is being filled with air, both the radius and the volume of the balloon are increasing. Related quantities are changing with respect to time.
In the example for balloon we can say that the rate of change in the volume (V) is related to the rate of change in radius (r). We say that dV/dt and dr/dt are related rates because V is related to radius r.

Example:

1. The air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm³/s. How fast is the radius of the balloon increasing when the radius is 36cm?

Using GAFSA in solving word problem.
Given:
volume = dV/dt = 100 cm³/s
radius = 36cm.

Ask:
Unknown is the rate of increase of the radius (dr/dt) when r = 36cm.

Formula:
Volume of sphere

V = 4/3 πr³

Solution:
Find the derivatives of both sides of the given formula.
V = 4/3 πr³

dV/dt = 3 (4/3)πr2               simplify
dV/dt = 4πr2 dr/dt              solve for the unknown quantity (dr/dt)
dr/dt = dV/dt / 4πr²        substitute the value of dV/dt and r
dr/dt = 100 / 4π (36)2     simplify
dr/dt = 100 / 4π (1296)   simplify
dr/dt = 100 / 5184π         simplify
dr/dt = 0.00614 cm/s

Answer:
Therefore the radius of the balloon is increasing at the rate of 0.00614 cm/s.

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Please comment below your student number with your favorite food.
(eg. #student102 Milktea and Pearl)

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Example no. 1
A sheet of cardboard measures 15cm by 7cm. Four equal squares are cut out of the corners and the sides are turned up to form an open rectangular box. Find the edge length of the squares that were cut out to give the box a maximum volume.

Let x be the height
y be the width
z be the length

based on the above illustration the constraint equations are;
the width = 7 - 2x
the length = 15 - 2x

The optimization equation is V = lwh
V = lwh                                   Substitute the given
= (15 - 2x) (7 - 2x) (x)        Distribute Property
= 4x ³ - 44x ² + 105x           Find the derivatives
V' (x) = 12x ² - 88x + 105        Set the derivatives equal to 0
12x ² - 88x + 105 = 0              Find the critical number/s using Factoring
(2x - 3) (6x - 35)=0                Use Principle of Zero Product
2x - 3 = 0              6x - 35 = 0
x = 3/2                   x = 35/6

since we have 2 critical numbers, we need to use second derivative test to find the maximum value.
V' (x) = 12x ² - 88x + 105
V''(x) = 24x - 88
V''(3/2) = 24(3/2) - 88
V''(3/2) = -52
V'' < 0

V' (x) = 12x ² - 88x + 105
V''(x) = 24x - 88
V''(35/6) = 24(35/6) - 88
V''(35/6) = 52
V'' > 0

NOTE:
1. If the second derivative is less than 0, then the critical number is a maximum.
2. If the second derivative is greater than 0, then the critical number is a minimum.

Therefore the maximum value is x = 3/2.

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Optimization Problem
In solving Optimization Problems, the first thing to do is to determined what is to be OPTIMIZE or MAXIMIZED. Then determine the constraint. The constraint is the quantity that is true regardless of the solution. It can be viewed as a condition that the problem must satisfy.

Example

The sum of two positive number is 42. Find these numbers if their product is a maximum.

Solution:

Let  x  - one of the two positive numbers

y  - the other positive number

x + y = 42    is the constraint equation to be satisfy

x + y = 42    Solve for the value of y
y = 42 - x

p = xy                 Optimization equation
p = xy                   Substitute the value of y
p = x ( 42 - x )       Multiply x inside the parenthesis
p = 42x - x2            Find the Derivatives
p’ = 42 - 2x            Set the derivatives equal to 0

42 - 2x = 0     Solve for x

-2x = -42

x = 21 (Critical number)

To test if the critical number is a maximum or minimum value, use the second derivative test.

p’ = 42 - 2x

p’’ = -2

p’’ < 0
Note:
1. If the second derivative is less than 0, then the critical number is a maximum.
2. If the second derivative is greater than 0, then the critical number is a minimum.

Therefore the maximum value occurs when x = 21

y = 41 - x     To find the value of y, Substitute the value of x

y = 42 - 21   Subtract

y = 21

The two numbers are 21 and 21

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Critical Number of a function

A number c in the domain of a function f said to be a critical number of f if either f’(c) = 0 or f’(c) does not exist.

Example:

1. f(x) = 2x ³  - x ⁴.

The first step is to find the derivatives of the given function.
f’(x) = 6x 2 - 4x ³

6x2 - 4x³ = 0                     Setting f’(x) equal to 0

2x²(3 - 2x) = 0                  Factoring

2x² = 0  or  3 - 2x = 0        Using the Principle of Zero Product

x = 0      or     x = 3/2

Therefore, the critical numbers are 0 and 3/2

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Please comment below your student number with your favorite color. (eg. #student1- Yellow)

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Based on the graph, we have the closed interval [a, e],

point e is the absolute maximum since it has the highest value on the graph

Point d is the absolute minimum

Point c is the local/relative maximum

Point b is the local/relative minimum

Note that if we draw a horizontal tangent line at c and d, and that is Fermat’s theorem, and according to the theorem if a function has a local maximum or local minimum value at some point and f’(c) exist then f’(c)= 0 same with f’(b)= 0.
Anytime you have first derivative function being equal to zero then those points are known as critical numbers. So b and c are critical numbers.

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Choose and answer 1 of the two items below.

1. For Critical Value
f(x) = 2x2 / x²+ 1.

2. For Optimization Problem (part 1)
Find a non-negative number that is less than or equal to 2, whose difference from its squares is a maximum.

Note: Write your answer in any kind of paper and comment it below.
(Don't be afraid to answer, I'm here to help you ❤️ 😊)

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The Extreme Value Theorem

If the function f is continuous on a closed interval I, then it has an absolute maximum value and an absolute minimum value on I.
The minimum and maximum value can be obtained either at the endpoints of the interval I or at the critical number of f in the interval.

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Exercise no. 2 ( Implicit Differentiation)

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Exercise no. 2 ( Implicit Differentiation)

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Choose and answer 1 of the five items below
1. y⁴ = 4xy
2. xy² - 2x²y = 4
3. 4x² + 9y² = 36
4. 100x² - 25y² = 100
5. x³ + 2y³ - 2x - 3y = 2

Note: Write your answer in any kind of paper and comment it below. Thank you!

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The First- Derivative Test for Relative Extrema

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The First- Derivative Test for Relative Extrema

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For any continuous function f that has exactly one critical value c in an open interval (a, b)

1. f has a relative minimum at c if f’(x) < 0 on (a, c) and f’(x) > 0 on (c, b). That is, f is decreasing to the left of c and increasing to the right of c.

2. f has a relative maximum at c if f’(x) > 0 on (a, c) and f’(x) < 0 on (c, b).

That is, f is increasing  to the left of c and decreasing to the right of c.

3. f has neither a relative maximum nor a relative minimum at c if f’ (x) has the same sign on (a, c) as on (c, b).

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HELLO STEM Students, Please be minded of the following house rules. Thank you!

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HELLO STEM Students, Please be minded of the following house rules. Thank you!

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Welcome to my BLOG,
I'm Mr. Christian H. Renegado, your online guide to help you learn Basic calculus.
Here with Padlet, you can learn a lot about basic calculus. Just seat back and relax. ENJOY!