CEVA'S THEOREM

Ceva's theorem is a criterion for the concurrence of cevians in a triangle 

CEVIAN: A cevian is a line segment or ray that extends from one vertex of a polygon ( usually a triangle) to the opposite side (or the extension of that side) 

In the given diagram CD is a cevian

Statement :
Let ABC be a triangle and let D,E,F be points on lines BC,CA,AB respectively . Lines AD,BE,CF are concurrent if and only if
(BD / DC) × (CE / EA) × (AF / FB) = 1

TRIGNOMETRIC FORM 

The trignometric form of Ceva's theorem states that AD,BE,CF concur if and only if 

(sin ∠BAD / sin ∠DAC) × (sin ∠CBE / sin ∠EBA) × (sin ∠ACF / sin ∠FCB) = 1

MENELAUS THEOREM 

IN THE GIVEN TRIANGLE 

(AD/BD) X (BE/CE) X (CF/AF) = 1 

THE SINE RULE

In any triangle ABC (acute or obtuse angled) : a/sinA = b/sinB = c/sinC 

MEAN VALUE THEOREM

Statement:

If a function f(x)f(x)f(x) is continuous on the closed interval [a,b][a, b][a,b] and differentiable on the open interval (a,b)(a, b)(a,b), then there exists at least one point c∈(a,b)c \in (a, b)c∈(a,b) such that:

f′(c)=f(b)−f(a)/b−a)​

✅ Conditions:

• f(x)) must be continuous on [a,b]
  
  

• f(x) must be differentiable on (a,b)
  
  

🔍 Interpretation:

There is at least one point ccc in the interval where the instantaneous rate of change (the derivative) is equal to the average rate of change over the entire interval.

Example:

Let f(x)=x2f(x) = x^2f(x)=x2, on interval [1,3][1, 3][1,3]

1. fff is continuous and differentiable everywhere.
   
   

2. Average rate of change:
   
   

f(3)−f(1)3−1=9−12=4\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 43−1f(3)−f(1)​=29−1​=4

3. Find c such that f′(c)=4f'(c) = 4f′(c)=4:
   
   

f′(x)=2x⇒2c=4⇒c=2f'(x) = 2x \Rightarrow 2c = 4 \Rightarrow c = 2f′(x)=2x⇒2c=4⇒c=2

So, at x=2x = 2x=2, the derivative equals the average rate of change

INTERGAL CALCULUS

If f(x) is a continuous function on the interval [a,b][a, b][a,b], and we define a function F(x)F(x)F(x) as:

F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dtF(x)=∫ax​f(t)dt

Then F(x) is differentiable, and:

F′(x)=f(x)F'(x) = f(x)F′(x)=f(x) 

Conditions:

• f(x) must be continuous on [a,b][a, b][a,b]
  
  

• F(x) is defined as the definite integral of f from a to x
  
  

🔍 Interpretation:

Differentiation and integration are inverse operations. If you first integrate a function and then differentiate it, you get the original function back.

Example:

Let f(t)=sin⁡(t)f(t) = \sin(t)f(t)=sin(t), and define:

F(x)=∫0xsin⁡(t) dtF(x) = \int_0^x \sin(t)\,dtF(x)=∫0x​sin(t)dt

Then, by the Fundamental Theorem:

F′(x)=sin⁡(x)F'(x) = \sin(x)F′(x)=sin(x) 

Fundamental Theorem of Calculus (Part 2)

Statement:

If f(x)f(x)f(x) is continuous on [a,b][a, b][a,b], and F(x)F(x)F(x) is any antiderivative of f(x)f(x)f(x), then:

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)∫ab​f(x)dx=F(b)−F(a)

📌 Example:

Evaluate ∫13x2 dx\int_1^3 x^2\,dx∫13​x2dx

• Antiderivative of x2x^2x2 is F(x)=x33F(x) = \frac{x^3}{3}F(x)=3x3​
  
  

• So,
  
  

∫13x2 dx=[x33]13=273−13=263\int_1^3 x^2\,dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}∫13​x2dx=[3x3​]13​=327​−31​=326​