2.1 Limits and continuity of a function

Limits of a Function

A limit describes the behavior of a function as its input approaches a particular value. It allows us to understand what value a function is approaching, even if it does not actually reach that value.

Formal Definition of a Limit:

lim⁡x→cf(x)=L\lim_{{x \to c}} f(x) = Lx→clim​f(x)=L

This means that as $x$ approaches $c$, the function $f(x)$ approaches the value $L$. The limit exists if and only if $f(x)$ approaches the same value $L$ from both the left-hand side and right-hand side of $c$.

• Left-hand limit: lim⁡x→c−f(x)=L\lim_{{x \to c^-}} f(x) = Lx→c−lim​f(x)=L
• Right-hand limit: lim⁡x→c+f(x)=L\lim_{{x \to c^+}} f(x) = Lx→c+lim​f(x)=L

If lim⁡x→c−f(x)=lim⁡x→c+f(x)=L\lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = Llimx→c−​f(x)=limx→c+​f(x)=L, the limit exists and equals $L$.

Example: Consider f(x)=x2−1x−1f(x) = \frac{x^2 – 1}{x – 1}f(x)=x−1x2−1​. At $x=1$, the function is undefined (division by zero), but we can compute the limit as $x$ approaches 1:

lim⁡x→1×2−1x−1=lim⁡x→1(x−1)(x+1)x−1=lim⁡x→1(x+1)=2\lim_{{x \to 1}} \frac{x^2 – 1}{x – 1} = \lim_{{x \to 1}} \frac{(x – 1)(x + 1)}{x – 1} = \lim_{{x \to 1}} (x + 1) = 2x→1lim​x−1x2−1​=x→1lim​x−1(x−1)(x+1)​=x→1lim​(x+1)=2

Even though $f(x)$ is not defined at $x=1$, the limit exists and equals 2.

Continuity of a Function

A function is continuous at a point $x=c$ if the following three conditions are satisfied:

1. The function is defined at $c$: $f(c)$ exists.
2. The limit exists at $c$: lim⁡x→cf(x)\lim_{{x \to c}} f(x)limx→c​f(x) exists.
3. The limit equals the function value: lim⁡x→cf(x)=f(c)\lim_{{x \to c}} f(x) = f(c)limx→c​f(x)=f(c).

If a function is continuous at every point in its domain, it is said to be continuous on its domain.

Example: Consider f(x)=x2f(x) = x^2f(x)=x2.

• The function is defined for all $x$.
• The limit as $x$ approaches any value $c$ is: lim⁡x→cx2=c2\lim_{{x \to c}} x^2 = c^2x→clim​x2=c2
• The function value is f(c)=c2f(c) = c^2f(c)=c2.

Since lim⁡x→cx2=c2=f(c)\lim_{{x \to c}} x^2 = c^2 = f(c)limx→c​x2=c2=f(c), the function is continuous for all real values of $x$.

Types of Discontinuities

1. Removable Discontinuity: Occurs when the limit exists, but the function is either undefined at that point or the function’s value does not match the limit. It can often be “fixed” by redefining the function at that point.
2. Jump Discontinuity: Occurs when the left-hand and right-hand limits exist but are not equal.
3. Infinite Discontinuity: Occurs when the function approaches infinity as $x$ approaches a particular point.

2.1.1 ε − δ definition of limit and continuity.

$\epsilon -\delta$ Definition of Limit and Continuity

The $\epsilon -\delta$ (epsilon-delta) definition is a formal and rigorous way to define the limit of a function. This method gives a precise description of what it means for a function to approach a particular value as its input gets close to some point.

$\epsilon -\delta$ Definition of a Limit

Let $f(x)$ be a function, and let $L$ be a real number. We say that the limit of $f(x)$ as $x$ approaches $c$ is $L$, written as:

lim⁡x→cf(x)=L\lim_{x \to c} f(x) = Lx→clim​f(x)=L

If for every $\epsilon >0$ (no matter how small), there exists a $\delta >0$ such that whenever $0<|x-c|<\delta$, it follows that $|f(x)-L|<\epsilon$.

• $\epsilon$ represents how close we want $f(x)$ to be to $L$.
• $\delta$ represents how close $x$ needs to be to $c$ to ensure $f(x)$ is within $\epsilon$ of $L$.

In simpler terms:

• For every small tolerance $\epsilon$, we can find a small distance $\delta$ such that when $x$ is within $\delta$ units of $c$ (but not equal to $c$), the value of $f(x)$ is within $\epsilon$ units of $L$.

Formal Statement:

$$\forall \epsilon >0,\exists \delta >0\text{ such that if }0<|x-c|<\delta ,\text{ then }|f(x)-L|<\epsilon .$$

Example:

Let’s prove that:

lim⁡x→2(3x+1)=7\lim_{x \to 2} (3x + 1) = 7x→2lim​(3x+1)=7

We need to show that for any $\epsilon >0$, there exists a $\delta >0$ such that if $0<|x-2|<\delta$, then $|(3x+1)-7|<\epsilon$.

1. Start with $|(3x+1)-7|$:

   $$|(3x+1)-7|=|3x-6|=3|x-2|$$

2. We want $3|x-2|<\epsilon$, so divide both sides by 3:

   ∣x−2∣<ε3|x – 2| < \frac{\varepsilon}{3}∣x−2∣<3ε​

Thus, we can choose δ=ε3\delta = \frac{\varepsilon}{3}δ=3ε​. Therefore, for any $\epsilon >0$, if ∣x−2∣<δ=ε3|x – 2| < \delta = \frac{\varepsilon}{3}∣x−2∣<δ=3ε​, it follows that $|(3x+1)-7|<\epsilon$. This proves that lim⁡x→2(3x+1)=7\lim_{x \to 2} (3x + 1) = 7limx→2​(3x+1)=7.

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$\epsilon -\delta$ Definition of Continuity

A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are satisfied:

1. $f(c)$ is defined (i.e., the function has a value at $c$).
2. lim⁡x→cf(x)\lim_{x \to c} f(x)limx→c​f(x) exists.
3. The limit equals the function value, lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c)limx→c​f(x)=f(c).

Using the $\epsilon -\delta$ definition, the function $f(x)$ is continuous at $c$ if for every $\epsilon >0$, there exists a $\delta >0$ such that:

$$\forall \epsilon >0,\exists \delta >0\text{ such that if }|x-c|<\delta ,\text{ then }|f(x)-f(c)|<\epsilon .$$

This means that the values of $f(x)$ get arbitrarily close to $f(c)$ as $x$ gets arbitrarily close to $c$, ensuring the function is smooth and without jumps at that point.

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Example of Continuity:

Let’s check if the function f(x)=x2f(x) = x^2f(x)=x2 is continuous at $x=1$.

1. The value of the function at $x=1$ is f(1)=12=1f(1) = 1^2 = 1f(1)=12=1.
2. The limit of $f(x)$ as $x$ approaches 1 is: lim⁡x→1×2=1\lim_{x \to 1} x^2 = 1x→1lim​x2=1

Now, for any $\epsilon >0$, we want to find $\delta$ such that if $|x-1|<\delta$, then ∣x2−1∣<ε|x^2 – 1| < \varepsilon∣x2−1∣<ε.

∣x2−1∣=∣(x−1)(x+1)∣|x^2 – 1| = |(x – 1)(x + 1)|∣x2−1∣=∣(x−1)(x+1)∣

We want:

$$|(x-1)(x+1)|<\epsilon$$

To control $|x+1|$, note that if $|x-1|<1$, then $0<x<2$, and thus $|x+1|<3$. So, we can require:

∣x−1∣⋅3<ε⇒∣x−1∣<ε3|x – 1| \cdot 3 < \varepsilon \quad \Rightarrow \quad |x – 1| < \frac{\varepsilon}{3}∣x−1∣⋅3<ε⇒∣x−1∣<3ε​

Thus, we can choose δ=ε3\delta = \frac{\varepsilon}{3}δ=3ε​. Therefore, f(x)=x2f(x) = x^2f(x)=x2 is continuous at $x=1$.

 

2.1.2 Evaluate the limit of the function

It seems like you want to evaluate the limit of a function but haven’t specified which function. Here are a few common examples of how to evaluate limits:

Example 1: Polynomial Limit

Evaluate the limit:

lim⁡x→3(2×2+5)\lim_{{x \to 3}} (2x^2 + 5)x→3lim​(2x2+5)

Solution: Substituting $x=3$:

=2(3)2+5=2⋅9+5=18+5=23= 2(3)^2 + 5 = 2 \cdot 9 + 5 = 18 + 5 = 23=2(3)2+5=2⋅9+5=18+5=23

So,

lim⁡x→3(2×2+5)=23\lim_{{x \to 3}} (2x^2 + 5) = 23x→3lim​(2x2+5)=23

Example 2: Rational Function Limit

Evaluate the limit:

lim⁡x→2×2−4x−2\lim_{{x \to 2}} \frac{x^2 – 4}{x – 2}x→2lim​x−2x2−4​

Solution: First, substitute $x=2$:

=22−42−2=00 (indeterminate form)= \frac{2^2 – 4}{2 – 2} = \frac{0}{0} \text{ (indeterminate form)}=2−222−4​=00​ (indeterminate form)

Factor the numerator:

=(x−2)(x+2)x−2= \frac{(x – 2)(x + 2)}{x – 2}=x−2(x−2)(x+2)​

Now cancel $x-2$ (valid for x≠2x \neq 2x=2):

$$=x+2$$

Now evaluate the limit:

lim⁡x→2(x+2)=2+2=4\lim_{{x \to 2}} (x + 2) = 2 + 2 = 4x→2lim​(x+2)=2+2=4

So,

lim⁡x→2×2−4x−2=4\lim_{{x \to 2}} \frac{x^2 – 4}{x – 2} = 4x→2lim​x−2x2−4​=4

Example 3: Trigonometric Limit

Evaluate the limit:

lim⁡x→0sin⁡xx\lim_{{x \to 0}} \frac{\sin x}{x}x→0lim​xsinx​

Solution: This is a well-known limit:

lim⁡x→0sin⁡xx=1\lim_{{x \to 0}} \frac{\sin x}{x} = 1x→0lim​xsinx​=1

 

2.1.3 Continuity of the function

2.1.3 Continuity of the Function

To determine the continuity of a function at a specific point, we need to check the following three conditions for the function $f(x)$ at the point $x=c$:

1. The function is defined at $c$: $f(c)$ exists.
2. The limit exists at $c$: lim⁡x→cf(x)\lim_{{x \to c}} f(x)limx→c​f(x) exists.
3. The limit equals the function value: lim⁡x→cf(x)=f(c)\lim_{{x \to c}} f(x) = f(c)limx→c​f(x)=f(c).

If all these conditions are met, the function is continuous at $x=c$. If the function is continuous at every point in its domain, it is said to be continuous on that domain.

Example: Analyzing Continuity of a Function

Let’s consider the function $f(x)$:

f(x)={x2if x<12if x=13−xif x>1f(x) = \begin{cases} x^2 & \text{if } x < 1 \\ 2 & \text{if } x = 1 \\ 3 – x & \text{if } x > 1 \end{cases}f(x)=⎩⎨⎧​x223−x​if x<1if x=1if x>1​

Step 1: Check Continuity at $x=1$

1. Is $f(1)$ defined?

   $$f(1)=2(\text{function is defined})$$

2. Does the limit exist at $x=1$?
   • Left-hand limit:

   lim⁡x→1−f(x)=lim⁡x→1−x2=12=1\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} x^2 = 1^2 = 1x→1−lim​f(x)=x→1−lim​x2=12=1

   • Right-hand limit:

   lim⁡x→1+f(x)=lim⁡x→1+(3−x)=3−1=2\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} (3 – x) = 3 – 1 = 2x→1+lim​f(x)=x→1+lim​(3−x)=3−1=2Since the left-hand limit (1) does not equal the right-hand limit (2), the overall limit does not exist at $x=1$:

   lim⁡x→1f(x) does not exist.\lim_{{x \to 1}} f(x) \text{ does not exist.}x→1lim​f(x) does not exist.

3. Does the limit equal the function value?
   • Since the limit does not exist, we cannot satisfy the third condition.

Conclusion: The function $f(x)$ is not continuous at $x=1$ because the limit does not exist at that point.

Checking Continuity at Another Point

Let’s check the continuity of the same function at $x=0$.

1. Is $f(0)$ defined?

   f(0)=02=0f(0) = 0^2 = 0f(0)=02=0

2. Does the limit exist at $x=0$?
   • Left-hand limit:

   lim⁡x→0−f(x)=lim⁡x→0−x2=02=0\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} x^2 = 0^2 = 0x→0−lim​f(x)=x→0−lim​x2=02=0

   • Right-hand limit:

   lim⁡x→0+f(x)=lim⁡x→0+x2=02=0\lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} x^2 = 0^2 = 0x→0+lim​f(x)=x→0+lim​x2=02=0Thus,

   lim⁡x→0f(x)=0.\lim_{{x \to 0}} f(x) = 0.x→0lim​f(x)=0.

3. Does the limit equal the function value?

   lim⁡x→0f(x)=0=f(0).\lim_{{x \to 0}} f(x) = 0 = f(0).x→0lim​f(x)=0=f(0).

Conclusion: The function $f(x)$ is continuous at $x=0$.

 

2.2 Derivatives of function

2.2.1 Definition and Geometrical Interpretation of Derivatives

Definition of the Derivative

The derivative of a function $f(x)$ at a point $x=a$ is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. Mathematically, the derivative f′(a)f'(a)f′(a) is given by:

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{{h \to 0}} \frac{f(a + h) – f(a)}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

• f′(a)f'(a)f′(a) is the derivative of the function $f(x)$ at the point $a$.
• $h$ is a small increment in $x$.
• The expression f(a+h)−f(a)h\frac{f(a + h) – f(a)}{h}hf(a+h)−f(a)​ represents the average rate of change of the function between the points $a$ and $a+h$.

Interpretation of the Derivative

1. Slope of the Tangent Line: The derivative at a point gives the slope of the tangent line to the graph of the function at that point. If you think of the function as a curve, the derivative tells us how steep the curve is at any given point.
2. Rate of Change: The derivative represents how much the function $f(x)$ changes as $x$ changes. If $f(x)$ describes a physical quantity (like position), then the derivative f′(x)f'(x)f′(x) describes the rate of change of that quantity (like velocity).

Geometrical Interpretation

To visualize the concept of the derivative, consider the following:

• Graph of the Function: Imagine a curve representing the function $f(x)$.
• Point of Interest: Pick a point $(a,f(a))$ on the curve.
• Secant Line: Draw a secant line connecting two points $(a,f(a))$ and $(a+h,f(a+h))$. The slope of this secant line is given by the average rate of change:

slope of secant=f(a+h)−f(a)h\text{slope of secant} = \frac{f(a + h) – f(a)}{h}slope of secant=hf(a+h)−f(a)​

• Tangent Line: As $h$ approaches 0 (i.e., as the second point approaches the point $(a,f(a))$), the secant line approaches the tangent line at $(a,f(a))$. The slope of the tangent line is the derivative f′(a)f'(a)f′(a).

In this image, the blue secant line represents the average rate of change between two points on the curve, while the red tangent line represents the instantaneous rate of change at point $a$.

Example:

Consider the function f(x)=x2f(x) = x^2f(x)=x2.

1. Calculate the Derivative:

   f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0x2+2xh+h2−x2h=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2xf'(x) = \lim_{{h \to 0}} \frac{(x + h)^2 – x^2}{h} = \lim_{{h \to 0}} \frac{x^2 + 2xh + h^2 – x^2}{h} = \lim_{{h \to 0}} \frac{2xh + h^2}{h} = \lim_{{h \to 0}} (2x + h) = 2xf′(x)=h→0lim​h(x+h)2−x2​=h→0lim​hx2+2xh+h2−x2​=h→0lim​h2xh+h2​=h→0lim​(2x+h)=2x

2. Interpret the Derivative:
   • At any point $x$, the derivative f′(x)=2xf'(x) = 2xf′(x)=2x gives the slope of the tangent line to the curve y=x2y = x^2y=x2 at that point. For example:
     • At $x=1$, f′(1)=2f'(1) = 2f′(1)=2, which means the slope of the tangent line at $(1,1)$ is 2.
     • At $x=2$, f′(2)=4f'(2) = 4f′(2)=4, which means the slope of the tangent line at $(2,4)$ is 4.

2.2.2 Derivatives of the functions (Algebraic only)

The derivative of a function describes how the function changes as its input changes. Below are the basic rules and formulas for finding the derivatives of common algebraic functions.

1. Power Rule

For any function of the form f(x)=xnf(x) = x^nf(x)=xn, where $n$ is a real number, the derivative is given by:

f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1

Example:

f(x)=x3  ⟹  f′(x)=3x2f(x) = x^3 \implies f'(x) = 3x^{2}f(x)=x3⟹f′(x)=3x2

2. Constant Rule

The derivative of a constant $c$ is zero:

f(x)=c  ⟹  f′(x)=0f(x) = c \implies f'(x) = 0f(x)=c⟹f′(x)=0

Example:

f(x)=5  ⟹  f′(x)=0f(x) = 5 \implies f'(x) = 0f(x)=5⟹f′(x)=0

3. Sum Rule

If $f(x)$ and $g(x)$ are differentiable functions, then the derivative of their sum is:

(f+g)′(x)=f′(x)+g′(x)(f + g)'(x) = f'(x) + g'(x)(f+g)′(x)=f′(x)+g′(x)

Example:

f(x)=x2+3x  ⟹  f′(x)=2x+3f(x) = x^2 + 3x \implies f'(x) = 2x + 3f(x)=x2+3x⟹f′(x)=2x+3

4. Difference Rule

If $f(x)$ and $g(x)$ are differentiable functions, then the derivative of their difference is:

(f−g)′(x)=f′(x)−g′(x)(f – g)'(x) = f'(x) – g'(x)(f−g)′(x)=f′(x)−g′(x)

Example:

f(x)=x2−4x  ⟹  f′(x)=2x−4f(x) = x^2 – 4x \implies f'(x) = 2x – 4f(x)=x2−4x⟹f′(x)=2x−4

5. Product Rule

If $u(x)$ and $v(x)$ are differentiable functions, then the derivative of their product is:

(uv)′=u′v+uv′(uv)’ = u’v + uv’(uv)′=u′v+uv′

Example:

f(x)=x2(3x)  ⟹  u=x2, v=3xf(x) = x^2(3x) \implies u = x^2, \, v = 3xf(x)=x2(3x)⟹u=x2,v=3x u′=2x, v′=3  ⟹  f′(x)=(2x)(3x)+(x2)(3)=6×2+3×2=9x2u’ = 2x, \, v’ = 3 \implies f'(x) = (2x)(3x) + (x^2)(3) = 6x^2 + 3x^2 = 9x^2u′=2x,v′=3⟹f′(x)=(2x)(3x)+(x2)(3)=6x2+3x2=9x2

6. Quotient Rule

If $u(x)$ and $v(x)$ are differentiable functions, then the derivative of their quotient is:

(uv)′=u′v−uv′v2\left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2}(vu​)′=v2u′v−uv′​

Example:

f(x)=x23x  ⟹  u=x2, v=3xf(x) = \frac{x^2}{3x} \implies u = x^2, \, v = 3xf(x)=3xx2​⟹u=x2,v=3x u′=2x, v′=3  ⟹  f′(x)=(2x)(3x)−(x2)(3)(3x)2=6×2−3x29x2=3x29x2=13u’ = 2x, \, v’ = 3 \implies f'(x) = \frac{(2x)(3x) – (x^2)(3)}{(3x)^2} = \frac{6x^2 – 3x^2}{9x^2} = \frac{3x^2}{9x^2} = \frac{1}{3}u′=2x,v′=3⟹f′(x)=(3x)2(2x)(3x)−(x2)(3)​=9x26x2−3x2​=9x23x2​=31​

7. Chain Rule

If $g(x)$ is a differentiable function and $f(x)=g(h(x))$ is a composite function, then the derivative is given by:

f′(x)=g′(h(x))⋅h′(x)f'(x) = g'(h(x)) \cdot h'(x)f′(x)=g′(h(x))⋅h′(x)

Example:

f(x)=(x2+1)3f(x) = (x^2 + 1)^3f(x)=(x2+1)3

Let g(u)=u3g(u) = u^3g(u)=u3 and h(x)=x2+1h(x) = x^2 + 1h(x)=x2+1:

g′(u)=3u2, h′(x)=2x  ⟹  f′(x)=3(h(x))2⋅h′(x)=3(x2+1)2⋅(2x)=6x(x2+1)2g'(u) = 3u^2, \, h'(x) = 2x \implies f'(x) = 3(h(x))^2 \cdot h'(x) = 3(x^2 + 1)^2 \cdot (2x) = 6x(x^2 + 1)^2g′(u)=3u2,h′(x)=2x⟹f′(x)=3(h(x))2⋅h′(x)=3(x2+1)2⋅(2x)=6x(x2+1)2

 

2.2.3 Maxima and Minima of Functions

Maxima and minima refer to the highest and lowest points on a function’s graph, respectively. These points are critical for understanding the behavior of functions, especially in optimization problems.

1. Definition

• Maximum: A function $f(x)$ has a local maximum at $x=c$ if $f(c)\ge f(x)$ for all $x$ in some neighborhood of $c$.
• Minimum: A function $f(x)$ has a local minimum at $x=c$ if $f(c)\le f(x)$ for all $x$ in some neighborhood of $c$.

Global Maximum/Minimum: A global maximum (or minimum) is the highest (or lowest) value of the function over its entire domain.

2. Finding Maxima and Minima

To find the local maxima and minima of a function, we typically follow these steps:

1. Find the Derivative: Calculate the derivative f′(x)f'(x)f′(x) of the function $f(x)$.
2. Set the Derivative to Zero: Solve f′(x)=0f'(x) = 0f′(x)=0 to find the critical points. These points are where the function’s slope is zero and could potentially be maxima or minima.
3. Test Critical Points: Use the second derivative test or the first derivative test to determine whether each critical point is a maximum, minimum, or neither.
   • Second Derivative Test:
     • If f′′(c)>0f”(c) > 0f′′(c)>0, then $f(x)$ has a local minimum at $x=c$.
     • If f′′(c)<0f”(c) < 0f′′(c)<0, then $f(x)$ has a local maximum at $x=c$.
     • If f′′(c)=0f”(c) = 0f′′(c)=0, the test is inconclusive.
   • First Derivative Test:
     • Analyze the sign of f′(x)f'(x)f′(x) on intervals around the critical points.
     • If f′(x)f'(x)f′(x) changes from positive to negative at $c$, $c$ is a local maximum.
     • If f′(x)f'(x)f′(x) changes from negative to positive at $c$, $c$ is a local minimum.

3. Example: Finding Maxima and Minima

Example Function: Let’s analyze the function f(x)=−x2+4xf(x) = -x^2 + 4xf(x)=−x2+4x.

1. Find the Derivative:

   f′(x)=−2x+4f'(x) = -2x + 4f′(x)=−2x+4

2. Set the Derivative to Zero:

   $$-2x+4=0⟹2x=4⟹x=2$$

3. Test Critical Points:
   • Second Derivative Test: f′′(x)=−2f”(x) = -2f′′(x)=−2 Since f′′(x)<0f”(x) < 0f′′(x)<0, $f(x)$ has a local maximum at $x=2$.
4. Find the Maximum Value:

   f(2)=−(2)2+4(2)=−4+8=4f(2) = -(2)^2 + 4(2) = -4 + 8 = 4f(2)=−(2)2+4(2)=−4+8=4Thus, the function has a local (and global) maximum at the point $(2,4)$.

 

2.3 Indefinite and Definite Integral

Integration is a fundamental concept in calculus that is essentially the reverse process of differentiation. It is used to calculate areas under curves, among many other applications.

1. Indefinite Integrals

An indefinite integral represents a family of functions whose derivative is the integrand. It is written in the form:

$$\int f(x)dx=F(x)+C$$

where:

• $\int$ denotes the integral sign.
• $f(x)$ is the integrand (the function being integrated).
• $F(x)$ is the antiderivative of $f(x)$ (the function whose derivative is $f(x)$).
• $C$ is the constant of integration, representing any constant value, since the derivative of a constant is zero.

Example: To find the indefinite integral of f(x)=3x2f(x) = 3x^2f(x)=3x2:

∫3×2 dx=x3+C\int 3x^2 \, dx = x^3 + C∫3x2dx=x3+C

2. Definite Integrals

A definite integral computes the accumulation of a quantity over a specified interval $[a,b]$. It is represented as:

∫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)

where:

• $F(x)$ is any antiderivative of $f(x)$.
• $a$ and $b$ are the limits of integration, with $a<b$.
• The result represents the net area under the curve of $f(x)$ from $x=a$ to $x=b$.

Example: To evaluate the definite integral of f(x)=3x2f(x) = 3x^2f(x)=3x2 from $x=1$ to $x=2$:

1. Find the antiderivative:

   F(x)=x3F(x) = x^3F(x)=x3

2. Apply the Fundamental Theorem of Calculus:

   ∫123×2 dx=F(2)−F(1)=23−13=8−1=7\int_1^2 3x^2 \, dx = F(2) – F(1) = 2^3 – 1^3 = 8 – 1 = 7∫12​3x2dx=F(2)−F(1)=23−13=8−1=7

3. Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that:

1. If $F$ is an antiderivative of $f$ on an interval $[a,b]$, 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)

2. If $f$ is continuous on $[a,b]$, then the function F(x)=∫axf(t) dtF(x) = \int_a^x f(t) \, dtF(x)=∫ax​f(t)dt is continuous on $[a,b]$ and differentiable on $(a,b)$, and:

   F′(x)=f(x)F'(x) = f(x)F′(x)=f(x)

4. Techniques of Integration

Common techniques for evaluating integrals include:

• Substitution: A method used to simplify the integrand by substituting a part of the expression with a new variable.

  Example:

  ∫(2x)ex2 dx(let u=x2, du=2x dx)\int (2x) e^{x^2} \, dx \quad \text{(let } u = x^2, \, du = 2x \, dx\text{)}∫(2x)ex2dx(let u=x2,du=2xdx) =∫eu du=ex2+C= \int e^u \, du = e^{x^2} + C=∫eudu=ex2+C

• Integration by Parts: A method based on the product rule for differentiation:

  $$\int udv=uv-\int vdu$$

• Partial Fractions: Used to integrate rational functions by breaking them down into simpler fractions.

2.3.1 Meanings of Integrals

Integrals have several interpretations and meanings in mathematics and applied fields. Here are the primary meanings of integrals:

1. Area Under a Curve

The most common interpretation of the definite integral is as the area under a curve defined by a function $f(x)$ between two points $a$ and $b$:

∫abf(x) dx\int_a^b f(x) \, dx∫ab​f(x)dx

• Geometric Interpretation: If $f(x)$ is positive over the interval $[a,b]$, the definite integral gives the exact area between the curve $y=f(x)$ and the x-axis from $x=a$ to $x=b$. If $f(x)$ is negative, the integral represents the area below the x-axis, which is considered negative.

2. Accumulation Function

The definite integral can be viewed as an accumulation function. It quantifies the total accumulation of a quantity represented by $f(x)$ over the interval $[a,b]$.

• Example: If $f(t)$ represents the rate of water flow into a tank over time, then ∫abf(t) dt\int_a^b f(t) \, dt∫ab​f(t)dt gives the total amount of water that has flowed into the tank between times $a$ and $b$.

3. Average Value of a Function

The integral can also be used to find the average value of a continuous function $f(x)$ over the interval $[a,b]$:

Average value=1b−a∫abf(x) dx\text{Average value} = \frac{1}{b – a} \int_a^b f(x) \, dxAverage value=b−a1​∫ab​f(x)dx

• This formula gives the average height of the function over the specified interval, which can be particularly useful in various applications, such as determining the average temperature over a period.

4. Solution to Differential Equations

Integrals play a crucial role in solving ordinary differential equations (ODEs). Many physical phenomena are modeled by ODEs, and integrating these equations can help find solutions.

• Example: The integral of a rate function can provide the original function describing the quantity of interest. For example, if dydt=k\frac{dy}{dt} = kdtdy​=k, integrating both sides with respect to $t$ gives $y(t)=kt+C$, where $C$ is the constant of integration.

5. Probability and Statistics

In probability theory, integrals are used to calculate probabilities and expected values.

• Probability Density Function (PDF): For a continuous random variable with a PDF $f(x)$, the probability that the variable falls within a certain interval $[a,b]$ is given by the definite integral:

P(a≤X≤b)=∫abf(x) dxP(a \leq X \leq b) = \int_a^b f(x) \, dxP(a≤X≤b)=∫ab​f(x)dx

• Expected Value: The expected value $E[X]$ of a continuous random variable can be computed using:

E[X]=∫−∞∞xf(x) dxE[X] = \int_{-\infty}^{\infty} x f(x) \, dxE[X]=∫−∞∞​xf(x)dx

 

2.3.2 Some Standard Integrals

In calculus, certain integrals frequently occur, and knowing their standard forms can greatly simplify the process of integration. Here’s a list of some common standard integrals:

1. Basic Power Functions

• Integral of a Power Function:

  ∫xn dx=xn+1n+1+C,n≠−1\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1∫xndx=n+1xn+1​+C,n=−1

• Integral of x−1x^{-1}x−1:

  ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1​dx=ln∣x∣+C

2. Exponential Functions

• Integral of the Exponential Function:

  ∫eax dx=1aeax+C\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C∫eaxdx=a1​eax+C

• Integral of the Natural Exponential Function:

  ∫ax dx=axln⁡a+C(a>0,a≠1)\int a^x \, dx = \frac{a^x}{\ln a} + C \quad (a > 0, a \neq 1)∫axdx=lnaax​+C(a>0,a=1)

3. Trigonometric Functions

• Integral of Sine and Cosine:

  $$\int \sin xdx=-\cos x+C$$ $$\int \cos xdx=\sin x+C$$

• Integral of Tangent:

  $$\int \tan xdx=-\ln |\cos x|+C=\ln |\sec x|+C$$

• Integral of Cosecant:

  $$\int \csc xdx=-\ln |\csc x+\cot x|+C$$

• Integral of Secant:

  $$\int \sec xdx=\ln |\sec x+\tan x|+C$$

• Integral of Cotangent:

  $$\int \cot xdx=\ln |\sin x|+C$$

4. Hyperbolic Functions

• Integral of Hyperbolic Sine:

  $$\int \sinh xdx=\cosh x+C$$

• Integral of Hyperbolic Cosine:

  $$\int \cosh xdx=\sinh x+C$$

• Integral of Hyperbolic Tangent:

  $$\int \tanh xdx=\ln |\cosh x|+C$$

5. Other Useful Integrals

• Integral of 1×2+a2\frac{1}{x^2 + a^2}x2+a21​:

  ∫1×2+a2 dx=1atan⁡−1(xa)+C\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C∫x2+a21​dx=a1​tan−1(ax​)+C

• Integral of 1a2−x2\frac{1}{\sqrt{a^2 – x^2}}a2−x2​1​:

  ∫1a2−x2 dx=sin⁡−1(xa)+C\int \frac{1}{\sqrt{a^2 – x^2}} \, dx = \sin^{-1} \left(\frac{x}{a}\right) + C∫a2−x2​1​dx=sin−1(ax​)+C

• Integral of 1×2+a2\frac{1}{\sqrt{x^2 + a^2}}x2+a2​1​:

  ∫1×2+a2 dx=ln⁡∣x+x2+a2∣+C\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln \left| x + \sqrt{x^2 + a^2} \right| + C∫x2+a2​1​dx=ln​x+x2+a2​​+C

 

2.3.3 Meaning of ∫ f(x)dx

The notation $\int f(x)dx$ represents the process of integration of the function $f(x)$ with respect to the variable $x$. This expression has several interpretations and meanings, which are crucial for understanding its applications in mathematics and science. Here are the key meanings:

1. Antiderivative

The most fundamental meaning of $\int f(x)dx$ is that it represents the antiderivative or primitive of the function $f(x)$. This means that:

$$\int f(x)dx=F(x)+C$$

where:

• $F(x)$ is a function such that F′(x)=f(x)F'(x) = f(x)F′(x)=f(x).
• $C$ is the constant of integration, accounting for the fact that there are infinitely many antiderivatives (differing by a constant).

Example: If $f(x)=2x$, then:

∫2x dx=x2+C\int 2x \, dx = x^2 + C∫2xdx=x2+C

2. Area Under the Curve

For a definite integral (with limits), the expression can represent the area under the curve of $f(x)$ from $x=a$ to $x=b$:

∫abf(x) dx\int_a^b f(x) \, dx∫ab​f(x)dx

This quantity is the net area between the graph of $f(x)$ and the x-axis over the interval $[a,b]$. If $f(x)$ is above the x-axis, the area is positive; if it is below, the area is negative.

3. Accumulation Function

The integral can also be viewed as an accumulation function. In this context, if $f(x)$ represents a rate of change (e.g., velocity), then $\int f(x)dx$ represents the total accumulation of the quantity over the interval.

For instance, if $f(t)$ is the rate of water flowing into a tank, then:

$$\int f(t)dt$$

gives the total volume of water that has flowed into the tank over a time interval.

4. Mean Value of a Function

The integral can be used to find the average value of a function over a certain interval $[a,b]$:

Average value of f=1b−a∫abf(x) dx\text{Average value of } f = \frac{1}{b-a} \int_a^b f(x) \, dxAverage value of f=b−a1​∫ab​f(x)dx

This formula gives the mean height of the function on the interval, which can be useful in various applications, including statistics and physics.

5. Connection to Differential Equations

In the context of differential equations, $\int f(x)dx$ is crucial in solving equations where $f(x)$ represents a rate of change. The integral helps recover the original function from its derivative.

6. Probability and Statistics

In probability, if $f(x)$ is a probability density function (PDF), then $\int f(x)dx$ over an interval gives the probability that a random variable falls within that interval.

 

2.3.4 Problems on finding definite integral

Finding definite integrals involves calculating the area under a curve between two specific points. Here are several problems with step-by-step solutions.

Problem 1: Basic Polynomial Function

Evaluate the definite integral:

∫13(2×3−4×2+3) dx\int_1^3 (2x^3 – 4x^2 + 3) \, dx∫13​(2x3−4x2+3)dx

Solution:

1. Find the Antiderivative:
   • Integrate the function:

   ∫(2×3−4×2+3) dx=2×44−4×33+3x=x42−4×33+3x+C\int (2x^3 – 4x^2 + 3) \, dx = \frac{2x^4}{4} – \frac{4x^3}{3} + 3x = \frac{x^4}{2} – \frac{4x^3}{3} + 3x + C∫(2x3−4x2+3)dx=42x4​−34x3​+3x=2x4​−34x3​+3x+C

2. Evaluate from 1 to 3:

   [x42−4×33+3x]13=(342−4(33)3+3(3))−(142−4(13)3+3(1))\left[ \frac{x^4}{2} – \frac{4x^3}{3} + 3x \right]_1^3 = \left( \frac{3^4}{2} – \frac{4(3^3)}{3} + 3(3) \right) – \left( \frac{1^4}{2} – \frac{4(1^3)}{3} + 3(1) \right)[2x4​−34x3​+3x]13​=(234​−34(33)​+3(3))−(214​−34(13)​+3(1)) =(812−36+9)−(12−43+3)= \left( \frac{81}{2} – 36 + 9 \right) – \left( \frac{1}{2} – \frac{4}{3} + 3 \right)=(281​−36+9)−(21​−34​+3) =(812−722+182)−(12−86+186)= \left( \frac{81}{2} – \frac{72}{2} + \frac{18}{2} \right) – \left( \frac{1}{2} – \frac{8}{6} + \frac{18}{6} \right)=(281​−272​+218​)−(21​−68​+618​) =(272)−(12+106)=272−(12+53)= \left( \frac{27}{2} \right) – \left( \frac{1}{2} + \frac{10}{6} \right) = \frac{27}{2} – \left( \frac{1}{2} + \frac{5}{3} \right)=(227​)−(21​+610​)=227​−(21​+35​) =272−36−106=272−136= \frac{27}{2} – \frac{3}{6} – \frac{10}{6} = \frac{27}{2} – \frac{13}{6}=227​−63​−610​=227​−613​To combine:

   =816−136=686=343= \frac{81}{6} – \frac{13}{6} = \frac{68}{6} = \frac{34}{3}=681​−613​=668​=334​

The final answer is:

∫13(2×3−4×2+3) dx=343\int_1^3 (2x^3 – 4x^2 + 3) \, dx = \frac{34}{3}∫13​(2x3−4x2+3)dx=334​

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Problem 2: Trigonometric Function

Evaluate the definite integral:

∫0π2sin⁡x dx\int_0^{\frac{\pi}{2}} \sin x \, dx∫02π​​sinxdx

Solution:

1. Find the Antiderivative:

   $$\int \sin xdx=-\cos x+C$$

2. Evaluate from 0 to π2\frac{\pi}{2}2π​:

   [−cos⁡x]0π2=−cos⁡(π2)−(−cos⁡(0))\left[ -\cos x \right]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) – (-\cos(0))[−cosx]02π​​=−cos(2π​)−(−cos(0)) $$=-0+1=1$$

The final answer is:

∫0π2sin⁡x dx=1\int_0^{\frac{\pi}{2}} \sin x \, dx = 1∫02π​​sinxdx=1

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Problem 3: Exponential Function

Evaluate the definite integral:

∫13e2x dx\int_1^3 e^{2x} \, dx∫13​e2xdx

Solution:

1. Find the Antiderivative:

   ∫e2x dx=12e2x+C\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C∫e2xdx=21​e2x+C

2. Evaluate from 1 to 3:

   [12e2x]13=12e2(3)−12e2(1)\left[ \frac{1}{2} e^{2x} \right]_1^3 = \frac{1}{2} e^{2(3)} – \frac{1}{2} e^{2(1)}[21​e2x]13​=21​e2(3)−21​e2(1) =12e6−12e2= \frac{1}{2} e^6 – \frac{1}{2} e^2=21​e6−21​e2

The final answer is:

∫13e2x dx=12(e6−e2)\int_1^3 e^{2x} \, dx = \frac{1}{2}(e^6 – e^2)∫13​e2xdx=21​(e6−e2)

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Problem 4: Area Between Curves

Evaluate the definite integral representing the area between the curves y=x2y = x^2y=x2 and $y=4$ from $x=0$ to $x=2$.

Solution:

1. Identify the Top and Bottom Functions: Here, $y=4$ is above y=x2y = x^2y=x2 on the interval $[0,2]$.
2. Set up the Integral:

   Area=∫02(4−x2) dx\text{Area} = \int_0^2 (4 – x^2) \, dxArea=∫02​(4−x2)dx

3. Find the Antiderivative:

   ∫(4−x2) dx=4x−x33+C\int (4 – x^2) \, dx = 4x – \frac{x^3}{3} + C∫(4−x2)dx=4x−3x3​+C

4. Evaluate from 0 to 2:

   [4x−x33]02=(4(2)−(2)33)−(4(0)−(0)33)\left[ 4x – \frac{x^3}{3} \right]_0^2 = \left( 4(2) – \frac{(2)^3}{3} \right) – \left( 4(0) – \frac{(0)^3}{3} \right)[4x−3x3​]02​=(4(2)−3(2)3​)−(4(0)−3(0)3​) =(8−83)−0=8−83=243−83=163= \left( 8 – \frac{8}{3} \right) – 0 = 8 – \frac{8}{3} = \frac{24}{3} – \frac{8}{3} = \frac{16}{3}=(8−38​)−0=8−38​=324​−38​=316​

The final answer is:

Area=163\text{Area} = \frac{16}{3}Area=316​

 

2.3.5 Area of plane regions

Calculating the area of plane regions involves finding the space enclosed by curves in a two-dimensional plane. The approach depends on the specific shapes and boundaries of the region. Here are several methods for finding areas of different plane regions.

1. Area Between Two Curves

To find the area between two curves, $y=f(x)$ (the upper curve) and $y=g(x)$ (the lower curve), from $x=a$ to $x=b$, the area $A$ can be calculated using the integral:

A=∫ab(f(x)−g(x)) dxA = \int_a^b (f(x) – g(x)) \, dxA=∫ab​(f(x)−g(x))dx

Example:

Find the area between the curves y=x2y = x^2y=x2 and $y=x+2$ from $x=0$ to $x=2$.

1. Identify the curves:
   • Upper curve: $y=x+2$
   • Lower curve: y=x2y = x^2y=x2
2. Set up the integral:

   A=∫02((x+2)−x2) dxA = \int_0^2 ((x + 2) – x^2) \, dxA=∫02​((x+2)−x2)dx

3. Integrate:

   A=∫02(x+2−x2) dx=∫02(−x2+x+2) dxA = \int_0^2 (x + 2 – x^2) \, dx = \int_0^2 (-x^2 + x + 2) \, dxA=∫02​(x+2−x2)dx=∫02​(−x2+x+2)dx

4. Calculate the integral:

   =[−x33+x22+2x]02= \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_0^2=[−3x3​+2x2​+2x]02​ =(−(2)33+(2)22+2(2))−(0)= \left( -\frac{(2)^3}{3} + \frac{(2)^2}{2} + 2(2) \right) – \left( 0 \right)=(−3(2)3​+2(2)2​+2(2))−(0) =(−83+2+4)=(−83+63+123)= \left( -\frac{8}{3} + 2 + 4 \right) = \left( -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} \right)=(−38​+2+4)=(−38​+36​+312​) =103= \frac{10}{3}=310​

The area between the curves is 103\frac{10}{3}310​ square units.

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2. Area of a Region Bounded by a Single Curve

To find the area under a curve $y=f(x)$ from $x=a$ to $x=b$:

A=∫abf(x) dxA = \int_a^b f(x) \, dxA=∫ab​f(x)dx

Example:

Find the area under the curve y=x2y = x^2y=x2 from $x=1$ to $x=3$.

1. Set up the integral:

   A=∫13×2 dxA = \int_1^3 x^2 \, dxA=∫13​x2dx

2. Integrate:

   A=[x33]13A = \left[ \frac{x^3}{3} \right]_1^3A=[3x3​]13​

3. Calculate the integral:

   =((3)33−(1)33)=(273−13)=263= \left( \frac{(3)^3}{3} – \frac{(1)^3}{3} \right) = \left( \frac{27}{3} – \frac{1}{3} \right) = \frac{26}{3}=(3(3)3​−3(1)3​)=(327​−31​)=326​

The area under the curve from $x=1$ to $x=3$ is 263\frac{26}{3}326​ square units.

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3. Area of a Region Bounded by Polar Curves

For regions defined in polar coordinates, the area $A$ can be calculated using the formula:

A=12∫θ1θ2r(θ)2 dθA = \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 \, d\thetaA=21​∫θ1​θ2​​r(θ)2dθ

where $r(\theta )$ is the polar function and θ1\theta_1θ1​ and θ2\theta_2θ2​ are the limits of integration.

Example:

Find the area of one loop of the rose curve defined by $r(\theta )=2\sin (2\theta )$.

1. Determine the limits:
   • The loop occurs when $r=0$. Setting $2\sin (2\theta )=0$ gives $\theta =0$ and θ=π2\theta = \frac{\pi}{2}θ=2π​.
2. Set up the integral:

   A=12∫0π2(2sin⁡(2θ))2 dθA = \frac{1}{2} \int_0^{\frac{\pi}{2}} (2 \sin(2\theta))^2 \, d\thetaA=21​∫02π​​(2sin(2θ))2dθ =12∫0π24sin⁡2(2θ) dθ=2∫0π2sin⁡2(2θ) dθ= \frac{1}{2} \int_0^{\frac{\pi}{2}} 4 \sin^2(2\theta) \, d\theta = 2 \int_0^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta=21​∫02π​​4sin2(2θ)dθ=2∫02π​​sin2(2θ)dθ

3. Use the identity sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1 – \cos(2x)}{2}sin2x=21−cos(2x)​:

   =2∫0π21−cos⁡(4θ)2 dθ=∫0π2(1−cos⁡(4θ)) dθ= 2 \int_0^{\frac{\pi}{2}} \frac{1 – \cos(4\theta)}{2} \, d\theta = \int_0^{\frac{\pi}{2}} (1 – \cos(4\theta)) \, d\theta=2∫02π​​21−cos(4θ)​dθ=∫02π​​(1−cos(4θ))dθ

4. Calculate the integral:

   =[θ−sin⁡(4θ)4]0π2=(π2−0)−(0−0)=π2= \left[ \theta – \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{2}} = \left( \frac{\pi}{2} – 0 \right) – \left( 0 – 0 \right) = \frac{\pi}{2}=[θ−4sin(4θ)​]02π​​=(2π​−0)−(0−0)=2π​

The area of one loop of the rose curve is π2\frac{\pi}{2}2π​ square units.