Welcome to Indefinite Integration!
Hey everyone! Get ready to explore one of the two major ideas in calculus: integration. Don't worry if the name sounds intimidating. We're going to break it down step-by-step. In this chapter, you'll learn what indefinite integration is and how to use it.
So, what's the big deal? Well, if differentiation is about finding the rate of change (like finding a car's speed from its position), then integration is the exact opposite! It's like having the speed and working backwards to find the distance travelled. It's a powerful tool for "undoing" differentiation and solving all sorts of real-world problems. Let's get started!
What is Indefinite Integration? The Reverse of Differentiation!
Recognise the concept of indefinite integration (Syllabus 7.1)
Think about this: if we differentiate the function $$f(x) = x^2$$, we get $$f'(x) = 2x$$. Simple enough, right?
Now, let's ask the reverse question: "What function, when differentiated, gives us $$2x$$?"
You'd probably say $$x^2$$, and you'd be right! This process of working backwards from the derivative to the original function is called integration (or sometimes anti-differentiation).
The Mysterious "+ C"
Here's a small puzzle. What's the derivative of these functions?
- $$y = x^2$$ --> $$ \frac{dy}{dx} = 2x $$
- $$y = x^2 + 5$$ --> $$ \frac{dy}{dx} = 2x $$ (since the derivative of a constant is 0)
- $$y = x^2 - 100$$ --> $$ \frac{dy}{dx} = 2x $$
They all have the same derivative! This means that when we go backwards, we don't know what the original constant was. It could have been 5, -100, or any other number.
To solve this, we always add a "constant of integration", which we write as "+ C", to our answer. This 'C' represents all the possible constants that could have been there.
So, the indefinite integral of $$2x$$ is actually $$x^2 + C$$.
Notation Time!
This is how we write an integral:
$$ \int f(x) \,dx = F(x) + C $$
- $$ \int $$ : This is the integral sign. It looks like a stretched-out 'S'.
- $$ f(x) $$ : This is the function you are integrating, called the integrand.
- $$ dx $$ : This tells us we are integrating "with respect to x". It's a crucial part of the notation, so don't forget it!
- $$ F(x) + C $$ : This is the result, the indefinite integral of $$f(x)$$.
Key Takeaway
Indefinite integration is the reverse process of differentiation. When we integrate, we find a "family" of functions, so we must always remember to add the constant of integration, + C.
Your Integration Toolkit: Basic Rules and Formulas
Understand basic properties and formulae (Syllabus 7.2, 7.3)
Just like with differentiation, you don't have to figure things out from scratch every time. Here are the essential rules that will be your best friends for this topic.
Rule 1: The Power Rule (The Superstar of Integration)
This is the most common rule you'll use. It's the reverse of the power rule for differentiation.
Formula: $$ \int x^n \,dx = \frac{x^{n+1}}{n+1} + C \quad (\text{as long as } n \neq -1) $$
Simple Steps:
- Add 1 to the power (n).
- Divide by the new power (n+1).
- Add C! (Never forget this!)
Example 1: Find $$ \int x^4 \,dx $$
Solution: The power is 4. New power is 4+1 = 5. Divide by 5.
$$ \int x^4 \,dx = \frac{x^5}{5} + C $$
Example 2 (Tricky Power): Find $$ \int \sqrt{x} \,dx $$
Solution: First, rewrite the root as a power! $$ \sqrt{x} = x^{1/2} $$.
New power is $$ \frac{1}{2} + 1 = \frac{3}{2} $$. Divide by $$ \frac{3}{2} $$ (which is the same as multiplying by $$ \frac{2}{3} $$).
$$ \int x^{1/2} \,dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C $$
Rule 2: The Constant Rule
Formula: $$ \int k \,dx = kx + C \quad (\text{where k is a constant}) $$
Example: $$ \int 8 \,dx = 8x + C $$
Rule 3: The Special Case (when n = -1)
What happens if we try the power rule on $$ \int x^{-1} \,dx $$, which is $$ \int \frac{1}{x} \,dx $$? We'd get $$ \frac{x^0}{0} $$, and we can't divide by zero! This case has its own special rule.
Formula: $$ \int \frac{1}{x} \,dx = \ln|x| + C $$
Did you know? We use the absolute value signs $$|x|$$ because the logarithm function is only defined for positive numbers, but the original function $$1/x$$ works for negative numbers too.
Rule 4: The Exponential Rule (The Easiest One!)
Remember how $$e^x$$ is its own derivative? Well, it's pretty much its own integral too!
Formula: $$ \int e^x \,dx = e^x + C $$
Putting It Together: Properties of Integration
What if we have more complex functions? We can combine the rules!
1. Constant Multiple Rule: You can pull a constant multiplier out of the integral.
$$ \int k \cdot f(x) \,dx = k \int f(x) \,dx $$
Example: $$ \int 5x^2 \,dx = 5 \int x^2 \,dx = 5 \left( \frac{x^3}{3} \right) + C = \frac{5x^3}{3} + C $$
2. Sum/Difference Rule: You can integrate a function term by term.
$$ \int (f(x) \pm g(x)) \,dx = \int f(x) \,dx \pm \int g(x) \,dx $$
Example: Find $$ \int (6x^2 - e^x + \frac{3}{x}) \,dx $$
Solution: Integrate each part separately.
$$ = \int 6x^2 \,dx - \int e^x \,dx + \int \frac{3}{x} \,dx $$
$$ = 6 \left( \frac{x^3}{3} \right) - (e^x) + 3 \left( \ln|x| \right) + C $$
$$ = 2x^3 - e^x + 3\ln|x| + C $$
Note: We only need to write one `+ C` at the end to represent all the constants combined.
Key Takeaway
Master the four basic formulas (Power Rule, Constant, 1/x, e^x). Use the properties to break down complicated expressions into smaller, manageable parts that you can integrate term by term.
Level Up: Integration by Substitution
Use integration by substitution (Syllabus 7.4)
Don't worry, this sounds harder than it is! It's basically the reverse of the Chain Rule. We use it when we have a function "nested" inside another function.
Analogy: Imagine you have a complicated sentence. You might replace a long, tricky phrase with a single simple word (like 'it') to understand the main structure, and then sub the phrase back in later. That's exactly what we do here!
The Step-by-Step Process:
- Choose 'u': Look for an "inside function". Let u equal this part. A good choice for 'u' is often something in brackets or in the denominator.
- Find 'du': Differentiate your 'u' with respect to x (find $$ \frac{du}{dx} $$) and then rearrange the equation to solve for $$dx$$.
- Substitute: Replace both the 'u' part and the 'dx' part in the integral. If you've done it right, all the 'x's should disappear, leaving a much simpler integral with only 'u's.
- Integrate: Solve the new, easy integral with respect to 'u'.
- Substitute Back: Replace 'u' with the original expression in 'x' to get your final answer. Don't forget + C!
Worked Example:
Find $$ \int 2x(x^2 + 5)^3 \,dx $$
Step 1: Choose 'u'. The "inside function" is clearly $$x^2 + 5$$.
Let $$ u = x^2 + 5 $$
Step 2: Find 'du'. Differentiate 'u'.
$$ \frac{du}{dx} = 2x $$
Rearrange to get a substitute for $$dx$$: $$ du = 2x \,dx $$ which means $$ dx = \frac{du}{2x} $$
Step 3: Substitute. Replace $$x^2+5$$ with $$u$$ and $$dx$$ with $$\frac{du}{2x}$$.
$$ \int 2x \cdot (u)^3 \cdot \frac{du}{2x} $$
Look! The $$2x$$ terms cancel out. This is a sign you chose the right 'u'! We are left with:
$$ \int u^3 \,du $$
Step 4: Integrate. This is just a simple power rule problem now.
$$ \int u^3 \,du = \frac{u^4}{4} + C $$
Step 5: Substitute Back. Replace 'u' with $$x^2 + 5$$.
Final Answer: $$ \frac{(x^2 + 5)^4}{4} + C $$
Common Mistakes to Avoid
- Forgetting to substitute back from 'u' to 'x'. Your final answer should be in terms of the original variable!
- Mixing up 'u' and 'du'. Keep your substitutions organised.
- Forgetting to add '+ C' at the end. It's an easy mark to lose!
Key Takeaway
Substitution turns a scary integral into a simple one. Choose 'u' (the inside part), find 'du', substitute, integrate, and substitute back. Practice helps you spot the right 'u' quickly!
Putting It All to Work: Applications of Indefinite Integration
Use indefinite integration to solve problems (Syllabus 7.5)
This is where we answer the question, "When will I ever use this?" The main application is finding a function when you know its rate of change (its derivative) and one specific value (an initial condition or a point on its curve).
Finding the Equation of a Curve from its Gradient
Remember that the gradient of a curve is given by its derivative, $$ \frac{dy}{dx} $$. If you are given the gradient function and a point the curve passes through, you can find the exact equation of the curve.
The Goal: Find the specific value of 'C'.
Example: The gradient of a curve is given by $$ \frac{dy}{dx} = 4x + 3 $$. If the curve passes through the point $$(2, 12)$$, find its equation.
Step 1: Integrate to find the general equation.
$$ y = \int (4x + 3) \,dx $$
$$ y = 4\left(\frac{x^2}{2}\right) + 3x + C $$
$$ y = 2x^2 + 3x + C $$
This is the "family" of curves. We need to find the one specific curve that goes through (2, 12).
Step 2: Use the given point to find C.
Substitute $$x=2$$ and $$y=12$$ into the equation.
$$ 12 = 2(2)^2 + 3(2) + C $$
$$ 12 = 2(4) + 6 + C $$
$$ 12 = 8 + 6 + C $$
$$ 12 = 14 + C $$
$$ C = -2 $$
Step 3: Write the final equation.
Now that we know C, we can write the specific equation.
Answer: $$ y = 2x^2 + 3x - 2 $$
Real-World Example: Motion
In physics, if you have the acceleration function $$a(t)$$, you can integrate it to get the velocity function $$v(t)$$. If you integrate the velocity function $$v(t)$$, you get the displacement function $$s(t)$$.
$$ v(t) = \int a(t) \,dt \quad \text{and} \quad s(t) = \int v(t) \,dt $$
Key Takeaway
Integration allows us to go from a "rate of change" back to the "total amount". Use the given information (like a point or an initial value) to solve for the constant 'C' and find a specific solution.