Basic Integration Rules

Integration reverses differentiation, so if you know how to differentiate, you're already halfway there! The fundamental rule is: ∫x^n dx = x^$n+1$/$n+1$ + C, where C is the constant of integration.

This constant appears because when we differentiate, any constant disappears - so integration must account for all possible constants. Think of it as covering all bases.

Let's see this in action: ∫x² dx becomes x³/3 + C. You can check this works by differentiating x³/3 back to x². For expressions with coefficients like ∫4x³ dx, you get 4x⁴/4 + C, which simplifies to x⁴ + C.

Quick Check: Always verify your integration by differentiating your answer - you should get back to the original expression!

Working with Multiple Terms and Constants

When integrating expressions with multiple terms, integrate each part separately and add them together. For ∫$2x² + 4x$ dx, you get 2x³/3 + 2x² + C.

Integrating constants requires a neat trick - remember that any number like 4 is actually 4x⁰. So ∫4 dx becomes 4x + C. This makes perfect sense when you think about it!

Expanding brackets before integration makes life much easier. For ∫x²$x+3$ dx, expand to get ∫$x³ + 3x²$ dx, then integrate to x⁴/4 + x³ + C.

The key is breaking down complex expressions into simpler parts that follow the basic rule. Once you've expanded or simplified, integration becomes straightforward.

Fractional Powers

Fractional powers follow exactly the same integration rule as whole numbers. Remember that x^$m/n$ = ⁿ√$x^m$, but you'll usually work with the fractional form.

For ∫x^(1/3) dx, add 1 to get x^(4/3), then divide by the new power: x^(4/3) ÷ (4/3) = (3/4)x^(4/3) + C. The arithmetic with fractions needs care, but the method stays the same.

More complex expressions like ∫$2x^(2/3) + (1/4)x^(1/2)$ dx just require you to handle each term separately. Take your time with the fraction arithmetic - that's usually where mistakes creep in.

Top Tip: Converting between fractional and radical forms can help you visualise what you're working with, but stick to fractions for calculations!

Preparing Expressions for Integration

Sometimes expressions aren't ready for integration straight away - you need to rewrite them using index rules first. This is especially important when dealing with denominators or square roots.

The key index rules are: x^$-a$ = 1/x^a, √$x^m$ = x^$m/2$, and the usual multiplication/division rules for powers. These let you convert awkward expressions into standard form.

For ∫$1/x³$ dx, rewrite as ∫x^(-3) dx, then integrate to get x^(-2)/(-2) + C = -1/(2x²) + C. The negative power becomes positive in the denominator.

Getting comfortable with these conversions is crucial - they turn impossible-looking integrals into routine applications of the basic rule.

Brackets and Quotients

Expanding brackets before integration saves you from headaches later. For ∫x²$x+1$ dx, expand to ∫$x³ + x²$ dx, giving you x⁴/4 + x³/3 + C.

With expressions like ∫√x$x + 3x²$ dx, first rewrite √x as x^(1/2), then expand to get ∫$x^(3/2) + 3x^(5/2)$ dx. Each term integrates separately using the standard rule.

Division by roots requires careful handling. For ∫$2x² - 5x$/√x dx, rewrite as ∫$2x^(3/2) - 5x^(1/2)$ dx after converting √x to x^(1/2) and simplifying each fraction.

The pattern is always the same: convert to standard form, then integrate term by term. Don't rush the algebra - accurate preparation makes integration straightforward.

More Complex Expressions

When dealing with square roots in denominators, convert everything to fractional powers first. This transforms messy-looking expressions into manageable ones that follow the basic integration rule.

For ∫√x$x + 3x²$ dx, you get ∫$x^(3/2) + 3x^(5/2)$ dx after expanding. This integrates to (2/5)x^(5/2) + (6/7)x^(7/2) + C - just watch your fraction arithmetic carefully.

Quotient expressions like ∫$2x² - 5x$/√x dx become much clearer when you separate the fraction: ∫2x^(3/2) dx - ∫5x^(1/2) dx. Each part integrates using the standard rule.

Success Strategy: Break complex expressions into simple pieces, convert to standard form, then integrate each piece separately!

Definite Integrals

Definite integrals have upper and lower bounds, giving you a specific numerical answer rather than a general function. The notation ∫$from a to b$ f(x) dx means you're finding the exact area or accumulated change.

The process involves finding the antiderivative F(x), then calculating F(b) - F(a). Notice there's no constant of integration needed - it cancels out when you subtract.

For ∫$from 1 to 2$ x² dx, first find the antiderivative: x³/3. Then evaluate $x³/3$ from 1 to 2, giving you 8/3 - 1/3 = 7/3.

This technique gives you precise numerical results, which is incredibly useful for finding areas, distances, and other real-world quantities.

Working with Definite Integrals

Complex definite integrals follow the same pattern: integrate first, then substitute the bounds. For ∫$from -2 to 3$ $x² + x + 1$ dx, you get $x³/3 + x²/2 + x$ evaluated from -2 to 3.

Substitute the upper bound: 27/3 + 9/2 + 3 = 9 + 4.5 + 3 = 16.5. Then subtract the lower bound result: (-8/3) + 4 - 2 = -8/3 + 2 = -2/3. So the final answer is 16.5 - (-2/3) = 16.5 + 2/3.

Integral equations can appear where you're given the definite integral's value and need to find an unknown bound. Set up the equation using the integration process, then solve algebraically.

The arithmetic can get messy, but the method stays consistent - integrate, substitute, subtract, and solve if needed.

Solving Integral Equations

When you're given that a definite integral equals a specific value, you can find unknown limits by setting up an equation. For ∫$from a to 6$ $2x + 3$ dx = 56, integrate first to get $x² + 3x$.

Substitute the bounds: (36 + 18) - $a² + 3a$ = 56. This simplifies to 54 - a² - 3a = 56, which rearranges to a² + 3a + 2 = 0.

Factoring the quadratic gives you $a + 2$$a + 1$ = 0, so a = -2 or a = -1. Both solutions are mathematically valid, but check which makes sense in the original context.

Problem-Solving Tip: Always check your solutions by substituting back into the original integral equation!

These problems combine integration skills with algebraic manipulation, making them excellent test questions that show your complete understanding.

Applications - Finding Areas

Integration's most visual application is calculating areas under curves. The definite integral ∫$from a to b$ f(x) dx gives you the exact area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

For the area enclosed by y = 4 - x² and the x-axis, first find where the curve meets the x-axis by solving 4 - x² = 0. This gives x = ±2, so you integrate from -2 to 2.

The area calculation becomes ∫$from -2 to 2$ $4 - x²$ dx = $4x - x³/3$ from -2 to 2. Evaluating gives you (8 - 8/3) - (-8 + 8/3) = 16 - 16/3 = 32/3 square units.

This technique works for any continuous curve, making integration incredibly powerful for solving real problems involving areas, volumes, and accumulation.