How to Do an Integral- Calculus Tutorial
What Integration Actually Is
Integration is finding the area under a curve. That's it. You have a function, you graph it, and integration tells you the total area between the curve and the x-axis over an interval.
There are two types: indefinite integrals (which give you a general function plus a constant) and definite integrals (which give you a specific numerical answer representing area).
If you're struggling with calculus, it helps to understand that integration is basically the opposite of differentiation. Where derivatives find rates of change, integrals accumulate quantities.
Essential Integration Formulas You Need to Memorize
These are your foundation. No calculator can save you if you don't know these cold:
- ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C when n ≠ -1
- ∫ eˣ dx = eˣ + C
- ∫ 1/x dx = ln|x| + C
- ∫ sin(x) dx = -cos(x) + C
- ∫ cos(x) dx = sin(x) + C
The +C is critical. For indefinite integrals, you always add the constant of integration. Forgetting it means you haven't finished the problem.
Basic Integration Techniques
Power Rule
The power rule handles polynomials and any term raised to a constant power. Increase the exponent by 1, then divide by the new exponent.
Example: ∫ x³ dx = x⁴/4 + C
U-Substitution
This is integration's version of the chain rule. You substitute a part of the integrand with a new variable to simplify the problem.
When to use it: You see a function nested inside another function, and you can identify a "u" where du matches the rest of the expression.
How to do it:
- Pick a "u" equal to the inner function
- Find du (differentiate u)
- Replace all instances of the inner function with u
- Replace dx with du using your relationship
- Integrate with respect to u
- Substitute back to x
Example: ∫ 2x(x² + 3)⁴ dx
Let u = x² + 3, then du = 2x dx
The integral becomes ∫ u⁴ du = u⁵/5 + C = (x² + 3)⁵/5 + C
Integration by Parts
Use this when u-substitution fails. It's based on the product rule for derivatives.
The formula: ∫ u dv = uv - ∫ v du
How to pick u: Use LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential). Pick u from earlier in the list.
Example: ∫ x eˣ dx
Let u = x, dv = eˣ dx
Then du = dx, v = eˣ
Answer: xeˣ - eˣ + C = eˣ(x - 1) + C
Integration Technique Comparison
| Technique | When to Use | Key Indicator |
|---|---|---|
| Power Rule | Simple powers of x | Single term, no composition |
| U-Substitution | Chain rule pattern inside | Inner function visible, derivative nearby |
| Integration by Parts | Product of different function types | LIATE applicable, no substitution works |
| Trig Substitution | Squares of trig functions or roots | √(a² - x²), √(a² + x²), √(x² - a²) |
| Partial Fractions | Rational functions with denominators that factor | Polynomial over polynomial, denominator factors |
Common Mistakes That Cost You Points
- Dropping the dx: Always include it. It's not optional notation.
- Forgetting +C: On indefinite integrals, this is half the answer.
- Wrong substitution direction: If your substitution makes things more complicated, try a different piece.
- Sign errors in integration by parts: The minus sign between uv and ∫vdu is easy to lose.
- Not checking your answer: Differentiate your result. You should get back the original integrand.
Getting Started: A Step-by-Step Example
Let's integrate: ∫ (3x² + 2x - 5) dx
Step 1: Break it into separate integrals
= ∫ 3x² dx + ∫ 2x dx + ∫ -5 dx
Step 2: Apply power rule to each
= 3(x³/3) + 2(x²/2) - 5x + C
Step 3: Simplify
= x³ + x² - 5x + C
Step 4: Verify by differentiating
d/dx(x³ + x² - 5x + C) = 3x² + 2x - 5 ✓
Definite vs Indefinite Integrals
Indefinite integrals give you a family of functions. Definite integrals give you a number.
For definite integrals, you evaluate at the upper and lower bounds and subtract:
∫ₐᵇ f(x) dx = F(b) - F(a)
Example: ∫₀² x² dx = [x³/3]₀² = (8/3) - (0) = 8/3
You don't add +C for definite integrals since it cancels out anyway.
When You're Stuck
If none of these techniques work:
- Try expanding the expression first (especially with products)
- Rewrite trig functions in different forms
- Complete the square on denominators
- Use a computer algebra system to identify the technique, then reverse-engineer how they got there
Some integrals don't have elementary antiderivatives. If you've exhausted standard techniques and nothing fits, you might be dealing with a function that requires special functions or numerical integration to evaluate.