Definite Integral Practice Problems- Worked Solutions
What You Need to Know Before You Start
Definite integrals calculate the exact area under a curve between two points. That's it. No philosophy, no metaphors—just signed area.
You need to know the Fundamental Theorem of Calculus to solve these. Evaluate the antiderivative at the upper limit, subtract the antiderivative at the lower limit. That's the entire process.
If you're struggling with basic antiderivatives, fix that first. These practice problems assume you can integrate standard functions.
Essential Formulas
These come up constantly. Memorize them or derive them fast:
- ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C (n ≠ -1)
- ∫eˣ dx = eˣ + C
- ∫1/x dx = ln|x| + C
- ∫sin(x) dx = -cos(x) + C
- ∫cos(x) dx = sin(x) + C
Practice Problems with Worked Solutions
Problem 1: Basic Polynomial
Evaluate: ∫₀² (3x² + 2x - 1) dx
Step 1: Find the antiderivative.
F(x) = 3(x³/3) + 2(x²/2) - x = x³ + x² - x
Step 2: Apply the Fundamental Theorem.
F(2) = 8 + 4 - 2 = 10
F(0) = 0
Answer: 10 - 0 = 10
Problem 2: Trigonometric Function
Evaluate: ∫₀^π sin(x) dx
Antiderivative of sin(x) is -cos(x).
-cos(π) - (-cos(0)) = -(-1) - (-1) = 1 + 1 = 2
Check this visually. One full hump of sin(x) from 0 to π has area 2. Makes sense.
Problem 3: Exponential with Substitution
Evaluate: ∫₁² 2x·e^(x²) dx
This looks ugly. Use u-substitution.
Let u = x². Then du = 2x dx.
The integral becomes ∫ u = 1 to 4 e^u du.
Antiderivative: e^u
e^4 - e^1 = e^4 - e
Answer: e⁴ - e
Problem 4: Rational Function (ln appears)
Evaluate: ∫₂⁵ (1/x) dx
Antiderivative: ln|x|
ln(5) - ln(2) = ln(5/2)
Answer: ln(5/2)
Problem 5: Definite Integral with Substitution (Changing Limits)
Evaluate: ∫₀¹ x·√(1-x²) dx
Let u = 1 - x². Then du = -2x dx, so x dx = -du/2.
When x = 0, u = 1. When x = 1, u = 0.
Integral becomes ∫₁⁰ √u · (-du/2) = (1/2) ∫₀¹ √u du
∫₀¹ u^(1/2) du = (1/2) · (2/3) u^(3/2) evaluated 0 to 1
(1/2) · (2/3) · (1 - 0) = 1/3
Problem 6: Integration by Parts
Evaluate: ∫₀¹ x·e^x dx
Use integration by parts: ∫u dv = uv - ∫v du
Let u = x, dv = e^x dx
Then du = dx, v = e^x
[x·e^x]₀¹ - ∫₀¹ e^x dx
= (1·e¹ - 0·e⁰) - (e¹ - e⁰)
= e - (e - 1)
= 1
Quick Reference: Integration Techniques
| Integral Type | Technique | Key Move |
|---|---|---|
| Polynomial only | Power rule | Increase exponent, divide |
| f(g(x))·g'(x) | u-substitution | Let u = inner function |
| Product of x and e^x or x and trig | Integration by parts | Pick u using LIATE |
| (ax+b)/(cx+d) | Division or u-sub | Simplify first |
| Even power of sin or cos | Identity reduction | Use sin²x = (1-cos2x)/2 |
Common Mistakes That Cost You Points
- Forgetting to change limits when doing u-substitution on definite integrals. Either change them or switch back to x before evaluating.
- Dropping the dx during substitution. Keep track of it.
- Wrong antiderivative for trig functions. Derivative of cos is -sin, not sin.
- Sign errors when distributing negatives. Check your arithmetic.
- Not simplifying final answers. e² - e³ is fine, but e²(1-e) might be cleaner.
Getting Started: Your Action Plan
Step 1: Identify the integral type. Is it basic? Does it need substitution? Parts?
Step 2: Choose your technique. Don't force u-sub if it's not a composite function.
Step 3: Work the antiderivative. Write every step if you're practicing.
Step 4: Evaluate at both limits. Subtract: F(b) - F(a).
Step 5: Check for negative answers. Area below the x-axis gives negative values.
Bottom Line
Definite integrals are mechanical. You learn the rules, you apply them, you get the answer. No talent required—only practice.
Work through 20 problems and you'll stop hesitating. Work through 50 and you'll finish exams with time to spare.
Go solve more problems.