Examples · Calculus
Integral Examples: 10 Worked Problems
Ten integrals, ordered so each one adds exactly one new idea to the last — and every antiderivative here is checked by differentiating it back.
Power rule, term by term
These worked examples move from the easiest shape to the ones that combine two ideas at once, and every one of them is an indefinite integral until the last, which adds limits to turn it into a definite integral instead. The first needs nothing but the power rule applied to each term separately.
- 3x³/3 + 5x²/2power rule on each term
- x³ + 5x²/2 + Cthe coefficient 3 cancels against the division by 3
Checking it: differentiate x³ + 5x²/2 and the power rule gives back 3x² + 5x exactly, which is the original integrand.
Negative exponents, including the awkward one
The power rule works below zero too, with one exception to watch for.
- x⁻¹/(−1)raise the exponent by one, to −1, then divide by it
- −x⁻¹ + Cwhich is −1/x + C
- x⁻²/(−2)raise the exponent to −2, divide by it
- −(x⁻²/2) + Cwhich is −1/(2x²) + C
A root, rewritten as a power
- ∫ x^½ dxa square root is a fractional power
- x^1.5 / 1.5power rule: raise by one, divide by the new exponent
- 2x^1.5 / 3 + Cdividing by 1.5 is the same as multiplying by 2/3
Standard functions with a linear inside
The next three each need the same extra step: divide by the coefficient of x inside the function.
- −cos(2x)the ordinary antiderivative of sin, applied to the inside
- −cos(2x) / 2 + Cdivide by the coefficient of x
- e^(3x)e to a linear power integrates to itself
- e^(3x) / 3 + Cdivide by the coefficient of x
- sin x + Cthe coefficient of x is 1, so nothing more is needed
A power of a linear expression, and the log case
- (2x + 1)⁶/6power rule: raise the exponent by one, divide by it
- (2x + 1)⁶ / 12 + Cthen divide by the coefficient 2 as well
- ln|2x + 1|the exponent −1 case: a logarithm, not the power rule
- ln|2x + 1| / 2 + Cdivide by the coefficient of x
A definite integral, start to finish
- F(x) = x³/3find the antiderivative first; the +C is dropped
- F(1) − F(0)the fundamental theorem: evaluate at the limits and subtract
- ⅓ − 0 = ⅓the area under y = x² from 0 to 1
A second one worth knowing: the definite integral of sin(x) from 0 to π is 2 — the sine curve's one full hump above the x-axis has an area of exactly 2, a fact that is easy to verify numerically and surprising the first time you see it.
Checking every result the same way
Differentiate the answer and compare it with the original integrand — the same linear-substitution and power-rule reasoning used above, run in reverse.
| Example | Antiderivative | Differentiate it back |
|---|---|---|
| 3x² + 5x | x³ + 5x²/2 | 3x² + 5x ✓ |
| sin(2x) | −cos(2x)/2 | sin(2x) ✓ |
| (2x+1)⁵ | (2x+1)⁶/12 | (2x+1)⁵ ✓ |
| e^(3x) | e^(3x)/3 | e^(3x) ✓ |
This is exactly the check the Integral Calculator on this site runs on itself before showing any answer — differentiate the candidate, compare it numerically to the original expression, and withhold the result if the two disagree. The exponential row confirms the same pattern holds for e to a linear power as it does for sin and for a pure power of x: whatever constant the antiderivative was divided by, differentiating reintroduces it exactly.
Every antiderivative found by the constant of integration's family — that is, every indefinite integral above — represents infinitely many functions differing only by that constant, and the check above works identically for all of them, since the +C differentiates to zero regardless of its value.
Questions about integral examples
What is the integral of 3x² + 5x?
x³ + 5x²/2 + C. Apply the power rule to each term, and the coefficient 3 cancels against the division by 3 in the first term.
What is the integral of x⁻²?
−x⁻¹ + C, which is the same as −1/x + C. Raise the exponent to −1, then divide by it.
Why does 1/(2x + 1) integrate to a logarithm?
Because it is the exponent −1 case, where the power rule would divide by zero. The logarithm formula takes over, and the coefficient of x still has to be divided out afterward.
How do I check an integral is correct?
Differentiate the result. If it reproduces the original integrand exactly, the integration is correct — this check never fails to catch a mistake.