Integration as the Reverse of Differentiation
Recover a function from its derivative by raising the power and dividing, and carry the constant of integration that differentiation destroyed. Use a known point to pin that constant down.
What a learner can do afterwards
- Integrate 6x² - 4x and check the answer by differentiating it
- Explain why general indefinite integrals include + c, why antiderivatives on one interval differ by a constant, and why c cancels in F(b) - F(a)
- Find the particular function whose derivative is 2x and which passes through (1, 5)
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You have used the power rule to find a derivative. Integration asks for a function with that derivative. For the Ridgeway tank, let V be volume and t be time. If water leaves at rate R with no inflow, dV/dt = -R. Recovering V requires integration and a known volume. Check an antiderivative by differentiating it; you can also evaluate that derivative to check a gradient.
For n ≠ -1, ∫xⁿ dx = xⁿ⁺¹/(n + 1) + c: raise the power and divide by the new power. Divide coefficients too; ∫5x³ dx = 5x⁴/4 + c. Integrate sums term by term, and ∫k dx = kx + c. Negative and fractional powers follow this rule where they are defined. For example, ∫x^(-3) dx = -x^(-2)/2 + c on either side of zero. Expand products first; multiplying antiderivatives does not integrate a product.
Integrate 6x² - 4x. The first term gives 6x³/3 = 2x³. The second gives -4x²/2 = -2x². Thus ∫(6x² - 4x) dx = 2x³ - 2x² + c. Differentiate to check: the two terms return 6x² and -4x, while c gives zero. For a fractional power, ∫x^(1/3) dx = (3/4)x^(4/3) + c because the new power is 4/3.
An indefinite integral names all antiderivatives on one interval. They differ only by a constant: if F' = G', then (F - G)' = 0, so F - G is constant there. For 2x, the family is x² + c. Differences such as F(b) - F(a) cancel c. An excluded point can split the domain into separate intervals, each with its own constant. For x⁻², the intervals x < 0 and x > 0 need not use the same c.
Suppose dy/dx = 2x and the curve passes through (3, 11). Integrate first: y = x² + c. Substitute the coordinates: 11 = 9 + c, so c = 2 and y = x² + 2. The point fixes the constant on this interval. To find another value, substitute its x-coordinate into the completed formula. At x = 4, y = 16 + 2 = 18.
For n ≠ -1, raise the power and divide by the new power. Add c for the antiderivative family on an interval. A known point fixes c. Differentiate to check.
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