Every lesson so far has asked the same question: given a function, what's its rate of change? This lesson flips the question around. Given a rate of change, can you recover the function it came from? A function F is called an antiderivativeof f if F'(x) = f(x) — differentiating F gets you back to f. The process of finding an antiderivative is called integration, written with the symbol ∫f(x) dx.
Here's the wrinkle: antiderivatives are never unique. Back in Lesson 0.2, you learned that the derivative of any constant is zero. Turn that fact around, and it means adding any constant to a function never changes its derivative. So if F(x) is one antiderivative of f(x), then F(x) + C is also an antiderivative of f(x), for absolutely any constant C.
You don't need a new rule from scratch here — you can reverse-engineer it from the power rule you already know. Lesson 0.2 established that d/dx(xⁿ⁺¹) = (n+1)xⁿ. Divide both sides by (n + 1), and you get d/dx[xⁿ⁺¹/(n+1)] = xⁿ. That means xⁿ⁺¹/(n+1) is an antiderivative of xⁿ — and adding the "+C" from the last section gives you the full family:
In plain language: raise the exponent by one, then divide by the new exponent. It's the power rule for derivatives, run in reverse.
Find ∫x³ dx, then verify the result by differentiating it.
Just like differentiation, integration is linear — constants pull outside the integral, and a sum of functions can be integrated term by term:
Find ∫(6x² − 4x + 3) dx.
An indefinite integral by itself only narrows things down to a family of functions. To pick out one specific member of that family, you need one more piece of information — a single known point the function passes through, called an initial condition.
Every curve shown here has the exact same derivative, f(x) = 2x — that's what it means to belong to the same antiderivative family, F(x) = x² + C. Drag C until the highlighted curve passes through the marked point.
Given F'(x) = 2x and F(2) = 5, find F(x).
This is Lesson 0.2's capstone, run in reverse. Given a constant acceleration and two initial conditions, integration rebuilds velocity and then position — with each initial condition solving for exactly one constant of integration.
An object has constant acceleration a(t) = −9.8 m/s², with v(0) = 15 m/s and x(0) = 2 m. Find v(t) and x(t).
Try the same process yourself with a different set of numbers.
An object has constant acceleration a(t) = 4 m/s², with v(0) = −2 m/s and x(0) = 0 m. Find v(t) and x(t).