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  2. AP Calculus
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Explain FTOC Part 1.

The derivative of the definite integral of a function with respect to its upper limit is the original function evaluated at that upper limit.

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Explain FTOC Part 1.

The derivative of the definite integral of a function with respect to its upper limit is the original function evaluated at that upper limit.

Explain FTOC Part 2.

The definite integral of a function from a to b is the difference between the antiderivative evaluated at b and the antiderivative evaluated at a.

What is the relationship between differentiation and integration according to the FTOC?

Differentiation and integration are inverse operations; one can 'undo' the other.

How do you evaluate ∫abf(x)dx\int_{a}^{b} f(x) dx∫ab​f(x)dx using FTOC Part 2?

  1. Find the antiderivative F(x) of f(x). 2. Evaluate F(b) and F(a). 3. Subtract F(a) from F(b): F(b) - F(a).

How do you find g′(x)g'(x)g′(x) if g(x)=∫axf(t)dtg(x) = \int_{a}^{x} f(t) dtg(x)=∫ax​f(t)dt?

Apply FTOC Part 1: g′(x)=f(x)g'(x) = f(x)g′(x)=f(x).

How do you handle a constant of integration when using FTOC Part 2?

The constant of integration cancels out when evaluating F(b) - F(a), so it's not necessary to include it.

How to find g′(x)g'(x)g′(x) if g(x)=∫ah(x)f(t)dtg(x) = \int_{a}^{h(x)} f(t) dtg(x)=∫ah(x)​f(t)dt?

Apply FTOC Part 1 and the chain rule: g′(x)=f(h(x))∗h′(x)g'(x) = f(h(x)) * h'(x)g′(x)=f(h(x))∗h′(x)

What does the Fundamental Theorem of Calculus, Part 1 state?

If g(x)=∫axf(t)dtg(x) = \int_{a}^{x} f(t) dtg(x)=∫ax​f(t)dt, then g′(x)=f(x)g'(x) = f(x)g′(x)=f(x).

What does the Fundamental Theorem of Calculus, Part 2 state?

If F(x)F(x)F(x) is an antiderivative of f(x)f(x)f(x), then ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a)∫ab​f(x)dx=F(b)−F(a).