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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.

What is the formula for FTOC Part 1?

g(x)=axf(t)dt    g(x)=f(x)g(x) = \int_{a}^{x} f(t) dt \implies g'(x) = f(x)

What is the formula for FTOC Part 2?

abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) dx = F(b) - F(a), where F is the antiderivative of f.

How do you evaluate abf(x)dx\int_{a}^{b} 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) if g(x)=axf(t)dtg(x) = \int_{a}^{x} f(t) dt?

Apply FTOC Part 1: 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) if g(x)=ah(x)f(t)dtg(x) = \int_{a}^{h(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)