Question 8.T.12: (Fundamental Theorem of Calculus) If F is differentiable on ...

(Fundamental Theorem of Calculus)

If F is differentiable on [a, b] and F^{\prime} ∈ \mathcal{R}(a, b) , then

\int_{a}^{b}{F^{\prime} (x)  dx} = F (b) − F (a) .        (8.17)

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

If P = \left\{x_{0}, x_{1}, …, x_{n}\right\} is any partition of [a, b] , then

F (b) − F (a) = \sum\limits_{i=0}^{n−1}{[F (x_{i+1}) − F (x_{i})]} .

Applying the mean value theorem to F on [x_{i}, x_{i+1}] , we can choose α_{i} ∈ (x_{i}, x_{i+1}) such that

F (x_{i+1}) − F (x_{i}) = F^{\prime}(α_{i}) (x_{i+1} − x_{i}) .

Consequently

F (b) − F (a) = \sum\limits_{i=0}^{n−1}{F^{\prime}(α_{i}) (x_{i+1} − x_{i})} = S \left(F^{\prime}, P, α \right) ,

where α = (α_{0}, α_{1}, . . . , α_{n−1}) is a mark on P.

Now choose a sequence (P_{k}) of partitions such that \left\|P_{k}\right\| → 0, and a corresponding sequence of marks(α_{k}) determined by the mean value theorem as above. We then have

F (b) − F (a) = S \left(F^{\prime}, P_{k}, α_{k} \right) ,  k ∈ \mathbb{N}.

Since F^{\prime} ∈ \mathcal{R}(a, b) , the right hand side converges to \int_{a}^{b}{F^{\prime} (x)  dx} as k → ∞.

Related Answered Questions

Question: 8.5

Verified Answer:

Since the function f (x) = x − x^{2}[/latex...
Question: 8.T.7

Verified Answer:

Let P = \left\{x_{0}, x_{1}, ..., x_{n}\rig...