# How do you evaluate the definite integral by the limit definition given #int (8-x)dx# from [0,8]?

The left Riemann sum is:

and then:

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To evaluate the definite integral ∫(8-x)dx from [0,8] using the limit definition, we first need to express the integral as a limit of Riemann sums. The definite integral of a function f(x) over the interval [a,b] is given by:

∫[a,b] f(x) dx = lim(n→∞) Σ[f(xi)]Δx

where Δx = (b - a)/n is the width of each subinterval, and xi is a sample point in the ith subinterval.

For our integral ∫(8-x)dx from [0,8], we have: a = 0, b = 8, and f(x) = 8 - x.

Therefore, Δx = (8 - 0)/n = 8/n.

We choose sample points xi in each subinterval [x_{i-1}, x_i], where x_i = a + iΔx. Thus, x_i = i(8/n) for i = 1, 2, ..., n.

The Riemann sum for our integral is given by: Σ[8 - x_i]Δx = Σ8 - i(8/n).

This simplifies to: Σ[8 - 8i/n]8/n = 8Σ(1 - i/n).

Now, we evaluate the limit as n approaches infinity: lim(n→∞) 8Σ(1 - i/n) = 8 lim(n→∞) Σ(1 - i/n).

Using the properties of summation, we have: = 8 lim(n→∞) [n(1 - 1/n) + (n-1)(1 - 2/n) + ... + (1)(1 - n/n)].

This is a telescoping series, and many terms cancel out, leaving: = 8 lim(n→∞) [n(1 - 1/n)].

Evaluating the limit, we get: = 8 * (8 - 0) = 64.

Therefore, the definite integral of (8-x)dx from [0,8] is 64.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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