Question

Use the definition of the definite integral (with right endpoints) to evaluate ∫ 2 5 ​ (4−2x)dx

Answer 1

To evaluate the definite integral using right endpoints, set up a Riemann sum by dividing the interval [2, 5] into smaller equal subintervals and evaluating the function at the right endpoint of each subinterval. Simplify the Riemann sum and take the limit as n approaches infinity to evaluate the definite integral.

Step-by-step explanation:

To evaluate the definite integral using right endpoints, we can use the definition of the definite integral:

∫ab f(x) dx = limn→∞ Σi=1n f(xi)Δx

In this case, we have ∫25 (4-2x) dx. We can set up a Riemann sum using right endpoints by dividing the interval [2, 5] into smaller equal subintervals and evaluating the function at the right endpoint of each subinterval.

Let's say we divide the interval into n subintervals. The width of each subinterval is Δx = (5-2)/n. The right endpoint of each subinterval is xi = 2 + iΔx, where i is the index of the subinterval.

Therefore, the Riemann sum becomes Σi=1n (4-2xi)Δx. To evaluate the definite integral, we need to take the limit as n approaches infinity. However, we can simplify the sum before taking the limit.

Substituting Δx = (5-2)/n and xi = 2 + iΔx into the sum, the Riemann sum becomes Σi=1n [4-2(2+i(5-2)/n)](5-2)/n. We can simplify this expression further by distributing and rearranging terms.

Once we have the simplified Riemann sum, we can take the limit as n approaches infinity to evaluate the definite integral.

Answer 2

To evaluate the definite integral of f(x) = 4 - 2x from x = 2 to x = 5, we partition the interval into n subintervals and sum the areas of rectangles at the right endpoints. The integral simplifies to -18/n × n(n+1)/2n which in the limit of n approaching infinity equals -9. Thus, the exact value of the integral is -9.

Step-by-step explanation:

To evaluate the definite integral of the function f(x) = 4 - 2x from x = 2 to x = 5 using the definition of the definite integral with right endpoints, we first partition the interval [2, 5] into n equally spaced subintervals and then sum the areas of the rectangles formed by the function values at the right endpoints of these subintervals.

Let's divide the interval [2, 5] into n subintervals, each of width Δx = (5 - 2)/n. The right endpoint of the ith subinterval is xi = 2 + i(5 - 2)/n. The function value at xi is f(xi) = 4 - 2xi.

The definite integral is then approximately:
∑_{i=1}^{n} f(xi)Δx = ∑_{i=1}^{n} (4 - 2(2 + i(3/n))) Δx, which simplifies to ∑_{i=1}^{n} (4 - 4 - 6i/n) Δx and further to -6/n ∑_{i=1}^{n} (i) Δx. By calculating the sum ∑_{i=1}^{n} i = n(n+1)/2, we get the integral approximately as -18/n × n(n+1)/2n = -9(n+1)/n. As n approaches infinity, this approximation becomes exact, and we find the exact value of the definite integral.

In the limit as n approaches infinity, -9(n+1)/n approaches -9. Therefore, the exact value of the definite integral ∫ 25 (4 - 2x) dx is -9.