integration of COS X with upper limit 2 and lower limit 0 . Evaluate by expressing as limit of sum

To evaluate the integral of cos(x) from 0 to 2 using the limit of a sum approach, we can break down the process into manageable steps. This method is rooted in the definition of the definite integral as the limit of a Riemann sum. Let’s walk through this together.

Setting Up the Riemann Sum

First, we need to divide the interval [0, 2] into n equal subintervals. The width of each subinterval, denoted as Δx, is given by:

Δx = (b - a) / n = (2 - 0) / n = 2/n

Choosing Sample Points

Next, we need to select sample points within each subinterval. A common choice is to use the right endpoint of each subinterval. The right endpoint for the k-th subinterval can be expressed as:

x_k = 0 + k * Δx = k * (2/n)

Constructing the Riemann Sum

The Riemann sum S_n can be expressed as:

S_n = Σ (from k=1 to n) f(x_k) * Δx

Substituting our function f(x) = cos(x) and the expressions for x_k and Δx, we have:

S_n = Σ (from k=1 to n) cos(k * (2/n)) * (2/n)

Expressing the Integral as a Limit

Now, we can express the definite integral as the limit of this Riemann sum:

∫ from 0 to 2 cos(x) dx = lim (n → ∞) S_n = lim (n → ∞) Σ (from k=1 to n) cos(k * (2/n)) * (2/n)

Evaluating the Limit

To evaluate this limit, we recognize that as n approaches infinity, the sum approaches the integral. The expression inside the sum becomes a Riemann sum that approximates the area under the curve of cos(x) from 0 to 2.

Using the Fundamental Theorem of Calculus

Instead of calculating the limit directly, we can also find the antiderivative of cos(x), which is sin(x). Thus, we can evaluate the definite integral using the Fundamental Theorem of Calculus:

∫ from 0 to 2 cos(x) dx = sin(x) | from 0 to 2 = sin(2) - sin(0)

Since sin(0) = 0, we have:

∫ from 0 to 2 cos(x) dx = sin(2)

Final Result

Therefore, the value of the integral of cos(x) from 0 to 2 is:

sin(2)

This approach not only demonstrates how to express the integral as a limit of a sum but also highlights the connection between Riemann sums and the Fundamental Theorem of Calculus. If you have any further questions or need clarification on any part of this process, feel free to ask!