Question

In Exercises \(19-30,\) evaluate the integral using antiderivatives, as in Example \(4 .\) \(\int_{\pi}^{2 \pi} \sin x d x\)

Short answer

The value of the integral \(\int_{\pi}^{2 \pi} \sin x d x\) is 2.

Step-by-step solution

Find the antiderivative

First, find the antiderivative of the function within the integral, \(\sin x\). The antiderivative of \(\sin x\) is \(-\cos x\). So the integral \(\int \sin x d x\) can be written as \(-\cos x + C\). Here, \(C\) is the constant of integration which will be zero since we're dealing with a definite integral.

Substitute the limits

Next, substitute the limits of the integral into \(-\cos x\). The limits are from \(\pi\) to \(2\pi\). So first substitute \(2\pi\) into \(-\cos x\) and then substitute \(\pi\). Calculate \(-\cos(2\pi)\) and \(-\cos\pi\). Then subtract the value when \(\pi\) was substituted from the value when \(2\pi\) was substituted. This gives, \(-\cos(2\pi) - -\cos\pi = 1 - (-1) = 2\).

Write the final answer

Now, write the final result. So, \(\int_{\pi}^{2 \pi} \sin x d x = 2\).

Key concepts

Antiderivatives

The concept of an antiderivative is fundamental in calculus. It relates directly to integration and, more specifically, to evaluating definite integrals. Essentially, an antiderivative is a function that 'reverses' the process of differentiation. In other words, if we have a function 'f' that is differentiated to get another function 'f', then an antiderivative of 'f' is a function 'F' such that when 'F' is differentiated, we get 'f' back.

When finding antiderivatives, we essentially ask: what function could we differentiate to obtain the given function? For our example, we consider the function \(\sin x\). Its antiderivative is \( - \cos x \) because when we differentiate \( - \cos x \), we get \(\sin x \). This relationship is a cornerstone in solving integrals because the antiderivative gives us the means to evaluate the area under the curve represented by 'f'.

One should remember that antiderivatives are not unique—since the derivative of a constant is zero, an antiderivative is traditionally expressed with a '+ C', where 'C' is any constant. In the context of definite integrals, however, this constant cancels out when we subtract the evaluations at the boundary limits.

Definite Integral Calculation

Calculating a definite integral is essentially a procedure to find the net area under a curve within certain bounds. To do this, we first find the antiderivative of the function, as we did with \(\sin x\), obtaining the antiderivative \( - \cos x\). The next step is to use the Fundamental Theorem of Calculus, which tells us that the definite integral from 'a' to 'b' of a function is equal to the difference between the values of its antiderivative evaluated at 'b' and 'a'.

The process looks like this:
1. Compute the antiderivative of the function.2. Evaluate the antiderivative at the upper limit of integration, which gives us one value.3. Evaluate the antiderivative at the lower limit of integration, which gives us another value.4. Subtract the second value from the first, and the result is the value of the definite integral.

Integration Limits

Integration limits are the values that bound the region under (or above, depending on the function) the curve for which we want to calculate the area when performing a definite integral. They are essentially the start and end points of the interval over which we integrate the function. In our exercise, the integration limits are \(\pi\) and \(2\pi\), denoting that we're interested in the area from \(\pi\) to \(2\pi\) under the curve of \(\sin x\).

Choosing the correct limits is crucial because they directly affect the outcome of the integration. It's also important to consider the behavior of the function between these limits; for example, if a function crosses the x-axis within the interval, the definite integral will accumulate positive and negative areas, which could cancel each other out partially or fully.