Question

.In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{0}^{\pi} \sin x d x$$

Short answer

The value of the definite integral of \( \sin x \) from \( 0 \) to \( \pi \) is 2.

Step-by-step solution

Identifying the Function

The function to be integrated is \( \sin x \) over the interval \( [0, \pi] \).

Applying Part 2 of the Fundamental Theorem of Calculus

Let \( F(x) \) be the antiderivative of \( \sin x \), which is \( -\cos x \). The fundamental theorem of calculus, part 2 states that the definite integral of a function from \( a \) to \( b \) is equal to the antiderivative of the function evaluated at \( b \) minus the antiderivative of the function evaluated at \( a \). Thus, the definite integral from 0 to \( \pi \) of \( \sin x \) is \( F(\pi) - F(0) = -\cos(\pi) - (-\cos(0)) \).

Evaluating the Antiderivative at the Limits of Integration

Substituting the limits of integration into the antiderivative expression, working on \( -\cos(\pi) - (-\cos(0)) \) the expression results to: \( (-(-1)) - (-1) = 2 \).

Key concepts

Definite Integral

Understanding the concept of a definite integral is crucial in calculus as it represents the accumulation of a function's values over a specific interval. In practical terms, it can be viewed as the area under the curve of a graph of the function between two points on the x-axis.

A definite integral has upper and lower limits, indicating the bounds of integration. Represented by the notation \( \int_a^b f(x) \, dx \), it is evaluated by finding the difference between the values of an antiderivative at the upper limit, \( b \) and the lower limit, \( a \). The exercise given, \( \int_0^{\pi} \sin x \, dx \), asks us to calculate the area under the sine curve from \( 0 \) to \( \pi \).

The key to solving a definite integral is first, to find an antiderivative of the function being integrated, then applying the limits of integration. Understanding this process helps one solve a vast range of problems in both pure and applied mathematics.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function \( F(x) \) whose derivative is the original function \( f(x) \). In simple words, if differentiating \( F(x) \) gives us \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).

Finding an antiderivative is essentially a process of reversing differentiation. If you know that the derivative of \( \cos x \) is \( -\sin x \), then it follows logically that an antiderivative of \( \sin x \) is \( -\cos x \). This is paramount for computing definite integrals using the Fundamental Theorem of Calculus, which bridges the gap between differentiation and integration.

For the exercise \( \int_0^{\pi} \sin x \, dx \), the antiderivative used is \( -\cos x \) which is crucial for evaluating the integral.

Sine Function Integration

Integrating the sine function, written as \( \sin x \), is a common task in calculus. The sine function oscillates between -1 and 1 and is periodic with a period of \( 2\pi \).

The integration of \( \sin x \) over one complete period \( 0 \) to \( 2\pi \) would result in zero, since the areas above and below the x-axis cancel each other out. However, when integrating from \( 0 \) to \( \pi \) as in the exercise, we are finding the area for half a period where \( \sin x \) remains positive.

Integration Over Half a Period

For \( 0 \) to \( \pi \), the sine function does not change sign, so this integral represents the area of one 'hump' of the wave. Calculating the integral \( \int_0^{\pi} \sin x \, dx \) using its antiderivative \( -\cos x \) results in a definite numerical value indicating this area without ambiguity.

Limits of Integration

The limits of integration determine the interval over which the function is to be integrated. In the context of definite integrals, they are the boundaries that encapsulate the area under the curve we’re interested in computing.

In the exercise, the limits of integration are \( 0 \) and \( \pi \) for the integral \( \int_0^{\pi} \sin x \, dx \). These limits are substituted into the antiderivative function. \( F(a) \) signifies evaluating \( F(x) \) at the lower limit, \( a \), and \( F(b) \) at the upper limit, \( b \).

Applying the Limits to the Antiderivative

The function \( -\cos x \) (the antiderivative of \( \sin x \) ) evaluated at \( \pi \) and \( 0 \) is calculated as \( F(\pi) - F(0) \) which is \( -\cos(\pi) - (-\cos(0)) \). Upon evaluating these cosines, we find \( -(-1) - (-1) \) which simplifies to \( 2 \). This final value represents the definite integral of \( \sin x \) over the range \( 0 \) to \( \pi \) and is the required solution.