Question

Find \(F\) as a function of \(x\) and evaluate it at \(x=2, x=5,\) and \(x=8\). $$ F(x)=\int_{0}^{x} \sin \theta d \theta $$

Short answer

The function is \( F(x) = -\cos(x) + 1 \). Thus, for \( x = 2 \), \( F(2) = -\cos(2) + 1 \), for \( x = 5 \), \( F(5) = -\cos(5) + 1 \) and for \( x = 8 \), \( F(8) = -\cos(8) + 1 \).

Step-by-step solution

Integrate the function within the integral

Starting with an definite integral for \( F(x) \): \( F(x) = \int_{0}^{x} \sin \theta d \theta \), apply the fundamental theorem of calculus where the integral of \( \sin \theta \) is \( -\cos \theta \). Thus \( F(x) = [-\cos \theta]_{0}^{x} \).

Evaluate the limits

Now replace the \( \theta \) with the upper and lower limit. The upper limit \( x \) and the lower limit \( 0 \). The expression now looks like: \( F(x) = -\cos(x) - (-\cos(0)) = -\cos(x) + \cos(0) \).

Evaluate the function at given points

Now evaluate the function at \( x = 2, x = 5 \) and \( x = 8 \). For \( x = 2 \): \( F(2) = -\cos(2) + 1 \) , for \( x = 5 \): \( F(5) = -\cos(5) + 1 \) and finally for \( x = 8 \): \( F(8) = -\cos(8) + 1 \)

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone concept in calculus that connects differentiation with integration. It essentially tells us that integration and differentiation are inverse processes.

There are two parts to this theorem. The first part ensures us that if we have a continuous function over an interval, then there exists an antiderivative that can be used to calculate the definite integral of the function over that interval. In practical terms, it allows us to evaluate the definite integral of a function by calculating the difference of its antiderivative at the bounds of the interval.

The second part states that if we have a function defined as an integral (like the function in our exercise), the derivative of this function is the original function that we are integrating. This means that if we have an integral of a function from a fixed point to a variable point, the derivative of this integral with respect to the variable is the original function. By using this theorem, the complex process of finding the area under the curve simplifies to just finding the antiderivatives and evaluating at specific points.

Integration of Trigonometric Functions

When integrating trigonometric functions, it is essential to know their basic antiderivatives. For instance, the integral of \( \text{sin}(x) \) is \( -\text{cos}(x) \), and the integral of \( \text{cos}(x) \) is \( \text{sin}(x) \). These integrals are relatively straightforward because sine and cosine functions are periodic and well-behaved.

However, other trigonometric integrals might involve using identities like the Pythagorean identity or angle addition formulas to simplify the integrand. The key is to rewrite the integrand in a form that makes the antiderivative more evident. For example, an integral that involves \( \text{tan}^2(x) \) might be easier to handle by expressing it as \( \text{sec}^2(x) - 1 \).

By knowing the basic antiderivatives and applying trigonometric identities, integrating these functions becomes a much less daunting task.

Evaluating Integrals at Specific Points

Evaluating integrals at specific points is like plugging in values into a function. Once the integration is done, and you have an antiderivative, you can simply substitute the x-values to find the area up to that point on the function's curve.

As shown in the exercise, first, the integral function is obtained using the fundamental theorem of calculus. Then, evaluating it at specific points such as \( x=2, x=5, \) and \( x=8 \) involves substituting these values into the antiderivative's expression. For trigonometric integrals, it is sometimes helpful to know standard trigonometric values, as they can simplify calculations.

Evaluating the integral at 0 is often straightforward, especially with trigonometric functions like sine and cosine, because their values at 0 are well known: \( \text{sin}(0) = 0 \) and \( \text{cos}(0) = 1 \). Hence, in the context of the exercise, where the lower limit was 0, the function simplifies significantly, leading to a simple application of the upper limit value to the antiderivative.