Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{-\pi / 2}^{\pi / 2}(\cos x-1) d x$$

Short answer

Question: Evaluate the definite integral of the function `f(x) = cos(x) - 1` over the interval `[-π/2, π/2]`. Answer: The definite integral of the given function over the specified interval is equal to 2.

Step-by-step solution

Finding the antiderivative of the given function

To begin, we need to find the antiderivative of `f(x) = cos(x) - 1`. We can do this by integrating each term separately: $$\int(\cos x - 1) dx = \int\cos x dx - \int1 dx$$ Now we need to find the integrals of `cos(x)` and `1`. The antiderivative of `cos(x)` is `sin(x)` and the antiderivative of `1` is `x`. So, $$\int(\cos x - 1) dx = \sin x - x + C$$ where `C` is the constant of integration.

Applying the Fundamental Theorem of Calculus

Now that we have found the antiderivative of the given function, we can apply the Fundamental Theorem of Calculus to evaluate the definite integral. The theorem states: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$ where `F(x)` is the antiderivative of `f(x)`. In our case, `F(x) = sin(x) - x`, `a = -π/2`, and `b = π/2`. Let's plug these values in to find the definite integral: $$\int_{-\pi / 2}^{\pi / 2}(\cos x-1) d x = (\sin(\pi / 2) - (\pi / 2)) - (\sin(-\pi / 2) - (-\pi / 2))$$

Evaluating the expression

Now we evaluate the expression: $$\int_{-\pi / 2}^{\pi / 2}(\cos x-1) d x = (1 - \pi / 2) - (-1 - \pi / 2) = 1 - \pi / 2 + 1 + \pi / 2$$ Finally, we can simplify the expression to find the value of the definite integral: $$\int_{-\pi / 2}^{\pi / 2}(\cos x-1) d x = 2$$ So, the definite integral of the given function over the specified interval is equal to 2.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical link between the concept of derivative and the concept of integration, both being foundational in the study of calculus. It serves as a bridge by stating that integration can be reversed by differentiation and vice versa. In essence, it says that if you have a continuous function f(x) over an interval [a, b], then once you find its antiderivative F(x), the definite integral of f(x) from a to b is simply the difference F(b) - F(a).

This theorem simplifies the process of finding definite integrals dramatically because it allows us to bypass the limiting process of Riemann sums. For students, understanding this theorem is vital because it's used to evaluate definite integrals without actually summing up infinite small pieces of areas under the curve. Rather, it becomes a simple task of subtracting two values of the antiderivative function.

Antiderivative

An antiderivative is essentially the reverse of taking a derivative. This is often referred to as the 'indefinite integral'. When we say the antiderivative of a function f(x), we are looking for another function F(x) such that F'(x) = f(x). In other words, if you were to take the derivative of F(x), you’d get the original function f(x).

Every antiderivative of a function is unique up to an additive constant, known as the constant of integration. That's because the process of differentiation wipes out constant terms; therefore, when reversing this process, you can't determine the original constant. This concept is crucial for solving definite integrals because once we find an antiderivative, we can apply the limits of our integral to get a precise answer.

Cosine Function Integration

Integration of trigonometric functions is a common task in calculus, and the cosine function is no exception. When integrating the cosine function cos(x), the antiderivative is the sine function sin(x). This is because the derivative of sine gives cosine. Therefore, when we integrate cos(x), we are reversing the derivative process and arriving at sin(x).

When integrating cos(x) for a definite integral, as in the provided exercise, we must evaluate at the upper and lower limits of the integral and take their difference. This particular operation transforms the process of finding the area under the curve of cosine into a straightforward evaluation of the sine function at specific points, showcasing how a solid grasp of antiderivatives makes the calculation of definite integrals much more manageable.