Question

Evaluate each definite integral to three significant digits. Check some by calculator. $$\int_{-2}^{2} x^{2} d x$$

Short answer

5.333

Step-by-step solution

Set up the Definite Integral

We are tasked with evaluating the definite integral of the function \(x^2\) from \(-2\) to \(2\). In mathematical terms, this is written as \(\int_{-2}^{2} x^{2} dx\). This represents the area under the curve of \(x^2\) between the limits of integration.

Apply Antiderivative Rules

To evaluate the definite integral, we first find the antiderivative (also known as the indefinite integral) of the function \(x^2\). The antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\) plus a constant 'C'. Therefore, the antiderivative of \(x^2\) is \(\frac{x^{3}}{3}+C\).

Evaluate the Antiderivative at the Bounds

We now compute the value of the antiderivative at the upper bound and subtract the value of the antiderivative at the lower bound. This gives us: \(\left[\frac{x^3}{3}\right]_{-2}^{2}\) which equals \(\frac{2^3}{3}-\frac{(-2)^3}{3}\).

Simplify the Expression

Simplifying the above expression gives \(\frac{8}{3} - \frac{-8}{3}\) which equals \(\frac{8}{3} + \frac{8}{3} = \frac{16}{3}\).

Calculate the Numerical Value

Converting the fraction \(\frac{16}{3}\) to a decimal yields approximately 5.333 to three significant digits.

Key concepts

Antiderivative

Understanding the antiderivative is critical for solving problems involving integral calculus. An antiderivative of a function is essentially the reverse process of taking a derivative. If you consider a simple power function, like \(x^2\), the derivative would tell us the slope of the tangent line at any given point on its curve. Conversely, the antiderivative gives us the original function whose rate of change (or derivative) is the function we started with.

Let's take the function \(x^2\) from our example. To find its antiderivative, we use a general rule: the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\), plus a constant 'C' since the derivative of any constant is zero. Following this rule, the antiderivative of \(x^2\) becomes \(\frac{x^3}{3} + C\). The 'C' represents all possible antiderivatives, but when we're dealing with definite integrals, it cancels out, so we don't need to worry about it here.

Integral Calculus

Integral calculus is a branch of mathematics focusing on finding the antiderivatives and solving for areas under curves. It's essentially the process of accumulation, where we're adding up many tiny pieces to find a total. In the problem \(\int_{-2}^{2} x^2 dx\), we use integral calculus to determine the continuous sum of infinitesimally small rectangles, each with height \(x^2\) and width \(dx\), from \(-2\) to \(2\).

This process consists of several steps. Firstly, as demonstrated in our example, we find the antiderivative of the function. We then evaluate this antiderivative at the boundaries of the integral, also known as the limits of integration, and subtract the lower boundary value from the upper boundary value. This calculation gives us the definite integral's value, which, in this case, is \(\frac{16}{3}\) or approximately 5.333.

Area under the Curve

The concept of the area under the curve is a fundamental idea in integral calculus. When we calculate a definite integral, such as \(\int_{-2}^{2} x^2 dx\), we're looking for the area bounded by the function \(x^2\), the x-axis, and the vertical lines at \(x = -2\) and \(x = 2\). The result is a geometric shape whose area is equal to our definite integral's value.

Visualization of the Area

In our exercise, the curve of \(x^2\) is symmetrical about the y-axis, and so the area from \(-2\) to \(0\) is the same as from \(0\) to \(2\). Since these areas are equal, they can be doubled for one interval to simplify calculations in some cases. However, it's not always the case with other functions, and full evaluation is necessary. By understanding the area under the curve, students grasp the concrete meaning behind the abstract calculations of integrals, making the concept less daunting.