Problem 35 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{-2}^{2}\left(x^{2}-4\right) d x$$

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Find the antiderivative of the integrand

To find the antiderivative of \((x^2 - 4)\), we can find the antiderivative of each term separately: $$\int(x^{2}-4) d x = \int x^{2} d x - \int 4 d x$$ Using the power rule for integration, the antiderivative of \(x^2\) is: $$\int x^{2} d x = \frac{1}{3}x^{3} + C_1$$ The antiderivative of 4 (a constant) is: $$\int 4 d x = 4x + C_2$$ Combining the two antiderivatives, we get: $$\int(x^{2}-4) d x = \frac{1}{3}x^{3} - 4x + C$$ where \(C = C_1 + C_2\).

Step 2: Evaluate the antiderivative at the limits of integration

Now we need to evaluate the antiderivative at the limits of integration, \(x = 2\) and \(x = -2\): $$F(2) = \frac{1}{3}(2)^{3} - 4(2) + C = \frac{8}{3} - 8 + C$$ $$F(-2) = \frac{1}{3}(-2)^{3} - 4(-2) + C = -\frac{8}{3} + 8 + C$$

Step 3: Subtract the antiderivative evaluated at \(x = -2\) from the value at \(x = 2\)

Finally, we subtract the antiderivative evaluated at \(x = -2\) from the value at \(x = 2\) using the Fundamental Theorem of Calculus: $$\int_{-2}^{2}(x^{2}-4) d x = F(2) - F(-2) = \left(\frac{8}{3} - 8 + C\right) - \left(-\frac{8}{3} + 8 + C\right)$$ Simplifying the expression, we get: $$\int_{-2}^{2}(x^{2}-4) d x = 16 - 16 = 0$$ Therefore, the integral of the given function is 0.

Key Concepts

Antiderivative

An antiderivative, put simply, is a function whose derivative is the original function you're working with. It's like going backwards from differentiating. By finding the antiderivative, you're essentially finding a function whose derivative leads to the given function. The notation for finding an antiderivative is generally \[\int f(x)\,dx\]where we aim to find a function \(F(x)\) such that \(F'(x) = f(x)\). Here, \(F(x)\) is the antiderivative of \(f(x)\).

For example, if you have \(f(x) = x^2 - 4\), then an antiderivative of this is \(F(x) = \frac{1}{3}x^3 - 4x + C\). The \(C\) signifies a constant of integration, which is essential because an infinite number of antiderivatives differ only by a constant. Understanding antiderivatives is crucial when working with integrals, especially in definite integral problems.

Definite Integral

The concept of a definite integral involves finding the net area under a curve within a given interval. Unlike an indefinite integral, which results in a family of functions, a definite integral results in a specific number. Think of it as summing an infinite number of infinitesimally small slices under the curve.

The definite integral of a function \(f(x)\) from \(a\) to \(b\) is denoted as \[\int_{a}^{b} f(x)\,dx\]According to the Fundamental Theorem of Calculus, a wonderful connection exists between antiderivatives and definite integrals. By evaluating the antiderivative \(F(x)\) at the upper limit \(b\) and subtracting the value at the lower limit \(a\), we obtain the result:

• Find the antiderivative of the function, say \(F(x)\)
• Evaluate \(F(b) - F(a)\)

In the example given for the integral of \(x^2 - 4\) from \(-2\) to \(2\), we ultimately found that the definite integral resulted in a value of 0. This indicates that the area above the x-axis equals the area below the x-axis within that range.

Power Rule for Integration

The power rule for integration is a pivotal technique for finding antiderivatives, especially for polynomials. It's somewhat similar to the power rule of differentiation, but in reverse. When you have a term like \(x^n\), the power rule for integration states that you can find its antiderivative using:\[\int x^n dx = \frac{x^{n+1}}{n+1} + C\]where \(n\) is not equal to \(-1\).

Let's break it down with an example from the problem: for \(x^2\), using the power rule gives us:

• Increase the exponent by 1: from 2 to 3.
• Divide by the new exponent: 3.
• Resulting antiderivative is \(\frac{1}{3}x^3 + C\)

This method works for any power of \(x\), making it an incredibly efficient tool in calculus. The presence of the constant \(+ C\) is a reminder that integrals provide us with a whole family of functions.