Problem 26 - Calculus Single Variable

Question

Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{-1}^{0} x^{99} d x $$

Step-by-Step Solution

Answer

The value of the definite integral \( \int_{-1}^{0} x^{99} \, dx \) is \(-\frac{1}{100}\).

Step 1: Identify the Problem Type

We need to evaluate the definite integral \( \int_{-1}^{0} x^{99} \, dx \) by finding an antiderivative of \( x^{99} \) and applying the Fundamental Theorem of Calculus.

Step 2: Find the Antiderivative

The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). For \( x^{99} \), we have \( \frac{x^{100}}{100} \). This is the antiderivative of \( x^{99} \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(x) = \frac{x^{100}}{100} \).

Step 4: Evaluate at the Bounds

Calculate \( F(0) \) and \( F(-1) \). We find \( F(0) = \frac{0^{100}}{100} = 0 \) and \( F(-1) = \frac{(-1)^{100}}{100} = \frac{1}{100} \).

Step 5: Compute the Definite Integral

Subtract the antiderivative evaluated at the lower bound from the antiderivative evaluated at the upper bound: \( F(0) - F(-1) = 0 - \frac{1}{100} = -\frac{1}{100} \).

Key Concepts

Antiderivative

An antiderivative, also referred to as an indefinite integral, is a function whose derivative is the given function. In the context of polynomials, such as the function \( x^{99} \), finding an antiderivative involves reversing the differentiation process. For polynomial functions \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} \). Let's consider why this works.

If you differentiate \( \frac{x^{n+1}}{n+1} \) with respect to \( x \), you get \( x^n \), which shows that \( \frac{x^{n+1}}{n+1} \) is indeed an antiderivative of \( x^n \). In our exercise, the function is \( x^{99} \). So its antiderivative becomes \( \frac{x^{100}}{100} \) because we increase the exponent by 1 and divide by this new exponent.

To summarize, finding an antiderivative is essentially a way of "undoing" the process of differentiation. This is a key step in solving definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is integral to connecting the concept of differentiation and integration. This theorem states that if a function \( F(x) \) is the antiderivative of \( f(x) \) over an interval \([a, b]\), then the definite integral of \( f(x) \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).

This means we can compute the area under the curve of \( f(x) \) between two points \( a \) and \( b \) by evaluating the antiderivative \( F(x) \) at these points and subtracting. This is extremely powerful because it allows us to calculate definite integrals using antiderivatives, making the process much simpler.

In the problem \( \int_{-1}^{0} x^{99} \, dx \), we first find the antiderivative \( F(x) = \frac{x^{100}}{100} \). Then, by the Fundamental Theorem, we compute \( F(0) - F(-1) \) to find the value of the definite integral. This approach is what simplifies many complex integral problems.

Polynomial Integration

Polynomial integration is one of the most straightforward types of integration. It involves finding the antiderivative of polynomial functions. These functions are sums of terms of the form \( ax^n \). To integrate a polynomial, you apply the antiderivative rule to each term individually.

Here's how it works:

• Increase the power of each term by one.
• Divide each term by this new power.
• Combine all the terms at the end.

This process gives you the antiderivative of the entire polynomial, which can then be used to evaluate definite integrals by applying the Fundamental Theorem of Calculus.

For example, the polynomial \( x^{99} \) integrates to \( \frac{x^{100}}{100} \). Then, you evaluate this expression at the given bounds of the integral. Polynomial integration is simple yet forms the foundation for integrating more complex expressions by breaking them down into simpler polynomial parts.