Problem 47 - Applied Calculus

Question

Evaluate each definite integral. $$ \int_{0}^{1}\left(x^{99}+x^{9}+1\right) d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to \(\frac{111}{100}\).

Step 1: Identify the Function to Integrate

The function we need to integrate is given by \(f(x) = x^{99} + x^9 + 1\). We will integrate this function over the interval from 0 to 1.

Step 2: Integrate Each Term Separately

The integral of a sum is the sum of the integrals, so integrate each term: \(\int (x^{99}) \, dx\), \(\int (x^{9}) \, dx\), and \(\int 1 \, dx\).

Step 3: Apply Power Rule for Integration

Using the power rule \(\int x^n\, dx = \frac{x^{n+1}}{n+1} + C\), we find: \(\int x^{99} \, dx = \frac{x^{100}}{100}\), \(\int x^9 \, dx = \frac{x^{10}}{10}\), and \(\int 1 \, dx = x\).

Step 4: Determine the Indefinite Integral

Combine the results to get the indefinite integral: \(F(x) = \frac{x^{100}}{100} + \frac{x^{10}}{10} + x\).

Step 5: Evaluate the Definite Integral

Evaluate the definite integral using the limits 0 and 1: \(\int_{0}^{1} x^{99} + x^9 + 1 \, dx = F(1) - F(0)\).

Step 6: Calculate the Values at the Bounds

Compute \(F(1) = \frac{1^{100}}{100} + \frac{1^{10}}{10} + 1 = \frac{1}{100} + \frac{1}{10} + 1\). Compute \(F(0) = \frac{0^{100}}{100} + \frac{0^{10}}{10} + 0 = 0\).

Step 7: Subtract to Find the Definite Integral

Subtract the two results: \(F(1) - F(0) = \left(\frac{1}{100} + \frac{1}{10} + 1\right) - 0 = \frac{1}{100} + \frac{1}{10} + 1\).

Step 8: Simplify the Result

Simplify \(\frac{1}{100} + \frac{1}{10} + 1\) by finding a common denominator and adding: \(\frac{1}{100} + \frac{10}{100} + \frac{100}{100} = \frac{111}{100}\).

Key Concepts

Power Rule for Integration

The Power Rule for Integration is a fundamental technique that simplifies the process of finding integrals of polynomial functions. This rule states that to integrate a function of the form \(x^n\), where \(n\) is any real number, you increase the exponent by 1, divide by the new exponent, and add a constant of integration, \(C\). Mathematically, it is expressed as:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)

Let's consider its application:

• For the function \(x^{99}\), applying the power rule gives \(\frac{x^{100}}{100} + C\).
• For \(x^9\), it becomes \(\frac{x^{10}}{10} + C\).
• For a constant value like \(1\), the integral is \(x + C\), considering it as \(x^0\).

These results can be combined to express the antiderivative of the original polynomial function. The power rule is not only applicable here but also an essential tool in calculus for simplifying integration.

Indefinite Integral

An indefinite integral represents a family of functions obtained by integrating a given function, without specified limits. In other words, it is the reverse operation of differentiation. If you differentiate the result of an indefinite integral, you should get the original function back. When you find the integral of \(f(x) = x^{99} + x^9 + 1\), you derive an antiderivative \(F(x)\), which is a continuous set of functions:

• \(F(x) = \frac{x^{100}}{100} + \frac{x^{10}}{10} + x + C\)

Here, \(C\) plays a crucial role as the constant of integration, accounting for any vertical shift in the graph of \(F(x)\). The power rule can seamlessly be applied to each term within a polynomial, giving a direct approach to obtaining the indefinite integral.

Evaluating Definite Integrals

Once we have an indefinite integral, we can evaluate the definite integral over a specific interval. This process is akin to finding the net area under the curve of the function between two points. The fundamental theorem of calculus bridges differentiation and integration, illustrating that a definite integral of a function from \(a\) to \(b\) is the difference \(F(b) - F(a)\).In our example, for the function \(f(x) = x^{99} + x^9 + 1\) evaluated from 0 to 1, we substitute these limits into the antiderivative obtained:

• First, calculate \(F(1) = \frac{1^{100}}{100} + \frac{1^{10}}{10} + 1 = \frac{1}{100} + \frac{1}{10} + 1\).
• Next, evaluate \(F(0) = \frac{0^{100}}{100} + \frac{0^{10}}{10} + 0 = 0\).

Finally, the definite integral is \(F(1) - F(0)\), which simplifies to \(\frac{111}{100}\). This result not only provides us with a precise area value but also showcases the beauty of integrating polynomial functions.