Problem 2 - Advanced Calculus

Question

For each of the following integrals, verify Corollary 7.20 by direct computation. a. \(\int_{0}^{1}\left(x^{2}+x^{3}\right) d x\) b. \(\int_{2}^{3}(x+1)^{2} d x\)

Step-by-Step Solution

Answer

The solution for the first integral \(\int_{0}^{1}\left(x^{2}+x^{3}\right) dx\) is \(\frac{1}{3} + \frac{1}{4} = \frac{7}{12}\), and for the second integral \(\int_{2}^{3}(x+1)^{2} dx\) is \(\frac{27}{3} - \frac{8}{3} + (9 - 4) + (3 - 2) = 9 - \frac{8}{3} + 5 + 1 = \frac{34}{3}\).

Step 1: Expand the integrand of the first integral

For the first integral \(\int_{0}^{1}\left(x^{2}+x^{3}\right) dx\), the integrand is already in expanded form, so there is no need to expand further. Proceed to integrate term by term.

Step 2: Integrate each term of the first integral separately

Integrate each term of the first integral separately using the power rule for integration \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\), where C is the constant of integration. For \(x^{2}\), the power is 2, and for \(x^{3}\), the power is 3.

Step 3: Apply the integration limits for the first integral

After integrating each term, apply the Fundamental Theorem of Calculus, which tells us to subtract the value of the antiderivative at the lower limit from the value at the upper limit. For each term: \(\int_{0}^{1}x^{2} dx = \frac{x^{3}}{3} \bigg|_0^1 \)and \(\int_{0}^{1}x^{3} dx = \frac{x^{4}}{4} \bigg|_0^1 \).

Step 4: Evaluate the definite integrals for the first integral

Evaluate the antiderivatives at the upper and lower limits, and find the difference for each term: \(\frac{1^{3}}{3} - \frac{0^{3}}{3}\) for \(x^2\) and \(\frac{1^{4}}{4} - \frac{0^{4}}{4}\) for \(x^3\).

Step 5: Expand the integrand of the second integral

For the second integral \(\int_{2}^{3}(x+1)^{2} dx\), expand \( (x+1)^2 \) using the binomial theorem or by multiplying \( (x+1)(x+1) \).

Step 6: Integrate each term of the second integral separately

After expanding \( (x+1)^2 \) to \( x^2 + 2x + 1 \) proceed to integrate each term separately using the power rule for integration.

Step 7: Apply the integration limits for the second integral

Apply the Fundamental Theorem of Calculus to the second integral by computing \(\int_{2}^{3}x^2 dx = \frac{x^3}{3} \bigg|_2^3\), \(\int_{2}^{3}2x dx = x^2 \bigg|_2^3\), and \(\int_{2}^{3}1 dx = x \bigg|_2^3\).

Step 8: Evaluate the definite integrals for the second integral

Evaluate the antiderivatives at the upper and lower limits, and sum the results for each term: \(\frac{3^{3}}{3} - \frac{2^{3}}{3}\) for \(x^2\), \(3^2 - 2^2\) for the \(2x\) term, and \(3 - 2\) for the constant term.

Key Concepts

Power Rule for Integration

Understanding the power rule for integration is key to solving many calculus problems, especially when dealing with polynomial expressions. When you come across an integral like \( \int x^n dx \), the power rule allows you to easily find the antiderivative. According to this rule, the integral of \( x \) raised to the power of \( n \) (where \( n \) is a real number and \( n eq -1 \)) is \( \frac{x^{n+1}}{n+1} \) plus the constant of integration, \( C \).

In the context of the given exercise, to integrate \( x^2 \) and \( x^3 \) between 0 and 1, you would use the power rule to obtain \( \frac{x^3}{3} \) for \( x^2 \) and \( \frac{x^4}{4} \) for \( x^3 \) before applying the definite integration limits. This method simplifies complex problems, ensuring that you can handle a wide variety of integrals with ease.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, and it is critical to calculating exact areas and solving real-world problems involving rates of change.

It states that if you have a continuous function \( f(x) \) on an interval \( [a, b] \), and \( F(x) \) is the antiderivative of \( f(x) \), then \( \int_a^b f(x) dx = F(b) - F(a) \). In other words, it allows us to evaluate the integral by finding the difference in values of the antiderivative at the upper and lower limits of integration.

During the solution process for our textbook problem, we utilize this theorem to compute the definite integral for both \( x^2 \) and \( x^3 \) terms by substituting the limits into the antiderivatives. This application reflects a practical and pivotal use of the theorem in solving definite integrals.

Binomial Theorem

The binomial theorem is a powerful tool for expanding expressions that are raised to a power, and it is particularly useful when dealing with polynomial integrals. The theorem provides a formula for expressing the expansion of a binomial \( (a + b)^n \) as a sum of terms of the form \( \binom{n}{k} a^{n-k}b^k \), where \( \binom{n}{k} \) represents the binomial coefficient.

For instance, in our second integral \( \int_{2}^{3}(x+1)^2 dx \), one could use the binomial theorem to expand \( (x+1)^2 \) into \( x^2 + 2x + 1 \) before applying the power rule for integration to each term separately. This not only simplifies the integrand but also sets the stage to apply the Fundamental Theorem of Calculus for evaluating the definite integral.