Problem 21 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^2_0 \biggl(\frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \biggr) \,dt \)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{14}{5} \).

Step 1: Identify the Integral

The integral given is a definite integral of the polynomial function \( \frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \) with respect to \( t \) from \( t=0 \) to \( t=2 \).

Step 2: Find the Antiderivative

To integrate term by term, we increase the power of each term by 1 and divide by the new power:\[ \int \left( \frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \right) \, dt = \frac{4}{5} \cdot \frac{t^4}{4} - \frac{3}{4} \cdot \frac{t^3}{3} + \frac{2}{5} \cdot \frac{t^2}{2} + C. \]Simplifying the coefficients, the antiderivative is:\[ \frac{1}{5}t^4 - \frac{1}{4}t^3 + \frac{1}{5}t^2 + C. \]

Step 3: Evaluate the Definite Integral

Substitute the upper limit (2) and the lower limit (0) into the antiderivative and find the difference:1. Substitute \( t=2 \): \[ \frac{1}{5}(2)^4 - \frac{1}{4}(2)^3 + \frac{1}{5}(2)^2 = \frac{1}{5}(16) - \frac{1}{4}(8) + \frac{1}{5}(4). \] Simplify: \( \frac{16}{5} - 2 + \frac{4}{5} = \frac{20}{5} - \frac{10}{5} + \frac{4}{5} = \frac{14}{5}. \)2. Substitute \( t=0 \): \[ \frac{1}{5}(0)^4 - \frac{1}{4}(0)^3 + \frac{1}{5}(0)^2 = 0. \]

Step 4: Calculate the Final Result

Subtract the value of the integral at the lower limit from the value at the upper limit:\[ \frac{14}{5} - 0 = \frac{14}{5}. \]

Key Concepts

Antiderivative

When working with integrals, the concept of an antiderivative is crucial. An antiderivative is essentially a function that reverses the process of differentiation. Given a function, finding its antiderivative consists of determining what original function could have been differentiated to yield the given function. For instance, when you see a polynomial like \( \frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \), you aim to find a function whose derivative matches this polynomial.
To find the antiderivative, you use the rule of increasing the exponent by one and then dividing the term by that new exponent. For example:

• The antiderivative of \( t^n \) is \( \frac{t^{n+1}}{n+1} \).

Once you compute the antiderivative, it usually includes a constant \( C \), which represents an infinite number of possible constants of integration. However, when dealing with definite integrals, this constant is eliminated when we substitute the limits of integration.

Polynomial Integration

Polynomial integration involves integrating expressions that are sums of power functions, like the polynomial \( \frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \). Integrating polynomials is typically one of the easiest integration tasks due to predictable patterns.
The key here is to integrate each term individually and apply the rule of powers—this results in a new polynomial with increased exponents. Here's a simplified view of the process:

• Increase each power by 1.
• Divide each coefficient by the new power.

For this exercise, you applied these rules to transform each term of the polynomial separately, and then combined these results to get the overall antiderivative. Working with polynomials prepares you well for more complex functions and serves as a building block for mastering integral calculus.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) links the concepts of differentiation and integration, two primary operations in calculus that may initially seem unrelated. It highlights that integration and differentiation are inverse processes.
According to the FTC, if you have found an antiderivative \( F(t) \) of a function \( f(t) \), then the definite integral of \( f(t) \) from \( a \) to \( b \) can be evaluated using:

• \( F(b) - F(a) \), where \( F \) is any antiderivative of \( f \).

This theorem simplifies evaluating definite integrals because once the antiderivative is known, calculating the exact value of the integral over any interval becomes a straightforward process of substitution. In the current exercise, using the FTC allowed us to evaluate \( \int_0^2 \left( \frac{4}{5}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \right) \,dt \) by simply substituting the limits into the antiderivative and finding the difference. Knowing how to apply the FTC efficiently is key to solving problems in integral calculus.