Problem 103 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{2}(2 t-1)(t+3) d t $$

Step-by-Step Solution

Answer

The evaluated definite integral is \( \frac{28}{3} \).

Step 1: Expand the Integrand

First, we need to expand the product within the integral: \((2t - 1)(t + 3)\). Apply the distributive property (FOIL method) to get:\[(2t \cdot t) + (2t \cdot 3) - (1 \cdot t) - (1 \cdot 3) = 2t^2 + 6t - t - 3 = 2t^2 + 5t - 3.\] So, the integral becomes:\[\int_{0}^{2} (2t^2 + 5t - 3) \, dt.\]

Step 2: Integrate Term by Term

Now integrate each term separately:1. \( \int 2t^2 \, dt = \frac{2t^3}{3} \)2. \( \int 5t \, dt = \frac{5t^2}{2} \)3. \( \int -3 \, dt = -3t \)Putting it all together, the indefinite integral is:\[ \frac{2t^3}{3} + \frac{5t^2}{2} - 3t + C. \]

Step 3: Evaluate the Definite Integral

To evaluate the definite integral from 0 to 2, substitute these limits into the antiderivative:1. Evaluate at \( t = 2 \): \[ \frac{2(2)^3}{3} + \frac{5(2)^2}{2} - 3(2) = \frac{16}{3} + 10 - 6. \]2. Evaluate at \( t = 0 \): \[ \frac{2(0)^3}{3} + \frac{5(0)^2}{2} - 3(0) = 0. \]3. Subtract the second from the first:\[ \left( \frac{16}{3} + 10 - 6 \right) - 0 = \frac{16}{3} + 10 - 6. \]

Step 4: Simplify the Result

Simplify the expression: \[ \frac{16}{3} + 10 - 6 = \frac{16}{3} + 4 = \frac{16}{3} + \frac{12}{3} = \frac{28}{3}. \]Therefore, the evaluated definite integral is equal to \( \frac{28}{3} \).

Key Concepts

Integrand Expansion

When dealing with definite integrals that involve products of expressions, it's often useful to first expand the integrand. This means we take our original expression \[ (2t - 1)(t + 3) \] and use algebraic techniques to rewrite it as a sum of simpler terms.
This process involves applying the distributive property, frequently using the "FOIL" method:

• First: Multiply the first terms in each binomial \( 2t \times t = 2t^2 \).
• Outside: Multiply the outer terms \( 2t \times 3 = 6t \).
• Inside: Multiply the inside terms \( -1 \times t = -t \).
• Last: Multiply the last terms \( -1 \times 3 = -3 \).

Once everything is combined, the expression becomes \( 2t^2 + 5t - 3 \).
This expanded form simplifies the integration process, as we can now integrate term by term.

Distributive Property

The distributive property is a fundamental algebraic rule that helps in simplifying complex expressions. It states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products.
In this exercise, the distributive property helps us expand the integrand \( (2t - 1)(t + 3) \) efficiently.
Each term within the parentheses is multiplied by each other term, thus allowing the complex expression to be written as a sum of simpler terms:

• Resulting in \( 2t^2 \) from the product \((2t \times t)\).
• Adding \( 6t \) from \((2t \times 3)\).
• Subtracting \( t \) from \((-1 \times t)\).
• Finally, subtracting \( 3 \) from \((-1 \times 3)\).

Working with expanded and simplified forms directly makes the process of integration easier and more straightforward.

Antiderivative

Finding the antiderivative is a crucial step in integrating a function. Once the integrand is expanded, as we have done here, we determine the antiderivative of each term separately.
The antiderivative, or the indefinite integral, undoes the process of differentiation. It's crucial for calculating definite integrals because it helps describe how a function accumulates area.
Here's how we find the antiderivative term by term from our expanded integrand \( 2t^2 + 5t - 3 \):

• The antiderivative of \( 2t^2 \) is \( \frac{2t^3}{3} \).
• For \( 5t \), it becomes \( \frac{5t^2}{2} \).
• And for \( -3 \), the antiderivative is \( -3t \).

When these terms are combined, they give us the antiderivative function \( \frac{2t^3}{3} + \frac{5t^2}{2} - 3t + C \), where \( C \) is the constant of integration which vanishes when evaluating definite integrals. This formula allows you to evaluate the definite integral by substituting the boundaries into this expression and finding their difference.