Problem 104 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The definite integral evaluates to 56.

Step 1: Set Up the Problem

The given integral is a definite integral of the function \((2+3t)^2\), integrated with respect to \(t\) over the interval from 0 to 2. Our task is to find the value of this integral.

Step 2: Expand the Square

We begin by expanding the square \((2+3t)^2\) using the formula \((a+b)^2 = a^2 + 2ab + b^2\). Here, \(a = 2\) and \(b = 3t\), so:\[(2+3t)^2 = 4 + 2 \times 2 \times 3t + (3t)^2 = 4 + 12t + 9t^2.\] This simplifies the integrand.

Step 3: Integrate Term by Term

Now, integrate each term separately:- The integral of \(4\) with respect to \(t\) is \(4t\).- The integral of \(12t\) with respect to \(t\) is \(\frac{12}{2} t^2 = 6t^2\).- The integral of \(9t^2\) with respect to \(t\) is \(\frac{9}{3} t^3 = 3t^3\).So, the antiderivative is:\[4t + 6t^2 + 3t^3.\]

Step 4: Evaluate the Definite Integral

Evaluate the antiderivative from 0 to 2, and calculate:\[\left[4t + 6t^2 + 3t^3\right]_0^2 = \left(4(2) + 6(2)^2 + 3(2)^3\right) - \left(4(0) + 6(0)^2 + 3(0)^3\right).\]This simplifies to:- \(8 + 24 + 24 = 56\) when substituting \(t = 2\).- Substitution for \(t = 0\) gives 0.Thus the result is \(56 - 0 = 56\).

Key Concepts

Integration Techniques

When tackling definite integrals, different integration techniques come into play. One common method is to rely on known formulas and techniques such as substitution or integration by parts. However, before diving into complex methods, it is crucial to see if simpler techniques like polynomial expansion can be applied.
Take, for example, the integral \( \int_{0}^{2}(2+3t)^{2} dt \). Here, expanding the square simplifies the integrand, allowing for straightforward term-by-term integration. It may seem less obvious at first, but starting with expansion can be the easiest path forward.
When faced with these types of problems, always consider:

• Directly applying fundamental integration rules.
• Identifying opportunities for simplification through algebraic manipulation.

These simple steps streamline the process and lead to more efficient solutions.

Polynomial Expansion

Often in calculus, particularly when integrating, simplifying the integrand makes the process much easier. Polynomial expansion is a valuable tool in these situations.
Consider the expression \( (2 + 3t)^2 \). Instead of tackling the integral directly, expanding this using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \) results in a polynomial: \( 4 + 12t + 9t^2 \).
This transformation:

• Makes each term separate and straightforward for integration.
• Reduces potential errors in computation.

Remember, spotting the chance to expand a polynomial can often shift a complex calculus problem into a manageable algebra problem.

Antiderivative Calculation

Once you have a simplified polynomial, finding the antiderivative term-by-term is straightforward. An antiderivative is essentially the opposite of taking a derivative—it's looking for a function whose derivative gives back the original polynomial terms.
In our example, the expanded expression \( 4 + 12t + 9t^2 \) leads to these antiderivatives:

• The antiderivative of \(4\) is \(4t\).
• For \(12t\), calculate \(\frac{12}{2}t^2 = 6t^2\).
• And for \(9t^2\), it's \(\frac{9}{3}t^3 = 3t^3\).

Consolidate them to the complete antiderivative: \(4t + 6t^2 + 3t^3\).
This process shows that by deriving each term individually, and then combining them, you can easily evaluate a definite integral.