Problem 97 - Calculus and its applications

Question

Evaluate. $$ \int_{2}^{5}(t+\sqrt{3})(t-\sqrt{3}) d t $$

Step-by-Step Solution

Answer

The evaluated integral is 30.

Step 1: Expand the Integrand

First, we need to expand the integrand \((t + \sqrt{3})(t - \sqrt{3})\) using the difference of squares formula: \((a+b)(a-b) = a^2 - b^2\). Therefore, we have \[t^2 - (\sqrt{3})^2 = t^2 - 3.\] The integrand simplifies to \[\int (t^2 - 3) \, dt.\]

Step 2: Set Up the Integral with Limits

With the integrand simplified, we now set up the definite integral from 2 to 5:\[\int_{2}^{5} (t^2 - 3) \, dt.\]

Step 3: Evaluate the Integral

We evaluate the integral by finding the antiderivative. The antiderivative of \(t^2\) is \(\frac{t^3}{3}\), and the antiderivative of \(-3\) is \(-3t\). This results in the expression:\[\left[ \frac{t^3}{3} - 3t \right]_{2}^{5}.\]

Step 4: Calculate the Definite Integral

Substitute the upper and lower limits into the antiderivative:\[\left(\frac{5^3}{3} - 3 \times 5\right) - \left(\frac{2^3}{3} - 3 \times 2\right)\]Calculate each term:\(\frac{125}{3} - 15\) simplifies to \(\frac{125}{3} - \frac{45}{3} = \frac{80}{3}\);\(\frac{8}{3} - 6\) simplifies to \(\frac{8}{3} - \frac{18}{3} = -\frac{10}{3}\).Together, we calculate:\[\frac{80}{3} - (-\frac{10}{3}) = \frac{80}{3} + \frac{10}{3} = \frac{90}{3} = 30.\]

Key Concepts

The Difference of Squares Formula

The difference of squares is an essential mathematical concept that helps simplify polynomial expressions. It's often used to expand and simplify expressions of the form \((a+b)(a-b)\). When you see this pattern, you can use the formula:

• \((a+b)(a-b) = a^2 - b^2\)

In our exercise, we start with the expression \((t + \sqrt{3})(t - \sqrt{3})\). By applying the difference of squares formula, we directly simplify this expression to:

• \(t^2 - (\sqrt{3})^2 = t^2 - 3\)

So, converting the multiplication into a subtraction of squares helps in simplifying the integrand effectively. This simplifies the evaluation of definite integrals, making the task much more manageable.

Understanding Antiderivatives

Antiderivatives play a crucial role in solving integrals. They are essentially the reverse process of differentiation. Given a function \(f(t)\), the antiderivative, denoted as \(F(t)\), is such that \( F'(t) = f(t) \).
For the function \(t^2\), its antiderivative is found by increasing the power of \(t\) by one and dividing by the new power, resulting in:

• \(F(t) = \frac{t^3}{3}\)

For a constant like \(-3\), its antiderivative is simply \(-3t\). This is because the derivative of \(-3t\) with respect to \(t\) returns the constant \(-3\) again.

• \(F(t) = -3t\)

Knowing these basics, we transformed the integrand \(t^2 - 3\) into its antiderivative: \(\frac{t^3}{3} - 3t\). This expression is key for evaluating definite integrals.

Evaluating a Definite Integral

Evaluating a definite integral involves integrating a function over a specific interval. The definite integral of a function \(f(t)\) from \(a\) to \(b\) is expressed as:

• \(\int_{a}^{b} f(t) \, dt\)

In our case, the function was simplified to \(t^2 - 3\), and we integrated between the limits 2 and 5.
After finding the antiderivative \( \left[ \frac{t^3}{3} - 3t \right] \), we evaluate the integral by:

• Substituting the upper limit and subtracting the evaluation of the lower limit:

\[\left(\frac{5^3}{3} - 3 \times 5\right) - \left(\frac{2^3}{3} - 3 \times 2\right)\]Each part simplifies to:

• Upper limit: \(\frac{125}{3} - 15\)
• Lower limit: \(\frac{8}{3} - 6\)

The result of these calculations ultimately gives us a definite integral value of 30. This demonstrates the complete process from setting up the integral to performing the arithmetic to find the area under the curve.