Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{1}^{2}\left(t^{5}-t^{3}+1\right) d t$$

Step-by-Step Solution

Answer

The value of the definite integral \(\int_{1}^{2}\left(t^{5}-t^{3}+1\right) dt\) is \(\frac{125}{12}\).

Step 1: Find the antiderivative

To evaluate the definite integral, we first need to find the antiderivative of the given function, \(f(t) = t^5 - t^3 + 1\). We'll use the power rule to obtain the antiderivative: Power rule: \(\int t^n dt = \frac{t^{n+1}}{n+1} + C\) Using the power rule for each term, we get: \(\int (t^5 - t^3 + 1)dt = \frac{t^6}{6} - \frac{t^4}{4} + t + C\)

Step 2: Apply the Fundamental Theorem of Calculus

Now, we'll apply the Fundamental Theorem of Calculus to find the definite integral between the bounds 1 and 2: \(\int_{1}^{2} (t^{5} - t^{3} + 1) dt = \left(\frac{t^6}{6} - \frac{t^4}{4} + t\right)\Big|_1^2\)

Step 3: Evaluate using the bounds

Plug in the upper bound, 2, and then subtract the result of plugging in the lower bound, 1: \(\left(\frac{(2)^6}{6} - \frac{(2)^4}{4} + 2\right) - \left(\frac{(1)^6}{6} - \frac{(1)^4}{4} + 1\right)\) Simplify the expression: \(= \left(\frac{64}{6} - 4 + 2\right) - \left(\frac{1}{6} - \frac{1}{4} + 1\right)\) Calculate the final result: \(= \left(\frac{128}{6}\right) - \left(\frac{1}{4}\right)\) \(= \frac{128 - 3}{12}\) \(= \frac{125}{12}\)

Step 4: Final answer

The value of the definite integral \(\int_{1}^{2} (t^{5} - t^{3} + 1) dt\) is \(\frac{125}{12}\).

Key Concepts

Antiderivative

An antiderivative is a function whose derivative is the original function we started with. Think of it as reverse engineering the process of differentiation.When you find the antiderivative of a function, you are discovering a formula that, when differentiated, returns the original function. For example, if we start with the function \( f(t) = t^5 - t^3 + 1 \), we need to find an expression \( F(t) \) such that \( F'(t) = f(t) \). This expression is the antiderivative.

• The antiderivative of \( t^5 \) is \( \frac{t^6}{6} \). We increase the power by one and divide by the new power.
• For \( t^3 \), the antiderivative is \( \frac{t^4}{4} \).
• The constant \( 1 \) becomes \( t \) since the derivative of \( t \) is \( 1 \).

By calculating these, we obtain the full antiderivative: \( \frac{t^6}{6} - \frac{t^4}{4} + t + C \), where \( C \) represents the constant of integration.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful tool that connects differentiation with integration. It has two main parts.First, it tells us that if we have an antiderivative of a function \( f \), we can evaluate the definite integral of \( f \) over an interval \([a, b]\) by calculating the difference \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f \). This allows us to calculate the area under the curve defined by \( f(t) \).In practice, this means:

• Find the antiderivative \( F(t) \) of the function you are integrating.
• Evaluate \( F \) at the upper bound of the integral to get \( F(b) \).
• Evaluate \( F \) at the lower bound to obtain \( F(a) \).
• Subtract \( F(a) \) from \( F(b) \) to find the value of the definite integral.

In the original exercise, we applied this theorem to calculate:\[ \int_{1}^{2} (t^{5} - t^{3} + 1) dt = \left(\frac{t^6}{6} - \frac{t^4}{4} + t\right)\Big|_1^2 \] with the bounds explicit in the calculation.

Power Rule

The power rule is a fundamental technique used in calculus for finding the antiderivative (or derivative, in reverse). It's very straightforward.The rule is: for any term \( t^n \), where \( n \) is a real number, the antiderivative is \( \frac{t^{n+1}}{n+1} + C \), assuming \( n eq -1 \). This rule helps break down complex polynomials into simple computations. For example, in the exercise, we used it to find:

• \( \frac{t^6}{6} \) as the antiderivative of \( t^5 \)
• \( \frac{t^4}{4} \) as the antiderivative of \( t^3 \)
• \( t \) as the antiderivative of the constant \( 1 \)

Working term by term and adding constants allows you to solve integrals with confidence. Remember this rule as it simplifies many integral problems involving polynomial expressions.