Problem 26 - University Calculus: Early Transcendentals

Question

Use known area formulas to evaluate the integrals in Exercises \(23-28\) $$\int_{a}^{b} 3 t d t, \quad 0

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{3}{2}(b^2 - a^2) \).

Step 1: Identify the Integral

We need to evaluate the definite integral \( \int_{a}^{b} 3t \, dt \). This integral represents the area under the curve \( y = 3t \) from \( t = a \) to \( t = b \).

Step 2: Calculate the Antiderivative

Find the antiderivative of the function \( 3t \). The antiderivative of \( 3t \) is \( \frac{3}{2}t^2 \), since the derivative of \( \frac{3}{2}t^2 \) gives \( 3t \).

Step 3: Evaluate the Definite Integral

Use the Fundamental Theorem of Calculus to evaluate the definite integral from \( a \) to \( b \). Substitute \( b \) and \( a \) into the antiderivative:\[\int_{a}^{b} 3t \, dt = \left[\frac{3}{2}t^{2}\right]_{a}^{b} = \frac{3}{2}b^2 - \frac{3}{2}a^2\]

Step 4: Simplify the Expression

Simplify the expression obtained from evaluating the definite integral:\[\frac{3}{2}b^2 - \frac{3}{2}a^2 = \frac{3}{2}(b^2 - a^2)\]

Step 5: Conclusion

The integral \( \int_{a}^{b} 3t \, dt \) evaluates to \( \frac{3}{2}(b^2 - a^2) \). This results from finding the area under the curve of \( y = 3t \) between \( t = a \) and \( t = b \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus that connects the concepts of differentiation and integration. It essentially states that if you have a continuous function on a closed interval, you can find the definite integral of that function using its antiderivative. This theorem simplifies the process of finding areas, making it very important for calculating definite integrals.
To understand its practical application, consider this: when we want to find the area under the curve of a function between two points, we first determine the antiderivative of the function. After finding this antiderivative, the next step involves evaluating it at the two endpoints of the interval and calculating the difference.

• Step 1: Find the antiderivative of the function in question.
• Step 2: Evaluate this antiderivative at the upper limit of your interval.
• Step 3: Subtract the antiderivative evaluated at the lower limit from the result in Step 2.

This sequence of procedures illustrates how differentiation and integration are two sides of the same coin, bestowing powerful tools to solve a myriad of mathematical problems.

Antiderivative

An antiderivative is essentially a "reverse derivative." When we compute the derivative of a function, we are finding how the function changes at any given point. Conversely, when we look for an antiderivative, we are searching for a function whose derivative gives us the original function.
In our example, we started with the function \(3t\). To find its antiderivative, we needed a function that when differentiated, returned \(3t\). By applying the basic rules of integration, we find that \(\frac{3}{2}t^2\) is indeed that function.

• The rule for finding the antiderivative of \(t\) raised to a power is to increase the power by one, then divide by the new power, and don't forget to multiply by the constant in front (such as 3 in this case).
• Thus, the antiderivative of \(3t\) is calculated as \(\frac{3}{2}t^2\).

Antiderivatives are powerful because they allow us to determine the original function properties from the derivative, which is central to solving integration problems.

Area under the Curve

One of the primary applications of integration is to compute the area under a curve. This concept is practical in many real-world applications, such as calculating the distance traveled over time or determining accumulated profit in economics. By integrating, or summing up infinitesimally small parts of the area under a curve, we can derive a total area over a given interval.
In the definite integration problem \(\int_{a}^{b} 3t \, dt\), we were asked to find the area under the line represented by \(y = 3t\) from \(t = a\) to \(t = b\). This scenario highlights the strength of definite integrals, which not only help determine the total area but also give meaning to complex figures by breaking them into manageable parts.

• First, identify the function and the two endpoints \(a\) and \(b\).
• Use the antiderivative and the Fundamental Theorem of Calculus to evaluate the area.
• The calculated result here, \(\frac{3}{2}(b^2 - a^2)\), represents the area under the curve from point \(a\) to \(b\).

Understanding the concept of the area under the curve is crucial, as it allows us to solve numerous real-life problems where accumulation is involved.