Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use the Fundamental Theorem of Calculus to calculate \(\int_{1}^{2} t^{3} d t\).

Step-by-Step Solution

Answer

The definite integral of \(t^{3}\) from 1 to 2 is \(\frac{15}{4}\).

Step 1: Find the Antiderivative

Before applying the Fundamental Theorem of Calculus, we need to know the antiderivative of \(t^{3}\). In general, the antiderivative of \(t^m\), where \(m\) is any number, is \(\frac{1}{m+1} t^{m+1}\). So the antiderivative of \(t^{3}\) is \(\frac{1}{4} t^{4}\).

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that the definite integral from a to b of a function is equal to the antiderivative evaluated at b minus the antiderivative evaluated at a. In this case, a is 1 and b is 2. We then get the formula \(\int_{1}^{2} t^{3} dt = \left.\frac{1}{4} t^{4}\right|_{1}^{2}\).

Step 3: Evaluate the Antiderivative at the Bounds of Integration

We now substitute the bounds of integration into our antiderivative to get the final answer: \(\frac{1}{4} \cdot 2^{4} - \frac{1}{4} \cdot 1^{4} = 4 - \frac{1}{4} = \frac{15}{4}\).

Key Concepts

Antiderivative

The concept of an antiderivative is crucial when dealing with integrals. An antiderivative of a function is another function whose derivative is the original function. This means if you differentiate the antiderivative, you get back the original function.
For a simple example, consider the function \( f(t) = t^3 \). Its antiderivative is \( F(t) = \frac{1}{4} t^4 \).
This is derived using the power rule for integration, which states that the antiderivative of \(t^m\), where \(m\) is a constant, is \(\frac{1}{m+1} t^{m+1}\).

• The antiderivative gives a general form, often including a constant \(+ C\), because differentiating a constant gives zero.
• While working with definite integrals, the constant \(+ C\) cancels out; hence we don't explicitly include it in calculations.

Definite Integral

The definite integral of a function provides the accumulated value between two specific points on a curve. It's not just about finding a general antiderivative, but rather determining the net "area" under a curve within a certain interval.
The Fundamental Theorem of Calculus connects antiderivatives and definite integrals. Specifically, it allows us to calculate a definite integral if we know an appropriate antiderivative. Using our earlier example, \(\int_{1}^{2} t^3 \, dt\), this theorem tells us we can find the difference between the antiderivative evaluated at the upper limit 2, and the lower limit 1.

• The theorem formula is \( F(b) - F(a) \) where \( F \) is the antiderivative of the function being integrated.
• This calculation results in a specific value, \(\frac{15}{4}\) in this case.

This interpretation often reflects areas, when the function stands for a consistent positive value over the interval.

Integration

Integration is the process used to compute the area under curves, antiderivatives, or accumulated quantities. It's an essential concept in calculus, used extensively in fields like physics, engineering, and economics.
There are two main forms of integration: indefinite and definite. Indefinite integration focuses on finding the general antiderivative and includes a constant of integration, \(+ C\). Meanwhile, definite integration determines the exact area or accumulated change between two points.

• Integration complements differentiation. Where differentiation slices up the curve to analyze small changes, integration synthesizes these slices to find total accumulation.
• Applied to our example \(\int_{1}^{2} t^3 \, dt \), integration allowed us to move back from the derivative function \(t^3\) to its antiderivative \(\frac{1}{4}t^4\).
• The bounds of integration fixed specific points, giving us a true measure of change or area between them.

Understanding integration as a tool for "summing up" changes makes it foundational to calculus and its applications.