Problem 7 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{4} \sqrt{t} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{16}{3} \).

Step 1: Identify the Function

The integral we need to evaluate is \( \int_{0}^{4} \sqrt{t} \, dt \). Here, the function under the integral sign is \( f(t) = \sqrt{t} \).

Step 2: Find the Antiderivative

To evaluate the integral, we need the antiderivative of \( \sqrt{t} \). Recall that \( \sqrt{t} = t^{1/2} \), so the antiderivative \( F(t) \) is found by increasing the exponent by 1 and dividing by the new exponent: \[ F(t) = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}. \]

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \[ \int_{a}^{b} f(t) \, dt = F(b) - F(a). \] For our integral, evaluate \( F(4) \) and \( F(0) \).

Step 4: Evaluate the Antiderivative at the Bounds

We now substitute the bounds into the antiderivative:- \( F(4) = \frac{2}{3} (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \).- \( F(0) = \frac{2}{3} (0)^{3/2} = 0 \).

Step 5: Calculate the Definite Integral

Subtract \( F(0) \) from \( F(4) \) using the results from the previous step to find:\[ \int_{0}^{4} \sqrt{t} \, dt = F(4) - F(0) = \frac{16}{3} - 0 = \frac{16}{3}. \]

Key Concepts

Definite Integral

The concept of a definite integral can be quite fascinating. A definite integral is essentially a number that represents the area under a curve defined by a function, over a specific interval. For example, if you have a curve described by the function \( f(t) \) from \( t = a \) to \( t = b \), the definite integral, denoted as \( \int_{a}^{b} f(t) \, dt \), provides a measure of the total area between the curve and the \( t \)-axis.
This measurement is not just the geometric area; the integral considers areas above the \( t \)-axis as positive and those below as negative.

• Integrals provide valuable tools to calculate various quantities, such as areas, volumes, and even more complex physical properties.
• The value of a definite integral depends on the limits \( a \) and \( b \), distinct from an indefinite integral, which outlines a family of antiderivatives without specific bounds.

Understanding definite integrals enables you to solve complex problems in mathematics and its applications, including physics and engineering.

Antiderivative

In the context of calculus, an antiderivative is a function that "undoes" the action of differentiation. If \( F(t) \) is an antiderivative of \( f(t) \), it means that \( F'(t) = f(t) \).
The process of finding an antiderivative is called integration, and it is pivotal when evaluating definite integrals.Finding the antiderivative of a function allows you to build a bridge from the derivative of the function to all possible original functions that could have led to that derivative. For instance:

• For the function \( f(t) = \sqrt{t} \), or \( t^{1/2} \), its antiderivative is found by increasing the exponent by 1, making it \( t^{3/2} \), and dividing by the new exponent. This results in \( F(t) = \frac{2}{3} t^{3/2} \).
• Understanding antiderivatives is vital, as they allow you to evaluate definite integrals using the Second Fundamental Theorem of Calculus.

Evaluating Integrals

Evaluating a definite integral involves a few crucial steps that link the concepts of antiderivatives and limits of integration. The Second Fundamental Theorem of Calculus plays a central role here.
This theorem states that if \( F \) is an antiderivative of \( f \) over an interval \([a, b]\), then the integral \( \int_{a}^{b} f(t) \, dt \) can be found by simply computing \( F(b) - F(a) \).
Here's how it unfolds using an example:

• First, identify the function you need to integrate, like \( \sqrt{t} \) in the example given.
• Next, find its antiderivative. For \( \sqrt{t} \), it is \( \frac{2}{3} t^{3/2} \).
• Then, evaluate this antiderivative at the upper and lower limits of your interval. Compute \( F(4) - F(0) \), which equals \( \frac{16}{3} - 0 = \frac{16}{3} \).

This process of substituting the limits and taking the difference provides the result of the integral, which simplifies complex expressions down to easily understandable quantities.