Chapter 4: Problem 7
Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{4} \sqrt{t} d t $$
Short Answer
Expert verified
The integral evaluates to \( \frac{16}{3} \).
Step by step solution
01
Identify the Function
The integral we need to evaluate is \( \int_{0}^{4} \sqrt{t} \, dt \). Here, the function under the integral sign is \( f(t) = \sqrt{t} \).
02
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: \[ F(t) = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}. \]
03
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 \( F(0) \).
04
Evaluate the Antiderivative at the Bounds
We now substitute the bounds into the antiderivative:- \( F(4) = \frac{2}{3} (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \).- \( F(0) = \frac{2}{3} (0)^{3/2} = 0 \).
05
Calculate the Definite Integral
Subtract \( F(0) \) from \( F(4) \) using the results from the previous step to find:\[ \int_{0}^{4} \sqrt{t} \, dt = F(4) - F(0) = \frac{16}{3} - 0 = \frac{16}{3}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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:
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:
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 \( \frac{16}{3} - 0 = \frac{16}{3} \).