Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{1}(x+\sqrt{x}) d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral: $$\int_{0}^{1}(x+\sqrt{x}) dx$$ Answer: $$\frac{7}{6}$$

Step 1: Understand the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function is continuous on the interval [a, b], and F(x) is an antiderivative of the function, then the definite integral of the function equals: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$

Step 2: Find the antiderivative of the given function

In this case, the given function is \((x + \sqrt{x})\). We'll need to find its antiderivative, that is, finding a function F(x) such that F'(x) = x + \sqrt{x}$. We will do this by finding the antiderivative of each term separately and then adding them. For the first term, x, the antiderivative is: $$\int x dx = \frac{1}{2}x^2$$ For the second term, \(\sqrt{x}\) or \(x^{\frac{1}{2}}\), the antiderivative is: $$\int x^{\frac{1}{2}} dx = \frac{2}{3}x^{\frac{3}{2}}$$ So, the antiderivative of \((x + \sqrt{x})\) is: $$F(x) = \frac{1}{2}x^2 + \frac{2}{3}x^{\frac{3}{2}}$$

Step 3: Apply the Fundamental Theorem of Calculus

Now, we will use the Fundamental Theorem of Calculus to evaluate the integral: $$\int_{0}^{1}(x+\sqrt{x}) dx = F(1) - F(0)$$ Compute the value of F(1) and F(0): $$F(1) = \frac{1}{2}(1)^2 + \frac{2}{3}(1)^{\frac{3}{2}} = \frac{1}{2} + \frac{2}{3}$$ $$F(0) = \frac{1}{2}(0)^2 + \frac{2}{3}(0)^{\frac{3}{2}}= 0$$ Substitute the values of F(1) and F(0) back into the equation: $$\int_{0}^{1}(x+\sqrt{x}) dx = (\frac{1}{2} + \frac{2}{3}) - 0$$

Step 4: Calculate the value of the definite integral

Now that we have the values of F(1) and F(0), we can compute the value of the definite integral: $$\int_{0}^{1}(x+\sqrt{x}) dx = \frac{1}{2} + \frac{2}{3} = \frac{3 + 4}{6} = \frac{7}{6}$$ The value of the definite integral is \(\frac{7}{6}\).

Key Concepts

Understanding the Definite Integral

A definite integral is a fundamental concept in calculus used to find the accumulated area under a curve within a specific interval. Think of it as summarizing an infinite number of tiny slices that make up an entire region on a graph.
It is denoted as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function being integrated.
This integral gives us a number that describes the net area between the function and the x-axis from \(x = a\) to \(x = b\).

The Fundamental Theorem of Calculus is critical for evaluating these integrals. It connects derivatives and integrals, allowing us to compute the area under a curve by using antiderivatives.

• The theorem tells us that the definite integral of a function from \(a\) to \(b\) can be found by calculating the difference between the values of an antiderivative evaluated at \(b\) and \(a\).

This process simplifies what could be a complicated sum of areas into straightforward arithmetic.

Exploring Antiderivatives

An antiderivative is a function whose derivative yields the original function. If you're familiar with derivatives, an antiderivative is essentially the reverse process.
For example, if the derivative of \( F(x) \) is \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).

Antiderivatives are crucial in finding definite integrals. Once you find the antiderivative of a function, you can use it to evaluate the integral using the Fundamental Theorem of Calculus.

• In our exercise, the function \((x + \sqrt{x})\) was broken into parts to find its antiderivative, resulting in \( F(x) = \frac{1}{2}x^2 + \frac{2}{3}x^{3/2} \).
• This expression is used to evaluate the integral from 0 to 1 by substituting the limits into the antiderivative function.

Finding antiderivatives involves understanding different rules and techniques, such as power rule, exponential rule, and more. These skills enable you to address various functions effectively.

Continuous Functions and Their Importance

A continuous function is a function without any breaks, jumps, or holes in its domain. It means that you can draw the function's graph without lifting your pencil off the paper.
This property is crucial because the Fundamental Theorem of Calculus relies on it.

Continuous functions ensure that integrals and antiderivatives are well-defined and behave predictably within the interval being considered.

• In the exercise, the function \(x + \sqrt{x}\) is continuous on the interval \([0, 1]\), which makes it suitable for applying the Fundamental Theorem.
• Without continuity, certain functions might lead to complications or undefined regions, disrupting the calculation process.

When tackling calculus problems, recognizing and verifying the continuity of functions can simplify solving integrals and ensure accurate results.