Problem 26 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. Sketch the graph of the integrand and shade the region whose net area you have found. $$\int_{0}^{1}(x-\sqrt{x}) d x$$

Step-by-Step Solution

Answer

Question: Find the net area between the curve y=(x-√x) and the x-axis from x=0 to x=1. Answer: The net area between the curve y=(x-√x) and the x-axis from x=0 to x=1 is $$-\frac{1}{6}$$ square units.

Step 1: Finding the antiderivative of the integrand

To find the antiderivative of the function (x-√x), we first have to rewrite it as (x-x^(1/2)). Now, we will find the antiderivative individually for both terms. For x: $$\int x dx = \frac{1}{2}x^{2} + C_1$$ For x^(1/2): $$\int x^{1/2} d x = \frac{2}{3}x^{3/2} + C_2$$ Adding both antiderivatives, we get: $$\int (x - x^{1/2}) d x = \frac{1}{2}x^2 - \frac{2}{3}x^{3/2} + C$$

Step 2: Evaluating the integral using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the definite integral is the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit: $$\int_{0}^{1} (x - x^{1/2}) d x = \left[\frac{1}{2}x^2 - \frac{2}{3}x^{3/2}\right]_0^1$$ Evaluating the antiderivative at the upper limit 1: $$\frac{1}{2}(1)^2 - \frac{2}{3}(1)^{3/2} = \frac{1}{2} - \frac{2}{3}$$ Evaluating the antiderivative at the lower limit 0: $$\frac{1}{2}(0)^2 - \frac{2}{3}(0)^{3/2} = 0$$ Subtracting the lower limit result from the upper limit result: $$\int_{0}^{1} (x - x^{1/2}) d x = \frac{1}{2} - \frac{2}{3} = \frac{-1}{6}$$

Step 3: Sketching the graph of the integrand and shading the region

Sketch the graph of the function y=(x-x^(1/2)) in the interval [0, 1]. The function is a curve that intersects the x-axis at x=0 and x=1. It dips below the x-axis between these two points since for certain x values in the interval (0, 1), x^(1/2) is greater than x. This causes the net area to be negative. Now shade the region between the curve y=(x-x^(1/2)) and the x-axis from x=0 to x=1. The shaded area represents the net area found, which is equal to $$-\frac{1}{6}$$ in this case. This negative value indicates that the curve lies below the x-axis in the interval [0, 1].

Key Concepts

Definite Integrals

A definite integral is a mathematical tool that allows us to calculate the net area under a curve between two specific points on the x-axis. It's defined as the integral of a function over a specific interval

• The limits of integration specify this interval and are written at the top and bottom of the integral symbol.
• The fundamental theorem of calculus links the concept of integration with differentiation.

To evaluate a definite integral, we find the antiderivative of the integrand, the function being integrated. The next step is to use the fundamental theorem of calculus, which involves calculating the difference between the values of this antiderivative at the specified upper and lower limits of integration. This gives the signed area under the curve that the integrand represents, which can be positive or negative depending on whether the curve lies above or below the x-axis.

Antiderivative

Antiderivatives are functions that reverse the process of differentiation. When we find an antiderivative of a function, we are essentially finding a function whose derivative would give us back the original function.

• The process is often referred to as 'integration,' and it involves using standard formulas and rules.
• For example, the antiderivative of x is \(\frac{1}{2}x^2\), since the derivative of \(\frac{1}{2}x^2\) is x.

When we integrate more complex functions, like polynomials or terms with roots, we consider each term separately. In the given exercise, we split the function \(x \sqrt{x}\) into two parts and integrated them individually. The sum of these two antiderivatives gives us the complete antiderivative of the whole function. The resulting expression can be used to find the value of the definite integral by applying the fundamental theorem of calculus.

Net Area

Net area refers to the total area between the curve of the integrand and the x-axis across a specified interval, considering the sign based on the curve's position.

• If the curve is above the x-axis on the interval, the area is positive.
• If below, the area is deemed negative.

The concept of net area is distinct from just the area because it accounts for the position of the function relative to the axis. In the exercise, the integrand \(x - x^{1/2}\) crosses below the x-axis, making the resulting integral \(-\frac{1}{6}\), a negative net area. Net area is crucial as it reflects real-world concepts such as distance and displacement, emphasizing directionality.