Problem 99 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{1}\left(x^{3}-x^{1 / 3}\right) d x $$

Step-by-Step Solution

Answer

The value of the integral is \(-\frac{1}{2}\).

Step 1: Understand the Integral

The integral to evaluate is \( \int_{0}^{1}(x^3 - x^{1/3}) \, dx \). We are to find the area between the curve \( x^3 \) and \( x^{1/3} \) from 0 to 1.

Step 2: Integrate each Term Separately

First, separate the integral into two parts: \( \int_{0}^{1} x^{3} \, dx - \int_{0}^{1} x^{1/3} \, dx \). Use the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) to evaluate each part.

Step 3: Integrate \( x^3 \)

Apply the power rule to \( x^3 \): \[ \int x^3 \, dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4} \]. Evaluate this from 0 to 1, giving \( \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \).

Step 4: Integrate \( x^{1/3} \)

Apply the power rule to \( x^{1/3} \): \[ \int x^{1/3} \, dx = \frac{x^{(1/3)+1}}{(1/3)+1} = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} \]. Evaluate this from 0 to 1, giving \( \frac{3}{4}(1^{4/3}) - \frac{3}{4}(0^{4/3}) = \frac{3}{4} \).

Step 5: Calculate the Final Result

Subtract the results from Step 3 and Step 4: \( \frac{1}{4} - \frac{3}{4} = -\frac{1}{2} \). Thus, the value of the definite integral is \(-\frac{1}{2}\).

Key Concepts

Calculus and Integration

Calculus is a branch of mathematics that deals with the study of change and motion. One of its main components is integration, which is used to calculate things like total size or quantity. In our problem, we use integration to find the area under a curve between two points.

Definite integrals, specifically, allow us to calculate the exact area between a curve and the x-axis over a specified interval. When we write \[\int_{a}^{b} f(x) \, dx\]we mean to calculate this area from the point \( a \) to \( b \). The result of the integral will give us a numerical value representing this area.

• It can be used to determine the total distance traveled given a velocity function over time.
• It helps in calculating total accumulated quantities, such as the total charge accumulated in a capacitor over time.

Understanding calculus through integration opens up a deeper comprehension of how mathematical models describe the world around us.

Applying the Power Rule

The power rule is a fundamental tool in calculus used for integrating functions of the form \( x^n \). The power rule states that:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]where \( C \) is the constant of integration in indefinite integrals. However, when dealing with definite integrals, as we are here, the constant \( C \) cancels out because we evaluate the antiderivative at two points and subtract.

In our exercise, we applied the power rule to two separate functions, \( x^3 \) and \( x^{1/3} \). By transforming each using the power rule, we found their antiderivatives:

• For \( x^3 \), we got \( \frac{x^4}{4} \).
• For \( x^{1/3} \), we derived \( \frac{3}{4}x^{4/3} \).

These transformations allow us to evaluate each integral over a given interval, providing a numerical result for the definite integral.

Area Under a Curve

Finding the area beneath a curve is a practical application of integration in calculus. This method is especially useful in determining the space between a function's graph and the x-axis over a particular range.

In the definite integral of our problem, we are trying to find the net area between two curves: \( x^3 \) and \( x^{1/3} \), from 0 to 1. Calculating each area separately and then subtracting one from the other helps us determine the total area enclosed.

• The positive area above the x-axis is given by the integral of \( x^3 \).
• The area below the curve \( x^{1/3} \) acts as a subtracted section that influences the overall result."

The final result, \(-\frac{1}{2}\), indicates that the net area between these curves is negative, meaning more area lies beneath the x-axis overall during this interval, leading to an important insight about the behavior of these functions between the specified limits.