Problem 21 - Algebra 2

Question

Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ y=-x^{4}+2 x^{3}+3 $$

Step-by-Step Solution

Answer

The area under the curve \(y = -x^{4}+2 x^{3}+3\) from \(0 \leq x \leq 1\) is 5.5 square units

Step 1: Understand the Integral Function

For a function \(y=f(x)\), the definite integral from a to b, represented as \(\int_{a}^{b} f(x) dx\), gives the area under the curve. The task here is to find the area under the curve \(y = -x^{4}+2 x^{3}+3\) from \(0 \leq x \leq 1\). This translates to finding \(\int_{0}^{1} -x^{4}+2 x^{3}+3 dx\)

Step 2: Apply the Power Rule of Integration

The power rule of integration states that the integral of \(x^n\) with respect to x is \(\frac{x^{n+1}}{n+1}\). Using this rule, the integral of \(-x^{4}+2 x^{3}+3\) is \(-\frac{x^{5}}{5} + \frac{x^{4}}{2} + 3x\). Therefore, \(\int_{0}^{1} -x^{4}+2 x^{3}+3 dx = \[-\frac{x^{5}}{5} + \frac{x^{4}}{2} + 3x\] |_0^1\)

Step 3: Evaluate the Definite Integral at the Limits

To evaluate \[-\frac{x^{5}}{5} + \frac{x^{4}}{2} + 3x\] |_0^1\), first substitute x = 1 to get \(-\frac{1^{5}}{5} + \frac{1^{4}}{2} + 3*1 = -\frac{1}{5} + \frac{1}{2} + 3\). Next, substitute x = 0 to get 0. Subtract the latter result from the former to get \(-\frac{1}{5} + \frac{1}{2} + 3 - 0 = -\frac{1}{5} + \frac{5}{10} + 3 = \frac{2.5}{5} + 3 = 2.5 + 3 = 5.5 \)

Key Concepts

Power Rule of Integration

Integration is a crucial concept in calculus, helping us find areas under curves, among other things. One of the fundamental tools in integration is the power rule. This rule is a straightforward way to integrate functions like polynomial terms. If you have a function of the form \(x^n\), integrating it using the power rule is easy. You simply increase the exponent by one and then divide by the new exponent. So, the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\). For example, to integrate \(-x^4\), you would get \(-\frac{x^5}{5}\). This step-by-step method helps transform complicated expressions into simpler antiderivatives. Recognizing and applying the power rule efficiently makes solving integration problems much more manageable.

• Increase Exponent: Add 1 to the exponent.
• Divide by New Exponent: Divide by the newly increased exponent.

Area Under the Curve

Understanding the concept of the "area under the curve" is vital in calculus. The definite integral of a function over a specified interval calculates this area. In our example, the function \(y = -x^4 + 2x^3 + 3\) needs evaluation from \(x = 0\) to \(x = 1\). When you integrate a function, you essentially sum up an infinite number of tiny rectangles under the curve from the starting point to the endpoint. This gives you the total area under the curve, which can represent various physical quantities, such as distance or probability. By using the power rule of integration, we find the antiderivative and then the area between the curve and the x-axis over the given domain.
It's an essential method for calculating physical areas and understanding complex data represented graphically.

Evaluation of Definite Integrals

After finding the antiderivative of a function, the next step is the evaluation of the definite integral. This process involves computing the value of the integral at the upper and lower limits and then subtracting these results. For the function \(-x^4 + 2x^3 + 3\), we evaluated from \(0\) to \(1\).

Steps to Evaluate:

• Calculate the antiderivative at the upper limit (plug \(x = 1\)).
• Calculate the antiderivative at the lower limit (plug \(x = 0\)).
• Subtract the lower limit result from the upper limit result.

In our case, this results in \(-\frac{1}{5} + \frac{1}{2} + 3\), which simplifies to \(5.5\). This gives us the area under the curve within our specified domain. Understanding this concept ensures accuracy in solving practical and theoretical integral problems.