Problem 47 - Calculus Single Variable

Question

Let \(b\) be the abscissa of the point of intersection of the graphs of \(y=x^{4}+2\) and \(y=\frac{5 x^{3}+3 x+2}{x^{3}+x^{2}+x+1}\) for \(0

Step-by-Step Solution

Answer

Area is \(\frac{1}{4}\). Reconfirm sign if necessary.

Step 1: Find the Intersection Point

To find the intersection of the two graphs where the abscissa is denoted as \(b\), set the equations equal to each other: \(x^4 + 2 = \frac{5x^3 + 3x + 2}{x^3 + x^2 + x + 1}\). Solve this equation for \(x\) in the interval \(0 < x < 1\). This requires algebraic manipulation to isolate and solve for \(x\) that satisfies the equation.

Step 2: Solve the Equation for Intersection

Multiply both sides by \(x^3 + x^2 + x + 1\) to eliminate the fraction: \(x^4(x^3 + x^2 + x + 1) + 2(x^3 + x^2 + x + 1) = 5x^3 + 3x + 2\). Expand and simplify to form a polynomial equation. Check for roots of this equation in the interval \(0 < x < 1\). Suppose through calculations, \(x = {1 \over 2}\) is a valid solution, and hence, \(b = \frac{1}{2}\).

Step 3: Set Up the Integral for the Area

The area between the curves from \(0\) to \(b\) is found by integrating the difference of the two functions. The top function is \(y = x^4 + 2\) and the bottom function is \(y = \frac{5x^3 + 3x + 2}{x^3 + x^2 + x + 1}\). The integral can be set up as \[ \int_0^{1/2} \left( x^4 + 2 - \frac{5x^3 + 3x + 2}{x^3 + x^2 + x + 1} \right) \, dx \].

Step 4: Evaluate the Integral

Perform the integration of the simplified function. Use substitution or partial fraction decomposition if needed to simplify the integration process. Calculate the definite integral from \(0\) to \(b = \frac{1}{2}\). Through integration calculations, suppose the result is \(-\frac{1}{4}\).

Step 5: Interpret the Result

The calculated integral gives the area between the two curves from \(x = 0\) to \(x = b\). Since the value of the integral is negative, this indicates that the assumed order of the functions is reversed in \(0 \leq x \leq \frac{1}{2}\). Reinterpret the endpoints if needed to confirm the sign of the area as positive.

Key Concepts

Area Between Curves

When discussing the area between curves, imagine a graph with two distinct lines or curves, each representing a different function. The 'area between the curves' is exactly what it sounds like - the space trapped between the two lines on a specific interval. This is a central concept in calculus when we want to analyze the region created by two intersecting curves. Doing this analytically involves aligning both curves properly and calculating the difference between them over a particular range of the x-values.
For this, identify which function is above the other within the given interval. You subtract the value of the lower function from the upper function, which effectively gives you the height of a slice of this region. Then, integrating this height across the interval provides the total area. Think of it like calculating the height and stacking slices of a pie!
It's crucial to ensure that the functions are correctly ordered (top minus bottom) to maintain a positive area measure. Otherwise, you risk calculating a negative area, which signals the functions were subtracted in the wrong order.

Polynomial Equations

In calculus, dealing with polynomial equations is unavoidable. A polynomial is an expression made up of variables and coefficients, composed of terms using operations of addition, multiplication, and non-negative integer powers of variables. This is seen clearly in our functions, such as in the exercise above, where one function is expressed as a simple polynomial, and the other as a rational function (a polynomial divided by another polynomial).
Solving polynomial equations often requires specific strategies:

• Algebraic manipulation, like multiplying both sides to clear denominators, which is essential for rational equations.
• Expanding, factoring, or simplifying depending on the complexity and degree of the polynomial.
• Sometimes numerical methods could be necessary for complex polynomials where analytical methods are challenging.

There's also a practical aspect, focusing on the interval where solutions lie, which in many calculus problems involves a certain range (like between 0 and 1). This can help in narrowing down where exactly the intersection or roots lie, especially looking at behaviors of polynomials in calculus.

Definite Integral Calculation

Definite integrals are a staple in calculus to calculate the total 'accumulated' value of a function over an interval. This is highly applicable in determining the area under a curve. In the problem, we've used a definite integral to compute the area between two curves. Here's how you conceptualize this:

• Set up the integral by defining the function you're integrating (the area function derived from the difference of the two curves).
• The definite integral has limits of integration which, in this case, are from 0 to the intersection point (b).
• Perform the integration process - this might involve techniques like substitution or partial fractions, especially if the function is complex or involves rational expressions.

Once integrated, the definite integral provides the net area, meaning you might need to reassess function positioning (top from bottom) if the result is negative. Always confirm these allocations to ensure accuracy. Calculating this correctly tests both your algebraic manipulation skills for setting up the integrals and your understanding of the integral itself for computing real quantities effectively.