Problem 65 - Calculus Single Variable

Question

In each of Exercises \(65-68\), a definite integral is given. Do not attempt to calculate its value \(V .\) Instead, find the extreme values of the integrand on the interval of integration, and use these extreme values together with the inequalities of line (5.3.5) to obtain numbers \(A\) and \(B\) such that \(A \leq V \leq B\). $$ \int_{0}^{2} \frac{1+x}{1+x^{4}} d x $$

Step-by-Step Solution

Answer

For the integral \( \int_{0}^{2} \frac{1+x}{1+x^4} \,dx \), the extreme values show that \( 0.352 \leq V \leq 2 \).

Step 1: Identify the Integrand

The integrand for the given definite integral is \( f(x) = \frac{1+x}{1+x^4} \). This is the function whose extreme values we need to find on the interval \([0, 2]\).

Step 2: Find Critical Points

To find the extreme values, we first find the derivative of \( f(x) \). Using the quotient rule, the derivative \( f'(x) \) is given by: \[ f'(x) = \frac{(1 \cdot (1 + x^4)) - ((1+x) \cdot 4x^3)}{(1+x^4)^2} = \frac{1 + x^4 - 4x^3 - 4x^4}{(1+x^4)^2} = \frac{1 - 4x^3 - 3x^4}{(1+x^4)^2} \]. Set \( f'(x) = 0 \) and solve for \( x \).

Step 3: Solve for Critical Points

From \( f'(x) = 0 \), solve the equation \( 1 - 4x^3 - 3x^4 = 0 \). Testing for values in the interval \([0, 2]\), assume potential roots since solving analytically is complex. Notice that strong inspection for simple cases shows that near \( x = 0 \) and \( x = 2 \) can contribute to extrema.

Step 4: Evaluate at Endpoints

Evaluate the function \( f(x) \) at the endpoints of the interval, which are \( x = 0 \) and \( x = 2 \). Compute: \( f(0) = \frac{1}{1} = 1 \) and \( f(2) = \frac{3}{17} \approx 0.176 \).

Step 5: Find Intervals for Min/Max

Evaluate \( f(x) \) around other critical regions using test points like \( x = 1 \) within \([0, 2]\) to check if other more accurate extrema are present, giving \( f(1) \approx 0.2 \). Compare these values: 1, 0.176, and 0.2 to find global extrema.

Step 6: Use Extreme Values with Inequality

Based on the extreme values found, the function has maxima at \( x=0 \) with \( f(0) = 1 \) and minima at \( x=2 \) with \( f(2) = 0.176 \). Using inequality, estimate \( A = 2 \times 0.176 = 0.352 \), and \( B = 2 \times 1 = 2 \). Therefore, \( A \leq V \leq B \) is equivalent to \( 0.352 \leq V \leq 2 \).

Key Concepts

Extreme Values

In calculus, extreme values refer to the maximum and minimum values of a function on a particular interval. They play a key role in assessing the behavior of the integrand over the interval of integration. To find these extreme values, we look for points in the function where the derivative is zero or undefined, known as critical points. For the integrand given in this exercise, an analysis of the derivative helps in identifying these vital points.- On the interval \([0, 2]\), we calculated the derivative of \(f(x) = \frac{1 + x}{1 + x^4}\) to find where it changes direction.\( f'(x) = \frac{1 - 4x^3 - 3x^4}{(1+x^4)^2} \). Setting this to zero pinpoints potential extrema.- By evaluating the function at critical points, endpoints, and test values, we determine the extreme values within the specified domain.- In this case, \(f(0) = 1\) provides the maximum and \(f(2) = 0.176\) gives the minimum value on the interval \([0, 2]\). Understanding extreme values allows us to establish bounds for the integral value.

Integrand

The integrand is the core function inside the integral, representing the quantity to be summed up over the interval. In our given problem, the integrand is \(f(x) = \frac{1+x}{1+x^4}\). This rational expression combines a linear term \(1 + x\) in its numerator and a quartic term \(1 + x^4\) in its denominator.- The behavior of the integrand significantly affects the integral's evaluation, especially its extreme values and zero crossings.- By identifying the integrand, it becomes easier to find the derivative and assess how the function behaves across the interval.- Since this integrand involves a higher degree polynomial in the denominator, it results in a bounded range of outputs, making the derivation of extreme values more intricate.

Interval of Integration

The interval of integration is crucial as it defines the domain over which we evaluate the definite integral. For our problem, the interval is \([0, 2]\). Understanding this interval allows us to focus on the function's behavior precisely between these two points.- The endpoints, 0 and 2, are critical in evaluating both the integrand and its derivative, as the behavior at these points determines the boundary values for the integral.- Within this range, we examine how the function changes, including any increases, decreases, or stationary points that might suggest an extremum.- Using this interval, we find the greatest and lowest values of the integrand to determine the integral's bounds, offering insight into how the function is distributed across the range.

Quotient Rule

The quotient rule is a technique in calculus used to differentiate functions that are the ratio of two differentiable functions. In this context, we applied it to determine the derivative of the integrand \(f(x) = \frac{1+x}{1+x^4}\).- Using the quotient rule involves taking the derivative of the numerator and the denominator separately: \(f'(x) = \frac{(1 \,\cdot\, (1 + x^4)) - ((1+x) \,\cdot\, 4x^3)}{(1+x^4)^2}\).- This rule is instrumental when the function's derivative must be accurately calculated to find critical points.- Understanding the quotient rule equips you to handle more complex integral problems, especially those involving rational functions, by simplifying the process of identifying extreme values.