Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(F(x)=\int_{0}^{x}\left(t^{4}+1\right) d t\). (a) Find \(F(0)\). (b) Let \(y=F(x)\). Apply the First Fundamental Theorem of Calculus to obtain \(d y / d x=F^{\prime}(x)=x^{4}+1\). Solve the differential equation \(d y / d x=x^{4}+1\). (c) Find the solution to this differential equation that satisfies \(y=F(0)\) when \(x=0\). (d) Show that \(\int_{0}^{1}\left(x^{4}+1\right) d x=\frac{6}{5}\).

Step-by-Step Solution

Answer

(a) 0; (b) y(x) = \(\frac{x^5}{5} + x\); (d) \(\frac{6}{5}\).

Step 1: Evaluate F(0)

To find \( F(0) \), substitute \( x = 0 \) into the integral \( F(x) = \int_{0}^{x} (t^4 + 1) \, dt \). This becomes:\[ F(0) = \int_{0}^{0} (t^4 + 1) \, dt = 0 \] Since the integral from a point to itself is zero, \( F(0) = 0 \).

Step 2: Apply the First Fundamental Theorem of Calculus

The First Fundamental Theorem of Calculus states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). Here, \( f(t) = t^4 + 1 \), so:\[ F'(x) = x^4 + 1 \] Thus, \( \frac{dy}{dx} = F'(x) = x^4 + 1 \).

Step 3: Solve the Differential Equation

To solve the differential equation \( \frac{dy}{dx} = x^4 + 1 \), integrate both sides with respect to \( x \):\[ y = \int (x^4 + 1) \, dx \]This results in:\[ y = \int x^4 \, dx + \int 1 \, dx = \frac{x^5}{5} + x + C \]where \( C \) is the constant of integration.

Step 4: Apply Initial Condition

According to the problem, \( y = F(0) \) when \( x = 0 \). From Step 1, we know \( F(0) = 0 \). Substitute \( x = 0 \) and \( y = 0 \) into the equation \( y = \frac{x^5}{5} + x + C \):\[ 0 = \frac{0^5}{5} + 0 + C \]Thus, \( C = 0 \). The solution becomes \( y = \frac{x^5}{5} + x \).

Step 5: Verify the Integral

We need to compute \( \int_{0}^{1} (x^4 + 1) \, dx \) and verify it equals \( \frac{6}{5} \). Integrate the expression:\[ \int_{0}^{1} (x^4 + 1) \, dx = \left[ \frac{x^5}{5} + x \right]_{0}^{1} = \left( \frac{1^5}{5} + 1 \right) - \left( \frac{0^5}{5} + 0 \right) = \frac{1}{5} + 1 = \frac{6}{5} \]This confirms the calculated integral is indeed \( \frac{6}{5} \).

Key Concepts

Definite Integrals

The concept of definite integrals is essential in understanding the area under the curve for a given function over a specified interval. The notation \( \int_{a}^{b} f(x) \, dx \) represents the definite integral of the function \( f(x) \) from \( a \) to \( b \). For our specific problem, the definite integral \( \int_{0}^{1} (x^4 + 1) \, dx \) calculates the area under the curve of \( x^4 + 1 \) between \( x = 0 \) and \( x = 1 \).

The calculation of a definite integral provides a numerical answer representing this area. In our example, the result of \( \frac{6}{5} \) tells us exactly how much area is enclosed under the function from 0 to 1. This ability to represent on-the-ground totals for expressions makes definite integrals powerful in applied mathematics.

Differential Equations

A differential equation is an equation that involves the derivatives of a function. In the exercise, we dealt with the differential equation \( \frac{dy}{dx} = x^4 + 1 \). Solving such an equation means finding the function \( y \) such that its derivative matches the given equation.

To solve, we integrate the function on the right-hand side with respect to \( x \), leading to \( y = \int (x^4 + 1) \, dx \). This results in the expression \( y = \frac{x^5}{5} + x + C \), where \( C \) is the constant of integration. The method of integrating both sides is a common approach to finding the general solution to a differential equation.

Initial Conditions

Initial conditions are specific values provided for a function or its derivatives at a particular point. They are crucial for finding the particular solution to a differential equation. Given initial conditions, we can determine the specific constants within the general solution.

In our problem, the initial condition was \( y = F(0) \) when \( x = 0 \), and we knew from Step 1 that \( F(0) = 0 \). Using these values allowed us to solve for \( C \), the arbitrary constant in our integrated solution. By substituting \( x = 0 \) and \( y = 0 \) into \( y = \frac{x^5}{5} + x + C \), we found that \( C = 0 \). This specific constant lets us accurately solve the differential equation with the given condition.

Integration Techniques

Integration is the process of finding an integral, which can be used to compute area, volume, and other quantities. In dealing with polynomials like \( x^4 + 1 \), straightforward integration methods can be applied directly.

In this case, finding \( y = \int (x^4 + 1) \, dx \) involved separating it into \( \int x^4 \, dx \) and \( \int 1 \, dx \). Then we integrated each part independently:

• \( \int x^4 \, dx = \frac{x^5}{5} \)
• \( \int 1 \, dx = x \)

Combining these results gave \( y = \frac{x^5}{5} + x + C \). These "basic integral forms" are crucial for efficiently solving problems involving polynomials.