Problem 39 - Calculus

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^5/5 + x; (c) y = x^5/5 + x; (d) 6/5.

Step 1: Evaluate F(0)

Given that \[ F(x) = \int_{0}^{x}(t^4 + 1) \, dt, \] we want to find \( F(0) \). Since the upper and lower limits of the integral are both 0, \[ F(0) = \int_{0}^{0}(t^4 + 1) \, dt = 0. \] Thus, \( F(0) = 0. \)

Step 2: Differentiate F(x) using the First Fundamental Theorem of Calculus

According to the First Fundamental Theorem of Calculus, if \[ y = F(x) = \int_{0}^{x}(t^4 + 1) \, dt, \] then \[ \frac{dy}{dx} = F'(x) = x^4 + 1. \] This gives us the derivative of the function \( F(x) \).

Step 3: Solve the Differential Equation dy/dx = x^4 + 1

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

Step 4: Apply the Initial Condition to Find C

We know from Step 1 that \( F(0) = 0 \). Therefore, at \( x = 0 \), \( y = 0 \). Substituting into the equation, \[ 0 = \frac{0^5}{5} + 0 + C, \] implies \[ C = 0. \] This gives the particular solution \[ y = \frac{x^5}{5} + x. \]

Step 5: Evaluate the Definite Integral

To show \( \int_{0}^{1} (x^4 + 1) \, dx = \frac{6}{5}, \) we evaluate the integral: \[ \int_{0}^{1} (x^4 + 1) \, dx = \left[ \frac{x^5}{5} + x \right]_{0}^{1}. \] Substitute the upper and lower limits: \[ = \left( \frac{1^5}{5} + 1 \right) - \left( \frac{0^5}{5} + 0 \right) = \frac{1}{5} + 1 = \frac{1}{5} + \frac{5}{5} = \frac{6}{5}. \] This confirms the given statement.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that helps in finding the area under a curve between two limits. In the given problem, the integral \( F(x) = \int_{0}^{x}(t^4 + 1) \, dt \) represents the area from \( t = 0 \) to \( t = x \) under the curve \( t^4 + 1 \). When solving for \( F(0) \), you find that the definite integral from 0 to 0 results in 0. This is because there is no interval between the same limits, which implies no area is covered.

Using the properties of definite integrals:

• The integral of a function over an interval \([a, a]\) is always zero.
• If you know the function you want to integrate, calculating its exact area between specific limits helps in understanding accumulated quantities, like distance or total growth.

In the context of this problem, understanding definite integrals lets you appreciate how functions accumulate over an interval and provides a foundational step in calculus.

Differential Equations

Differential equations involve relationships between functions and their derivatives, characterizing the rates at which something changes. In this problem, we encounter the differential equation \( \frac{dy}{dx} = x^4 + 1 \). Solving such equations can determine how one quantity varies over time or space in relation to another.

When solving \( \frac{dy}{dx} = x^4 + 1 \), the solution involves integrating the function. By integrating, we undo the differentiation process, returning to the original function that describes \( y \). Through integration, you arrive at:

• \( y = \int (x^4 + 1) \, dx \)
• The result: \( y = \frac{x^5}{5} + x + C \) is our general solution with integration constant \( C \) representing any initial condition yet to be applied.

This process highlights the importance of differential equations in modeling various realistic scenarios, from physics to economics, capturing how systems change dynamically over time.

Initial Conditions

Initial conditions are crucial in solving differential equations as they help identify a particular solution. They set specific criteria that the solution must satisfy at a certain point. In practice, we might know the condition of a system at the start ("initial"), and want our solution to confirm this known fact. Applying initial conditions helps tailor our general solution to fit actual, observable data.

In the exercise, after solving the differential equation, the initial condition \( F(0) = 0 \) was applied, helping us determine the appropriate constant \( C \) in the solution. Here's how it works:

• Rewind the function by setting \( x = 0 \), corresponding to the start point.
• The condition \( y = 0 \) dictates \( C = 0 \), thus finalizing the specific solution: \( y = \frac{x^5}{5} + x \).

This ensures our solution accurately represents the situation from the start, adding value to the mathematical model's predictions.