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 \(\int_{0}^{1}\left(x^{4}+1\right) d x=\frac{6}{5}\).

Short answer

(a) F(0) = 0. (b) y = \(\frac{x^5}{5}+x\). (c) The solution is \(y=\frac{x^5}{5}+x\) with C=0. (d) The integral evaluates to \(\frac{6}{5}\).

Step-by-step solution

Evaluating F(0)

To find \( F(0) \), we substitute \( x = 0 \) into the integral function: \( F(x) = \int_{0}^{x}(t^4 + 1) \, dt \). So we compute: \[ F(0) = \int_{0}^{0}(t^4 + 1) \, dt = 0 \]This is because the integral from 0 to 0 of any function results in 0.

Applying the First Fundamental Theorem of Calculus

Using the First Fundamental Theorem of Calculus, if \( F(x) = \int_{0}^{x}(t^4 + 1) \, dt \), then the derivative \( F'(x) = dF/dx \) is equal to \( t^4 + 1 \) evaluated at \( t = x \). Therefore, \[ F'(x) = x^4 + 1 \].

Solving the Differential Equation

We solve the differential equation \( dy/dx = x^4 + 1 \) by integrating both sides:\[ y = \int (x^4 + 1) \, dx \].This gives us: \[ y = \frac{x^5}{5} + x + C \], where \( C \) is the constant of integration.

Determining the Constant C

Since \( y = F(0) \) when \( x = 0 \), and we found that \( F(0) = 0 \), substitute these into the equation:\[ 0 = \frac{0^5}{5} + 0 + C \].Thus, \( C = 0 \). Therefore, the solution is \[ y = \frac{x^5}{5} + x \].

Evaluating Definite Integral \( \int_{0}^{1}(x^4 + 1) \, dx \)

To show \( \int_{0}^{1}(x^4 + 1) \, dx = \frac{6}{5} \), evaluate the integral:\[ \int_{0}^{1}(x^4 + 1) \, dx = \left[ \frac{x^5}{5} + x \right]_{0}^{1} \].Calculate this by evaluating the antiderivative at 1 and subtracting the evaluation at 0:- For \( x = 1 \), \( \frac{1^5}{5} + 1 = \frac{1}{5} + \frac{5}{5} = \frac{6}{5} \).- For \( x = 0 \), \( \frac{0^5}{5} + 0 = 0 \).Thus, \( \int_{0}^{1}(x^4 + 1) \, dx = \frac{6}{5} - 0 = \frac{6}{5} \).

Key concepts

Definite Integral

Definite integrals provide a way to calculate the area under a curve over a specific interval. In our problem, the function is \(t^4 + 1\), and we are asked to evaluate the integral from 0 to 1. This is written as \(\int_{0}^{1} (x^4 + 1) \, dx\). A definite integral has two boundaries: a lower limit (0 in this case) and an upper limit (1 in this case).

To calculate a definite integral, we first need to find the antiderivative of the function. This is a function whose derivative is the integrand \((x^4 + 1)\). In this context, the antiderivative is \(\frac{x^5}{5} + x\).

Next, we evaluate this antiderivative at the upper and lower limits and subtract the results:

• At \(x = 1\), the antiderivative becomes \(\frac{1^5}{5} + 1 = \frac{6}{5}\).
• At \(x = 0\), it becomes \(0\). The area under the curve is then \(\frac{6}{5} - 0 = \frac{6}{5}\).

This gives the total accumulated value of the function from 0 to 1.

Differential Equations

Differential equations involve functions and their derivatives. They help describe various physical phenomena, such as population growth or cooling laws, through rates of change. In this exercise, we have the differential equation \(dy/dx = x^4 + 1\), which we need to solve.

The solution involves reevaluating the equation to find the function \(y\) by reversing the process of differentiation, that is by integration. By integrating \(x^4 + 1\) with respect to \(x\), we obtain:

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

where \(C\) is the constant of integration that we find in another step.

This integral provides us with a family of solutions that satisfy the original differential equation.

Constant of Integration

When integrating a function, an important final step is adding the constant of integration, denoted as \(C\). This constant accounts for all the possible vertical shifts in the graph of the antiderivative because the derivative of a constant is zero, making it invisible in the original differential equation.

In our specific case, after finding \(y = \frac{x^5}{5} + x + C\), we used the initial condition \(y = F(0) = 0\) when \(x = 0\) to solve for \(C\).

• Substituting these values gave us: \(0 = \frac{0^5}{5} + 0 + C\), which simplifies to \(C = 0\).

Thus, the specific solution becomes \(y = \frac{x^5}{5} + x\).

Finding the correct \(C\) is crucial as it allows us to find the specific solution to a problem within a family of solutions.