Problem 10 - Calculus

Question

If \(v(x)=d f / d x,\) what constants \(C\) make \(1-10\) true? $$ \int_{0}^{2} v(x) d x=C \int_{0}^{1} v(2 t) d t . $$

Step-by-Step Solution

Answer

C = 2.

Step 1: Understand the Problem

We are given that the derivative \(v(x)\) is equal to \(\frac{df}{dx}\). We need to find the constant \(C\) such that the integral from 0 to 2 of \(v(x)\) is equal to \(C\) times the integral from 0 to 1 of \(v(2t)\) with respect to \(t\).

Step 2: Change of Variables for Right Integral

Perform a change of variables for the integral on the right side. Let \( u = 2t \), then \( du = 2dt \) or \( dt = \frac{1}{2} du \). When \( t = 0, u = 0 \) and when \( t = 1, u = 2 \). Thus the integral changes to: \[ \int_{0}^{1} v(2t) \, dt = \frac{1}{2} \int_{0}^{2} v(u) \, du. \]

Step 3: Substitute Change of Variable Result into Equation

Substitute the result of the variable change back into the original equation: \[ \int_{0}^{2} v(x) \, dx = C \times \frac{1}{2} \int_{0}^{2} v(u) \, du. \] This implies \( \int_{0}^{2} v(x) \, dx = \frac{C}{2} \int_{0}^{2} v(x) \, dx \).

Step 4: Solve for Constant C

Equating both sides of the equation \( \int_{0}^{2} v(x) \, dx = \frac{C}{2} \int_{0}^{2} v(x) \, dx \), we can divide by \( \int_{0}^{2} v(x) \, dx \) (assuming it is non-zero), yielding 1 = \( \frac{C}{2} \). Solving for \( C \), we find \( C = 2 \).

Step 5: Conclusion

The constant \( C \) that satisfies the integral equation is \( C = 2 \). Substitute back to double-check and confirm if \( C = 2 \) works correctly for the given integrals.

Key Concepts

Change of Variables

The technique of "change of variables" is a powerful method used in calculus to simplify the process of integration. Imagine navigating a complex puzzle; a change of variables is like finding a simpler path through that puzzle. The key is substituting a tricky part of the integral with a new, easier expression.
In the given exercise, to simplify the integration of \(v(2t)\), we let \( u = 2t \). This substitution helps convert the variable \(t\) to \(u\) so that handling the limits of integration and the integrand becomes more straightforward. Consequently, the derivative of \(u\) is \(du = 2dt\), and solving for \( dt \) gives us \( dt = \frac{1}{2}du \).

• This change adjusts the limits of integration: when \(t\) equals 0, \(u\) will also be 0; and when \(t\) equals 1, \(u\) becomes 2.
• The integral \( \int_{0}^{1} v(2t) \, dt \) changes to \( \frac{1}{2} \int_{0}^{2} v(u) \, du \), which now shares the same form as the left-side integral.

By finding the right transformation using the change of variables, you can simplify complex integrals into more manageable forms.

Definite Integrals

Definite integrals provide a numerical value that can describe quantities like area under a curve or total change. Calculating a definite integral from a function tells you the cumulative effect over a particular interval. This concept is helpful in measuring everything from speed over time to the amount of a resource consumed.
In our context, we're dealing with the definite integral \( \int_{0}^{2} v(x) \, dx \), representing the accumulation of function \(v(x)\) between the limits of 0 and 2.

• The problem involves two definite integrals: the original integral of \(v(x)\) from 0 to 2, and the modified integral of \(v(2t)\) from 0 to 1, which after substitution becomes from 0 to 2, just like the original.
• By analyzing these integrations, we match them in form through the change of variables, allowing direct comparison and evaluation.

Understanding definite integrals is key to interpreting the impact or quantity captured over a specified interval.

Integration Techniques

Integration techniques encompass various methods used to evaluate integrals, whether indefinite or definite. By mastering these techniques, solving complex integrals can become a more intuitive and manageable task.
In this exercise, we used the change of variables—a technique crucial for simplifying integrals that involve compositions or transformations of functions. It changes the form of the integrand, allowing us to solve the integral more easily.

• Besides changing variables, other integration techniques include parts, partial fractions, and trigonometric substitution—each suited to different types of problems.
• Here, once the substitution was made, solving for the constant \(C\) involved equalizing the transformed definite integrals on both sides, which made the solution straightforward.

Through applying the right technique, such as variable change, we efficiently derive solutions to integral equations, determining constants and relationships between them.