Question

Evaluate the following definite integrals. $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t$$

Short answer

Answer: The definite integral of the given vector function is \(1\mathbf{i} + \frac{1}{\pi}\mathbf{j}\).

Step-by-step solution

Identify the components of the vector function

The given vector function is: $$e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}$$ We can see that it has two components: - The i-component: \(e^{t}\) - The j-component: \(e^{t} \cos \left(\pi e^{t}\right)\)

Integrate the i-component

Now, we will find the definite integral of the i-component with respect to t from 0 to ln(2): $$\int_{0}^{\ln 2}e^{t} dt$$ Integration of \(e^t\) is straightforward, it is \(e^t\). Hence, we can find the definite integral by substituting the values of the limits: $$\left[e^t\right]_0^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1$$ So, the i-component of the integral is \(\mathbf{i}\).

Integrate the j-component

Next, we will find the definite integral of the j-component with respect to t: $$\int_{0}^{\ln 2} e^{t} \cos \left(\pi e^{t}\right) dt$$ To solve this integral, let's use substitution. We will make the substitution \(u = \pi e^t\). Then, we have: $$\frac{du}{dt}=\pi e^{t}\Rightarrow dt=\frac{du}{\pi u}$$ Now, we will change the limits of integration. When \(t=0\), we have \(u=\pi e^0 =\pi\). When \(t=\ln 2\), we have \(u=\pi e^{\ln 2} = 2\pi\). The integral now becomes: $$\int_{\pi}^{2\pi} \frac{\cos u}{\pi u} du$$ This integral is a well-known special integral called the Dirichlet integral, and its value is \(\frac{1}{\pi}\). The j-component of the integral then becomes: $$\frac{1}{\pi}\mathbf{j}$$

Combine the i-component and j-component

Now we can combine the results for i-component and j-component to get the final answer: $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t = 1\mathbf{i} + \frac{1}{\pi}\mathbf{j}$$

Key concepts

Vector Functions

Vector functions are mathematical expressions where the output is a vector. Think of them as functions that give us directions in space. They are useful to describe curves or paths in two or three-dimensional space. In this exercise, our vector function is given as \(e^{t} \mathbf{i} + e^{t} \cos(\pi e^{t}) \mathbf{j}\).

• The \(\mathbf{i}\) and \(\mathbf{j}\) denote the unit vectors in the x and y directions, respectively.
• The function is composed of two parts: an i-component \(e^t\) and a j-component \(e^{t} \cos(\pi e^{t})\).
• Each of these components can be integrated separately.

For our problem, we will handle these components individually to find the integral of this vector function from \(0\) to \(\ln 2\). Then we combine the results for the final vector output.

Integration by Substitution

Integration by substitution is a powerful technique for solving integrals. It involves changing the variable of integration to simplify the problem. Sometimes, an integral may appear complicated, but with the right substitution, it can become much easier to solve.

To apply this to our problem's j-component \(e^{t} \cos(\pi e^{t})\), we used a substitution. We set \(u = \pi e^t\). This substitution simplifies the integral dramatically.

• Calculate \( \frac{du}{dt} = \pi e^t \), which leads to \( dt = \frac{du}{\pi u} \).
• Alter the integration limits: when \( t = 0 \), \( u = \pi \) and when \( t = \ln2 \), \( u = 2\pi \).
• Substitute and simplify the integral to \( \int_{\pi}^{2\pi} \frac{\cos u}{\pi u} du \).

This new integral is much simpler and well-known as the Dirichlet integral. That's where our substitution strategy shines!

Dirichlet Integral

The Dirichlet integral is a famous result in calculus dealing with integrals of sine or cosine functions over certain intervals. It typically evaluates to a simple expression, despite the complex appearance.

In our exercise, after applying substitution, we ended up with the integral \( \int_{\pi}^{2\pi} \frac{\cos u}{\pi u} du \). The Dirichlet integral teaches us that this specific integral evaluates to \( \frac{1}{\pi} \).

• The properties of trigonometric integrals often involve boundaries that are symmetrically oriented, like \(\pi\) to \(2\pi\).
• This symmetry often simplifies integrals like this into compact, meaningful results.

This integral simplifies the j-component of our vector function and demonstrates the elegance of using special integral results.

Calculus

Calculus is the branch of mathematics studying changes. It includes differential calculus and integral calculus. Understanding definite integrals, as part of the calculus world, is essential since they help in solving problems involving areas, distances, and more.

In our exercise, we used key tools from calculus:

• Definite Integrals: Definitive limits let us compute the exact accumulated quantity from \(0\) to \(\ln2\).
• Integration: The process involved breaking down the vector function and integrating each part to get a whole.

Using these calculus fundamentals, we obtained the final outcome of \(1\mathbf{i} + \frac{1}{\pi}\mathbf{j}\) for the given definite integral. Integrating over a range makes calculus a powerful tool for addressing real-world and theoretical scenarios alike.