Question

Evaluate the following definite integrals. $$\int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t$$

Short answer

Question: Evaluate the definite integral of the vector-valued function \(∫_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t\). Answer: The definite integral evaluates to \((e^2+1)\mathbf{i} + (2e^2+2)\mathbf{j} + (-e^2-1)\mathbf{k}\).

Step-by-step solution

Separate the integral for each component

First, let's separate the integral into three individual integrals for each component (i, j, and k): $$\int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t = \int_{0}^{2} t e^{t} \mathbf{i} dt + \int_{0}^{2} 2t e^{t} \mathbf{j} dt - \int_{0}^{2} t e^{t} \mathbf{k} dt$$

Integrate the i component

Now, we will integrate the i component: $$\int_{0}^{2} t e^{t} \mathbf{i} dt$$ To do this, we will use the integration by parts formula, which states that: $$\int u dv = uv - \int v du$$ Let's take \(u=t\) and \(dv=e^t dt\). Find the derivatives and antiderivatives of them: $$du=dt$$ $$v=e^t$$ Now, applying the integration by parts formula: $$\left[t e^t - \int e^t dt\right]_{0}^{2} \mathbf{i} = \left[t e^t - e^t\right]_{0}^{2} \mathbf{i}$$ Evaluate the expression at the limits: $$(2e^2 - e^2) \mathbf{i} - (0 - 1) \mathbf{i} = e^2 \mathbf{i} + \mathbf{i}$$

Integrate the j component

Next, integrate the j component: $$\int_{0}^{2} 2t e^{t} \mathbf{j} dt$$ We can take the constant 2 outside the integral: $$2 \int_{0}^{2} t e^{t} \mathbf{j} dt$$ This integral is similar to the one we did for the i component, but with an extra constant. We can use the result from Step 2, and just multiply by 2: $$2(e^2 \mathbf{j} + \mathbf{j}) = 2e^2 \mathbf{j} + 2\mathbf{j}$$

Integrate the k component

Finally, integrate the k component: $$-\int_{0}^{2} t e^{t} \mathbf{k} dt$$ This integral is equivalent to the one we did for the i component but with a negative sign. We can use the result from Step 2 and add a negative sign: $$-(e^2 \mathbf{k} + \mathbf{k}) = -e^2 \mathbf{k} - \mathbf{k}$$

Combine the results

Now, we will combine the results from Steps 2, 3, and 4 to get the final answer: $$(e^2 \mathbf{i} + \mathbf{i}) + (2e^2 \mathbf{j} + 2\mathbf{j}) + (-e^2 \mathbf{k} - \mathbf{k})$$ This simplifies to: $$\boxed{(e^2+1)\mathbf{i} + (2e^2+2)\mathbf{j} + (-e^2-1)\mathbf{k}}$$

Key concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is particularly useful when one of the functions is easily differentiable, and the other is easily integrable. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] Let's break down the process:

• Choose two functions within the integral: one to differentiate (\(u\)) and the other to integrate (\(dv\)).
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\).
• Substitute into the formula: \(uv - \int v \, du\) to solve the integral.
• Finally, evaluate using the limits if dealing with definite integrals.

In our exercise, for the integral \( \int_0^2 t e^t \, dt \), \(u\) is chosen as \(t\), making \(du = dt\), and \(dv\) as \(e^t \, dt\), yielding \(v = e^t\). Applying the formula helps simplify the process, ultimately leading to integration of each component of our vector expression.

Vector Calculus

Vector calculus extends the calculations you can apply in calculus to functions involving vectors. In our exercise, we deal with an integral of a vector function. This means we are integrating each component of the vector separately. Vector calculus is particularly helpful when dealing with force fields, velocity vectors, and multi-dimensional data. Let's look at the main points of handling vector calculus in integration:

• Separate each vector component and perform the integration independently. In our example:
  • For component \( \mathbf{i}, \) evaluate \( \int_0^2 t e^t \, dt \)
  • For component \( \mathbf{j}, \) evaluate \( 2 \int_0^2 t e^t \, dt \)
  • For component \( \mathbf{k}, \) evaluate \( -\int_0^2 t e^t \, dt \)
• After calculating each component, combine them back into a single vector result.

Vector calculus allows you to effectively manage the dimensions and directions that scalar calculus alone cannot handle, providing a more comprehensive understanding of physical and mathematical systems.

Using Limits for Evaluation

When an integral is defined over a finite range, it is known as a definite integral. This requires evaluation over specific limits, typically applied to produce a numerical result from a symbolic integration process. The steps for using limits to evaluate a definite integral start once the indefinite integral is solved. Here’s how to proceed:

• Complete the integration steps for the indefinite integral.
• Apply the limits by substituting the upper limit into the antiderivative and then subtracting the value found by substituting the lower limit.
• In vector integrals, remember to apply limits to each component separately, as we did for \( e^2 \mathbf{i} + \mathbf{i} \), \( 2e^2 \mathbf{j} + 2\mathbf{j} \), and \( -e^2 \mathbf{k} - \mathbf{k} \).

Limits convert the result from a general form to a specific value based on the range of the integration, which can be crucial in problems involving physics and engineering, where real-world boundaries are essential.