Question

Evaluate the following definite integrals. $${\int }_0^{\ln 2}(e^{-t}\mathbf{i}+2e^{2t}\mathbf{j}-4e^t\mathbf{k})dt$$

Short answer

Question: Evaluate the definite integral of the vector-valued function given by $${\int }_0^{\ln 2}(e^{-t}\mathbf{i}+2e^{2t}\mathbf{j}-4e^t\mathbf{k})dt$$. Answer: The definite integral of the given vector-valued function is $$\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}$$.

Step-by-step solution

Identify the components of the vector-valued function

The given vector-valued function is represented as: $$(e^{-t}\mathbf{i}+2e^{2t}\mathbf{j}-4e^t\mathbf{k})$$ Which has three components: - i component: $e^{-t}$ - j component: $2e^{2t}$ - k component: $-4e^t$

Integrate each component separately

We now need to integrate each component of the vector-valued function over the given interval [0, ln(2)]. i component integral: $${\int }_0^{\ln 2}{e}^{-t}dt$$ j component integral: $${\int }_0^{\ln 2}2e^{2t}dt$$ k component integral: $${\int }_0^{\ln 2}-4e^{t}dt$$

Evaluate the integrals of each component

Now, we evaluate each integral: i component: $${\int }_0^{\ln 2}{e}^{-t}dt=-e^{-t}{|}_0^{\ln 2}=-(\frac{1}{2}-1)=\frac{1}{2}$$ j component: $${\int }_0^{\ln 2}2e^{2t}dt=e^{2t}{|}_0^{\ln 2}=(4-1)=3$$ k component: $${\int }_0^{\ln 2}-4e^{t}dt=-4e^t{|}_0^{\ln 2}=-4(2-1)=-4$$

Combine the results

Now that we have evaluated each component, we can put them back together into a vector: $$\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}$$ So, the final answer is: $${\int }_0^{\ln 2}(e^{-t}\mathbf{i}+2e^{2t}\mathbf{j}-4e^t\mathbf{k})dt=\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}$$

Key concepts

Vector-Valued Functions

Vector-valued functions are an extension to functions which normally map to real numbers. Instead of mapping inputs to a single real number, vector-valued functions map inputs to vectors. In our exercise, we use a function that assigns a vector in three-dimensional space to the real number t.

This type of function is particularly useful in physics and engineering to describe quantities that possess both magnitude and direction, such as velocity or force. When integrating a vector-valued function, we treat each component of the vector separately as shown in our step-by-step solution. This means applying integration techniques to the i, j, and k components individually, using relevant integration rules, which is crucial to arrive at the correct answer.

Integration Techniques

Integration techniques are various methods used to compute the integral of a function. In the step-by-step solution provided, the integral of each component of the vector-valued function is found using basic antiderivatives since the components are exponential functions.

It's important to recognize the types of functions you're dealing with to apply the correct technique. For exponential functions of the form e^kt, where k is a constant, the antiderivative is straightforward. If you're facing more complex functions, such as those involving polynomial, trigonometric, or logarithmic functions, you might need to use techniques like substitution, integration by parts, or partial fractions.

Exponential Functions

Exponential functions, like the ones in our vector-valued function (e.g., e^−t, 2e^2t, and −4e^t), have the general form f(t) = ae^kt, where e is the base of natural logarithms. These functions feature prominently in the modeling of growth and decay processes in real-world scenarios, including population dynamics, finance, and physics.

The integrals of exponential functions are relatively simple to compute, as demonstrated in our steps. Recognizing the pattern of exponential growth or decay is key, and the definite integral of such functions over an interval gives us the accumulated value, which can represent anything from total growth to the distance traveled by an object.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration and plays a critical role in our understanding of calculus. Essentially, it states that if f is continuous on [a, b] and F is the antiderivative of f on [a, b], then the definite integral of f from a to b is given by F(b) - F(a).

In our problem’s solution, we apply this theorem when we find the antiderivatives of the exponential functions and evaluate them at the upper and lower limits of the integration (from 0 to ln(2)). This evaluation allows us to find the vector's total change over the interval, reflecting a combination of magnitudes and directions from the point of view of physics, culminating in the calculation of the net effect or total displacement.