Question

Evaluate the following definite integrals. $${\int }_1^4(6t^2\mathbf{i}+8t^3\mathbf{j}+9t^2\mathbf{k})dt$$

Short answer

Question: Evaluate the definite vector integral $${\int }_1^4(6t^2\mathbf{i}+8t^3\mathbf{j}+9t^2\mathbf{k})dt$$. Answer: The definite vector integral is equal to $$120\mathbf{i}+496\mathbf{j}+180\mathbf{k}$$.

Step-by-step solution

Identify the component functions

Identify the functions multiplying the unit vectors i, j, and k: - i-component function: $6t^2$ - j-component function: $8t^3$ - k-component function: $9t^2$

Integrate the i-component function

Using the power rule for integration, integrate the i-component function: $$\int (6t^2)dt=2t^3+C$$ Now, we'll find the definite integral: $${\left[2t^3\right]}_1^4=2(4^3)-2(1^3)=120$$

Integrate the j-component function

Using the power rule for integration, integrate the j-component function: $$\int (8t^3)dt=2t^4+C$$ Now, we'll find the definite integral: $${\left[2t^4\right]}_1^4=2(4^4)-2(1^4)=496$$

Integrate the k-component function

Using the power rule for integration, integrate the k-component function: $$\int (9t^2)dt=3t^3+C$$ Now, we'll find the definite integral: $${\left[3t^3\right]}_1^4=3(4^3)-3(1^3)=180$$

Combine the results

Combine the results from all integrals, incorporating the corresponding unit vectors i, j, and k: $${\int }_1^4(6t^2\mathbf{i}+8t^3\mathbf{j}+9t^2\mathbf{k})dt=120\mathbf{i}+496\mathbf{j}+180\mathbf{k}$$

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on multivariable functions and involves both differentiation and integration of vectors. It is especially important in physics and engineering as it deals with quantities that have both magnitude and direction, like velocity and force.
In the exercise provided, we dealt with a vector function of the form $6t^2\mathbf{i}+8t^3\mathbf{j}+9t^2\mathbf{k}$. This vector consists of three separate functions, each associated with one of the unit vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$.

• The $\mathbf{i}$ component describes the rate of change along the x-axis, $6t^2$.
• The $\mathbf{j}$ component describes the change along the y-axis, $8t^3$.
• The $\mathbf{k}$ component describes the change along the z-axis, $9t^2$.

Integrating a vector function means that each of its components is integrated separately. The solution then combines the definite integrals of these independent components to express the overall vector in terms of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$.

Integration Techniques

Integration is a key concept in calculus used to find areas, volumes, central points, and many useful things. With vector calculus, we encounter integration techniques much like in standard calculus, but we apply them component-wise to vector fields.
For our specific exercise, the definite integrals are computed over the interval from 1 to 4. Definite integrals give us the net area under the curve from one point to another, and in vector calculus, this means finding the accumulated change of each vector component over the interval.

• Step-by-step component-wise integration: We integrate each part of the vector separately as if they were independent functions.
• Additive property: After finding the definite integral for each component (i.e., $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$), we sum these results together.
• Substitution of limits: After integrating, we substitute the bounds into the integration result to obtain the definite integral.

Make sure to closely follow these principles when tackling problems involving vector functions and integration. It ensures that each component is treated appropriately, eventually leading to solving the entire vector function.

Power Rule in Integration

The power rule in integration is one of the fundamental techniques used to solve integrals, and it is particularly useful for polynomials. The power rule states that for any function of the form $t^n$, its integral is $\frac{t^{n+1}}{n+1}+C$, where $C$ is the integration constant.
In our exercise, this rule was applied separately to each component of the vector function. Let's break it down:

• For the $i$ component, $6t^2$, the power rule gives $2t^3+C$.
• For the $j$ component, $8t^3$, the integration gives $2t^4+C$.
• For the $k$ component, $9t^2$, it results in $3t^3+C$.

It's important to recognize that the power rule simplifies integration significantly for polynomial expressions, turning a potentially complex task into a straightforward calculation. When dealing with definite integrals, like in the exercise, the constant $C$ ultimately cancels as we find the difference between the upper and lower bounds of integration. Thus, focusing on the power rule can make evaluating vector integrals much simpler and efficient.