What conclusion can be drawn about the convergence of the series $\sum_{n=2}^{\infty}\frac{1}{n(\ln(n))^p}$ when $p>1$ using an appropriate test?

Converges by Comparison to a p-series with $p > 1$

Diverges by Direct Comparison to a harmonic series

Converges by Limit Comparison to a geometric series

Inconclusive without further information as both tests fail

If we have two vector-valued functions defined by $\vec{s}_1(t)=\langle t,t^{3}\rangle$ and $\vec{s}_2(t)=\langle t^{4}, t\rangle$, which statement correctly describes their antiderivatives?

They can be found by integrating each component separately.

They can only be found if they describe linear motion.

Only one antiderivative exists for both vectors since they share a common variable.

They do not have antiderivatives because their components are not exponential functions.

Question #4: If $\vec{f}(x)=\langle x + x² , \arctan(x)\rangle$, what is $\int_{-\pi/4}^{\pi/4} \vec{f}(x) dx$?

$\langle \pi^2/8 + \pi^2/16 , \pi/4 \rangle$

$\langle \pi^2/16 + \pi^2/32 , \pi/2 \rangle$

$\langle x+x^3/3 , \arctan(x)/x \rangle | (-\pi/4)^{\pi/4}$

$\langle x^3/3 , \tan(x) \rangle | (-\pi/4)^{\pi/4}$

Which technique could potentially reduce error when numerically approximating an integral involving rapid changes in direction within a vector-valued function?

Adaptive quadrature which adjusts interval sizes dynamically based upon rate of change within intervals.

Constant-size partitioning across entire range regardless of rate of change.

Using higher-order Taylor Polynomials that do not adjust intervals but might offer more precise local approximations.

Utilizing lower-order Newton-Cotes formulas that may have larger errors with rapid changes.

How is the parametric vector-valued function $\mathbf{s}(t) = \left(2t, e^{-\sin(t)}\right)$ integrated with respect to $t$ from $\pi$ to $\beta$?

$\left(\frac{1}{6}(\beta^2-\pi^2), \sin(\pi)-\sin(\beta)\right)$

$\left(\frac{1}{6}(\beta^2-\pi^2), \sin(\beta)-\sin(\pi)\right)$

$\left(\frac{1}{3}(\beta^2-\pi^2), \sin(\pi)-\sin(\beta)\right)$

$\left(\frac{1}{3}(\beta^2-\pi^2), \sin(\beta)-\sin(\pi)\right)$

What does the integral from a to b of v(t) represent in the context of a parametric function?

The displacement of the function

The acceleration of the function

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9.5 Integrating Vector-Valued Functions

2 min read•january 25, 2023

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By now, you should have a good understanding of vector functions. It’s alright if you don’t, for a refresher you can read our 9.4 study guide which talks about defining and differentiating vector-valued functions. This will help you understand this study guide and new concept easier!

Integrating these vector-valued functions allows us to go backward when relating the movement of a particle to its equation. For example, let's say we have a vector-valued function that gives the acceleration of a particle moving in space. A question asks us to determine the velocity of said particle. This means we need ✨integration✨

💭 Integrating Vector-Valued Functions

It’s good to first recognize the format of your function. It is either written in two or three dimensions; we want to know which one so that we are not confused about which terms should be paid attention to when integrating.

A two-dimension vector function might look like this:

r(t)=f(t)i+g(t)j

or like this:

r(t)=⟨f(t),g(t)⟩

These both mean the same thing! 🪄

❓ How do we integrate?

Luckily, we integrate each component of a vector function separately. No need to make things more complicated than they need to be! You look at each component as if it was its own problem. Let’s go through an example just to see how the College Board might ask a question like this.

✏️ Integrating Vector-Valued Functions Walkthrough

We are given:

r(t)=⟨t^2,\sin(t),e^t⟩

We know we are going to need to integrate this vector-valued function, so let’s re-write this including integrals on each component.

\int{r(t)}dt=⟨∫t^2dt,∫\sin(t)dt,∫e^tdt⟩

You can solve each of these integrals separately. If you need help with integrals, check out our guide 6.14 Selecting Techniques for Antidifferentiation. Although it says for AB, it will apply to this!

Our final answer will be:

R(t)=⟨\frac13t^3+C1,−\cos(t)+C2,e^t+C3⟩

Please note that C1, C2, and C3 just represent different possible constants and I added them for clarity. For these types of problems, you will typically get definite integrals so that there is less ambiguity. Follow your typical steps for solving these problems!

📝 Additional Practice Problem

Let’s try a word problem now (note this is a calculator-okay question). This problem (#9) is from the flipped math calculus website. All credit to them.

At time 𝑡 ≥ 0, a particle moving in the 𝑥𝑦-plane has a velocity vector given by.

v(t)=⟨3t^2,3⟩

If the particle is at point (1,2) at time 𝑡 = 0, how far is the particle from the origin at time 𝑡 = 2?

Try the problem on your own first! If you are stuck read through our explanation below!

🏁 Solution to Practice Problem

Step one is integrating the function. As we’ve discussed, you will integrate each of the components of the vector function separately. Before applying conditions, this is what you should have for your integrated vector function:

x(t)=∫3t^2dt=t^3+Cx

y(t)=∫3dt=3t+Cy

The Cx and Cy constants are just placeholders. We are going to figure out those numbers now:

Since we are given the position for t=0, we can use that information to work backward and find our constants.

x(t):1=0^3+Cx\therefore Cx=1

y(t):2=3(0)+Cy\therefore Cy=2

Now that we know the values for our constants let's put those into our integrated vector functions:

⟨t^3+1,3t+2 ⟩

Now we’re done with the hard parts! From here on it’s just simple math to finish off the problem. We will plug in t=2 to our equation to find the position of the particle at that point. This gives us

(9,8)

Finally, we’ll use the handy distance formula to find how far this point is from the origin. Since this would be a question in the calculator section, you’ll want to plug this into your calculator and give a number rounded to three decimal places rather than leave it in radical form.

\sqrt{(9-0)^2+(8-0)^2}

You should get 12.04 units as your final answer!

Why integrate vector-valued functions?

With these types of problems, you want to have a good understanding of what you can do with the functions. Vector functions often tell you the movement or position of a particle. This means the integrated function can give the total area moved over time or similar. Don’t worry too much about the nitty-gritty, but practice recognizing graphs for regular functions and matching them to their differentiated or integrated functions. You can get practice with this in our guide 9.6! Good luck and you’ve got this. 😁

Image Courtesy of Giphy

Key Terms to Review (13)

Cartesian Coordinates

: Cartesian coordinates are a system used to locate points on a plane using two perpendicular lines called axes (usually x and y). The x-coordinate represents the horizontal position, while the y-coordinate represents the vertical position.

Change of variables

: Change of variables refers to substituting one variable with another variable or expression in order to simplify an integral or equation.

Definite Integral

: A definite integral is a mathematical tool used to calculate the exact area between a function and the x-axis over a specific interval.

Displacement

: Displacement refers to the change in position of an object from its initial point to its final point, taking into account both distance and direction.

Green's Theorem

: Green's theorem relates a double integral over a region in the plane to a line integral around its boundary. It establishes an equivalence between these two types of integrals.

Indefinite Integral

: An indefinite integral is the reverse process of differentiation. It represents a family of functions that have the same derivative.

Integrals

: Integrals are mathematical tools used to find the area under a curve or to calculate the accumulation of quantities over a given interval.

Integration by Parts

: Integration by parts is a technique used to find the integral of a product of two functions. It involves using the formula ∫u dv = uv - ∫v du, where u and v are chosen functions.

Line Integrals

: A line integral measures the total effect along a curve or path, taking into account both direction and magnitude.

Polar Coordinates

: Polar coordinates are a two-dimensional coordinate system used to locate points in space using radial distance (r) and angular displacement (θ) from a reference point called the pole.

Real-valued functions

: Real-valued functions are mathematical functions that take real numbers as inputs and produce real numbers as outputs.

Substitution

: Substitution is an integration technique used to simplify complex integrands by replacing variables with new ones that make integration easier.

Vector-valued functions

: Vector-valued functions are functions that output vectors instead of scalars. They take in a parameter (usually denoted as t) and produce a vector with multiple components.