Please help with math question

Question

The velocity of a particle moving along the x-axis is given by v(t)=−sin(t−π/4). Which of the following gives the total distance traveled by the particle over the time interval 0≤t≤π ?

A particle moves along the x -axis with velocity given by v(t)=(t−1)e^(1−t) for time t≥0 . If the particle is at position x=3 at time t=0 , which of the following gives the position of the particle at time t=2?

The base of a solid is the region in the first quadrant bounded by the x- and y-axes and the graph of y=(x−2)^2/2(x+1). For the solid, each cross section perpendicular to the x-axis is a rectangle whose height is three times its width in the xy-plane. What is the volume of the solid?

Answer 1

These are three separate problems, and all three come down to the same big idea, which is that the integral of a rate gives you the accumulated amount. Let me go through them one at a time, because each one has its own trap.

The first one, total distance. This is the classic place students lose points. Total distance is not the same thing as displacement. Displacement is just the integral of the velocity, but total distance is the integral of the absolute value of the velocity, because going backward still counts as distance traveled:

total distance = ∫₀^π |v(t)| dt

So before integrating we have to find where the velocity changes sign. Setting -sin(t - π/4) = 0 gives t - π/4 = 0, so t = π/4. Checking a point on each side, v(0) = -sin(-π/4) = √2/2 which is positive, and v(π) = -sin(3π/4) = -√2/2 which is negative. So the particle moves right until t = π/4, then turns around and moves left. We split the integral there and flip the sign on the second piece:

∫₀^π/4 v dt = 1 - √2/2 ≈ 0.293

∫_π/4^π (-v) dt = 1 + √2/2 ≈ 1.707

Adding those, the √2/2 terms cancel and the total distance is exactly 2. If you notice, the displacement would have been -√2 ≈ -1.41, which is a completely different number, so this is a real distinction and not a technicality.

The second one, position from velocity. Here we use the accumulation form, where the new position is the old position plus the net change:

x(2) = x(0) + ∫₀² v(t) dt = 3 + ∫₀² (t - 1)e^1-t dt

Notice we want displacement this time, not distance, since the question asks where the particle is, so no absolute value. The integral needs integration by parts, and it works out very cleanly. Taking the part that gets differentiated as t - 1 and the rest as e^1-t, you end up with:

∫ (t - 1)e^1-t dt = -te^1-t

That is worth checking by differentiating it, which I did: the derivative of -te^1-t is -e^1-t + te^1-t = (t - 1)e^1-t, so it is right. Evaluating from 0 to 2 gives -2e^-1 - 0 = -2/e. So:

x(2) = 3 - 2/e ≈ 2.264

The particle actually ends up to the left of where it started, which makes sense because the velocity is negative for all of t < 1 and only turns positive after that.

The third one, volume by cross sections. For these you never need a formula, you just need the area of one slice. The slice is a rectangle whose width is the height of the region, which is y, and whose height is three times that, so 3y. The area of one cross section is:

A(x) = (y)(3y) = 3y²

and the volume is the integral of that across the base, which runs from x = 0 to x = 2 since the curve meets the x-axis at x = 2:

V = ∫₀² 3y² dx

Now I have to flag something about how your function got typed, because it changes the answer. You wrote y = (x-2)²/2(x+1), which can be read two ways. If it means ((x-2)²/2)(x+1), then squaring and expanding gives a polynomial and the integral comes out to exactly 312/35, about 8.914. If it means (x-2)² divided by 2(x+1), the integral involves a natural log and comes out to about 3.012, which is not a number any textbook would choose as an answer. Since AP answers are almost always clean, the first reading is the intended one, and 312/35 is your answer. Both readings do give the same region corners, y = 2 at x = 0 and y = 0 at x = 2, so you cannot tell them apart from the picture, only from the arithmetic coming out clean.

One general note, the first two problems say "which of the following gives," so they were multiple choice and the choices did not come through when you posted. The values above are what to match against your options, and for those two the answer choices are often left as an unevaluated integral, so look for ∫₀^π |v(t)| dt on the first one and 3 + ∫₀² v(t) dt on the second.

So, our final answers are 2 for the total distance, 3 - 2/e or about 2.264 for the position, and 312/35 or about 8.914 for the volume.