The Fundamental Theorem of Calculus Exercises

Answer

We're pretending that Jen went 55 mph during the first 10 minutes, 50 mph during the next 10 minutes, and so on.

Let's rewrite the table so that t is measured in hours instead of minutes:

Using the nice formula for taking a right hand sum, we estimate that Jen travelled

Since the velocity function is decreasing, on each sub-interval we're underestimating Jen's velocity. She wasn't going 55 mph for the first ten minutes—she was going faster than 55 mph except at the very end of the sub-interval. This means we under-estimated her distance on the first sub-interval. Similarly, we underestimated her distance on all the other sub-intervals. This means 34.2 miles is an under-estimate of the distance she travelled.

To get a more accurate estimate of how far Jen travelled during that hour, we can average the left- and right- hand sums. We estimate Jen travelled

When you're given information about someone's velocity and asked to estimate the distance they've travelled, if the problem doesn't specifically tell you what sort of sum to use, take both the left-hand and right-hand sum and average them. This average, also known as the trapezoid sum, is usually the best estimate you can get when you only have a few scattered values of the velocity.

Be Careful: Remember to include units in your answers.

Answer

There are 3 sub-intervals, each of length 3.

a) LHS(3) = [14.5 + 14 + 13.8](3)

= 126.9 feet

b) RHS(3) = [14 + 13.8 + 13](3)

= 122.4 feet

c) 

Answer

There are 6 sub-intervals, each of length 5.

a) LHS(6) = [27 + 25 + 20 + 10 + 3 + 2](5)

= 435 feet

b) RHS(6) = [25 + 20 + 10 + 3 + 2 + 0](5)

= 300 feet

c) 

Answer

Since the problem doesn't specify what type of sum to use, we'll use a trapezoid sum.

Answer

Since the problem doesn't specify what type of sum to use, we'll use a trapezoid sum.

When we're trying to figure out how far a cheetah runs in one minute, our estimates get more accurate as we measure the velocity more frequently. If all we know is that the cheetah was going 0 feet per second at the beginning of the minute and 50 feet per second at the end of the minute, our best guess is that the cheetah went about 25 feet per second for the whole minute, for a total of

25(60) = 1500 feet.

If we measure the velocity every ten seconds, then we get a better idea of what the cheetah's velocity is doing:

Measuring the velocity every 10 seconds lets us see that the velocity goes above 50 feet per second.

If we measure the velocity every 5 seconds, we get an even better idea of what's going on:

Answer

Answer

We're going to need to know the area between the graph of v(t) and the t-axis on each time interval, so let's just figure those areas out now:

(a) From t = 0 to t = 10 the pink elephant travels

(b) From t = 10 to t = 25 the pink elephant travels

(c) From t = 25 to t = 30 the pink elephant travels

(d) There are two ways to figure out how far the elephant travels from t = 0 to t = 30. One way is to look at the graph again, add up all the areas, and get

The other way is to remember the property of integrals that says

The integrals on the right-hand side were calculated in (a)-(c), and now we just add them all up:

Answer

(a) From t = 0 to t = 2 the bug moves

as we can see from the graph of v(t):

If the bug started at -4 on the number line and moved 8 units, at t = 2 the bug is now at 4 on the number line.

(b) From t = 2 to t = 3 the bug moves

as we can see from the graph of v(t):

Since the bug was at 4 when t = 2 and has moved 4 more units, when t = 3 the bug is at 8 on the numberline.

(c) From t = 3 to t = 5 the bug moves

as we can see from the graph of v(t):

When t = 3 the bug was at 8, and the bug has moved 4 more units, so when t = 5 the bug is at 12.

Answer

When t = 0 the cat is one foot above the ground:

(a) The area between the graph of v(t) and the t-axis on [0,3] is 6:

This means from time 0 to 3 seconds the cat climbed

Since the cat started 1 foot above the ground, when t = 3 it is 7 feet above the ground.

(b) The area between the graph of v(t) and the t-axis on [3,6] is 11:

This means from t = 3 to t = 6 the cat climbed

Since the cat was 7 feet above ground when t = 3, when t = 6 the cat is

7 + 11 = 18 feet above ground.

(c) The area between the graph of v(t) and the t-axis on [6,8] is 12:

This means from t = 6 to t = 8 the cat climbed

Since the cat was 18 feet above ground when t = 6, when t = 8 the cat is

18 + 12 = 30 = feet above ground.

(d) The area between the graph of v(t) and the t-axis on [8,10] is 12:

This means from t = 8 to t = 10 the cat climbed

Since the cat was 30 feet above ground when t = 8, when t = 10 the cat is

30 + 12 = 42 feet above ground.