Answer:
true
Step-by-step explanation:
Sec theta = 1/cos theta = 1/0.5 = 2
Option D is the correct answer.
For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:

Suppose

At the same time, we can write

Note that
(just reverse the sum). Let's pair the first terms of
and
, and the second, and the third, and so on:

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

But
, as we pointed out, so
