2. Rival Mathemon trainers, Jane and Jenny, are
trapped in the forest and surrounded by wild
Mathemon agitated by Jenny’s constant
disturbance of peace through a series of
attacks. Jane is a witness of this abuse and
she does her best to protect the wild
Mathemon by challenging Jenny to a battle to
stop her vicious ways.
4. Jenny used a DIAGRAM!
Let R be the shaded region bounded by the graphs of
y = 6x , y = e and the vertical line x=1.
2x
The foe’s Revoloo’s DEF has risen!
5. Jane used an INT BOOST!
Gralinte’s intelligence has risen!
6. The foe’s Revoloo used QUESTION!!
(a) Find the area of the enclosed region R.
It’s a critical hit! Gralinte takes 12 damage!
7. Gralinte used POINTS OF INTERSECTION!
One of the definitions of integration can be
known as finding the area underneath a
curve. In this case, R is the area that needs to
be integrated.
But since R is not infinitely continuous and
occurs on a closed interval, the points of
intersection must be found in order for the
area to be integrated.
8. To find the points of intersection, make both
functions equal to each other and find the
values where these functions meet.
There are two ways that you can do this.
Manually, you can find the value(s) of x, or
you can use the intersect function on your
calculator. In this specific case, it is
integrated to 1, since it was given that it is
enclosed by the vertical line x=1.
9. The point of intersection occur at approx.
x=0.1081. But since that number is quite
complex, we can assign a letter to that
number. So we can let T=0.1081.
The foe’s Revoloo took 9 damage!
Revoloo is paralyzed!
10. Gralinte used INTEGRATION!
dt
Now knowing the intersections, we can now use those as our
interval from which we are going to integrate and find the area
for. Remember that we let T = 0.1081 since we wouldn’t want to
waste our time rewriting such complex numbers all the time.
Solving for this using our calculators we should get….
A≈1.2398 units²
Revoloo took 6 damage!
11. “Hmph, you think you’re so tough Janey Poo?
See if you can handle this next attack! Go
get’em Revoloo!”
12. Revoloo used QUESTION!
(b) Find the volume of the solid generated
when R is revolved about the x-axis.
It’s a critical hit! Gralinte takes 17 damage!
13. Gralinte used INTEGRATION!
In every math problem lies a pattern. In this case, when we
revolve something around the x-axis or a horizontal line, the
pattern is in the general formula of how to revolve shapes
around this axis.
The volume for revolution around a horizontal line or the x-
axis is equal to the integral of the area of a circle, which is
πr2. This is because the cross sections of the area itself is in
the shape of a circle as it revolves around the specific axis of
rotation.
14. Since the area is consisted of
two different curves, the cross
section looks like a washer.
The upper function will be the
bigger radius R, and the lower
function will be the smaller
radius, r. As seen here:
15. In general, we should know that to find the volume of a
washer type question, we will need to take the area of the
large circle and the area of the small circle and then
subtract them from each other. From there, just integrate
along the interval, or through the intersections.
dx
ITS SUPER EFFECTIVE! Revoloo takes 24
damage!
16. Revoloo used DUST!
Gralinte has been blinded! Gralinte takes 2
damage!
“Gralinte, that’s enough! Come back!
Go, Derivee!”
17. Derivee used ANTI DIFFERENTIATE!
Before we begin anti differentiating, let’s first simplify the
integral to make it easier.
dx
From here, we can use simple anti differentiation rules to anti
differentiate.
18. We can now finish the question by applying our knowledge of
the fundamental theorem of calculus, which is that the integral
of a derivative = total change in the parent function along the
same interval. Previously, we found the parent function so now
all that’s left is to solve. Our approximate answer should be:
3
V 2 . 8073 units
Revoloo took 12 damage!
19. “You think you’re so great? There’s no way I’d
let you win! Wasn’t it obvious that I was just
playing around? Well, no more fooling
around! Take this!”
20. Revoloo used QUESTION!
(c)The region R is the base of a solid. For this
solid, each cross section perpendicular to the x-
axis is a rectangle whose height is 5 times the
length of its base in region R. Find the volume of
this solid.
Derivee is crushed by the sudden
overwhelming attack! Derivee takes 19
damage!
21. Derivee used LIST!
The question has already given us two key things which is
the base and height of the rectangle as shown here:
2x
height 5L L 6x e
Like the previous part, we should also be able to find the
volume of the solid by taking the area of said solid and
integrating it. The interval will remain the same since that is
where the actual solid starts and ends.
A lwh
Revoloo takes 10 damage! Revoloo is stunned
and is in critical condition!
23. Gralinte used INTEGRATION!
As stated earlier, to find the volume all we have to do is
integrate the area of the rectangle. Let’s first take a look at the
rectangle so that we know how to substitute in our values.
24. The width, as you can see, is basically the infinitely small values of
“x” as you go along the interval. Since it is so small, we just leave it
as “dx”.
Now knowing all our values, we can now integrate:
Then solve:
