WORKSHEET 15 SOLUTIONS
MATH 1300
Goal: To review some of the basic concepts from the course.
1. Find the area of the finite plane region bounded by the curve y = x3 and the tangent line to
this curve at (1, 1).
The tangent line to y = x3 at the point (1, 1) has slope
the point (1, 1). Thus its equation is
y − 1 = 3(x − 1)
or
dy
|
dx x=1
= 3x2 |x=1 = 3 and passes through
y = 3x − 2.
We need to find the other point of intersection of this line and y = x3 , so we set them equal:
3x − 2 = x3 .
This implies that x3 − 3x + 2 = 0. Now, x = 1 is a solution to this equation (why?), so (x − 1) is a
factor of x3 − 3x + 2. Using long division or just factoring we get
x3 − 3x + 2 = (x − 1)(x2 + x − 2) = (x − 1)(x − 1)(x + 2).
Thus, the other point of intersection is when x = −2, the point (−2, −8). Thus the area is
Z 1
3
= You finish.
x
−
(3x
−
2)
dx
−2
Can you explain why we put the absolute value sign around the integral? Is it needed?
2. A parabola opens downward and crosses the x−axis at x = 0 and x = A, with A > 0. Suppose
the area enclosed by the parabola’s arch and the x−axis equals 1.
(a) Find the equation of the parabola (its coefficients will depend on A).
We know the equation is of the form y = −ax2 + bx + c. However, since y = 0 when x = 0 and
when x = A, we know this factors as y = −ax(x − A) = −ax2 + aAx. We can find a in terms of A
since:
A
Z A
a 3 a 2 1 3
2
−ax + aAx dx = − x + Ax = a
1=
A .
3
2
6
0
0
Therefore a =
1
1 3
A
6
=
6
.
A3
(b) What is the highest point on the parabola in terms of A?
Now, y = −
6
6 2
x
+
x, so
A3
A2
12
6
dy
= − 3x + 2.
dx
A
A
dy
Thus dx
= 0 when x = A2 . Since this is a parabola opening downwards, we know x =
to a max. When x = A2 , we have
6
y=− 3
A
3
so ( A2 , 2A
) is the answer.
2
6 A
A
3
+ 2
=
,
2
A
2
2A
A
2
corresponds
3. Find the area between f (x) = sin x and g(x) = (sin x)2 for 0 ≤ x ≤ π. Hint. (sin x)2 =
1
(1 − cos(2x)).
2
We seem to need to know if f (x) and g(x) cross each other in the interval 0 ≤ x ≤ π, so we seek x
such that (sin x)2 = sin x. This holds if and only if (sin x)2 − sin x = 0 or rather sin x(sin x − 1) = 0.
Thus, sin x = 0, which tells us x = kπ for k ∈ Z, or sin x = 1, which tells us x = π2 + kπ for k ∈ Z.
This mean that for 0 ≤ x ≤ π the curves meet when x = 0, x = π2 , and x = π. However we know
that for 0 ≤ x ≤ π, (sin x)2 ≤ sin x, because sin x is always between 0 and 1 and squaring such a
number will yield a number at most as big as the original. Thus, the curves meet at x = π2 but do
not cross. Therefore the area can be computed by a single integral:
Z π
sin x − sin2 x dx
Area =
Z0 π
1
=
sin x − (1 − cos 2x) dx
2
Z0 π
1
− + sin x + cos 2x dx
=
2
0
= You Finish.