MATH 101 V2A
January 16th – Practice problems
Hints and Solutions
Z
x
1. Solve the integral equation f (x) = 1 + 4
f (t)dt. (Hint: Differentiate the equation
3
and then solve the resulting differential equation.)
Z
x
f (t)dt, namely
Solution: The FTC tells us how to differentiate
3
d
dx
Z
x
f (t)dt = f (x).
3
So, differentiating the right side of the equation gives us
Z x
d
f (t)dt = 0 + 4 · f (x) = 4f (x),
1+4
dx
3
d
and differentiating the left side of the equation will just give us dx
(f (x)) = f 0 (x).
0
Therefore, we should have that f (x) = 4f (x). Last term we saw that solutions to this
differential equation are of the form f (x) = Ae4x . If we substitute this into our original
equation we’ll get that
Z
x
Ae
4x
Ae4t dt.
=1+4
3
Since the function
A 4t
e
4
is an antiderivative of Ae4t , we get that
Z x
A 4t x
4t
4x
Ae dt = 1 + 4 ·
Ae = 1 + 4
e
4
3
3
4x
12
= 1 + Ae − Ae .
This means that Ae12 = 1, and so A = e−12 . Therefore the solution is f (x) = e4x−12 .
Z
2. Evaluate
b
tdt using the Fundamental Theorem of Calculus. Then confirm your
0
answer two ways, by using Euclidean geometry and the definition of integral.
Solution: The function 12 t2 is an antiderivative of f (t) = t, so the FTC tells us that
Z
0
b
1 2 b
1
tdt = t = b2 .
2 0 2
Z
b
tdt represents the area of
If you draw the graph of the function, you’ll notice that
0
a right-triangle of height b and base b. From Euclidean geometry we know that such a
triangle has area 12 b2 , so our second method gives us the same result.
Z 1
1
We used the definition of integral in class to show that
tdt = . Hint: Modify the
2
0
Z b
tdt.
argument from class to evaluate
0
2x−x2
Z
cos
3. Find the global extrema of f (x) =
0
Hint: Let
Z
x
G(x) =
cos
0
1
1 + t2
1
1 + t2
dt, if any exist.
dt.
Then f (x) = G(2x − x2 ), so, using the Chain Rule, f 0 (x) = G0 (2x − x2 )· (2 − 2x). The
1
1
0
2
. Therefore
FTC tells us that G0 (x) = cos 1+x
2 , so G (2x − x ) = cos
1+(2x−x2 )2
0
0
2
f (x) = G (2x − x ) · (2 − 2x) = cos
1
1 + (2x − x2 )2
· (2 − 2x).
Use this calculation and the tools we developed last term to show that f (x) has a
global maximum at x = 1 and no global minimum.
2