Solutions to Practice Problems

Math 113, Calculus II
Test 1: Review Solutions
3. Suppose you know that a certain function f is twice differentiable and that its graph over
[0, 2] is given in the figure on your review sheet. As you see, the printer was sloppy and spilled
a lot of ink on the graph. Decide, if possible, whether each of the following definite integrals
is positive, equal to zero, or negative.
(a) ∫ 2
0 f ′′ (x)dx
(b) ∫ 2
0 f ′ (x)dx
(c) ∫ 2
0 f(x)dx
Solution: By the Fundamental Theorem of Calculus (Part I), we know that ∫ 2
0 f ′′ (x)dx =
f ′ (2) − f ′ (0) because f ′ (x) is an antiderivative of f ′′ (x). Now f ′ measures the slope of the
curve, and we see that the slope of f is negative when x = 2 (because the curve is decreasing
there!) so that f ′ (2) < 0. Similarly, since f is increasing when x = 0, f ′ (0) > 0, so
f ′ (2) − f ′ (0), whatever the actual value of the numbers may be, must be negative as it is a
negative number (f ′ (2)) MINUS a positive number (f ′ (0)). Thus,
∫ 2
0
f ′′ (x)dx < 0.
Using the Fundamental Theorem of Calculus (Part I) again, ∫ 2
0 f ′ (x) = f(2)−f(0) since
f(x) is an antiderivative of its derivative, f ′ (x). Looking at the graph of y = f(x), we see
that f(2) is a small positive number while f(0) is a large positive number, so that f(2) − f(0)
must then be negative. Therefore,
∫ 2
0
f ′ (x)dx < 0.
Lastly, we can see a fair bit of the graph, and the portion of it to the left of the ink spot is all
well above the x-axis and by itself contributes a positive area larger than the total area of the
ink blot BELOW the x-axis. Thus, as ∫ 2
0
f(x)dx measures the signed area of f(x) over the
interval [0, 2] and the part of this area to the left of the ink blot is larger than any negative
amount the ink blot could be covering and the part to the right of the blot is again positive,
we conclude that the signed area must be positive, so
4. If f is a continuous function, find f(4) if
∫ 2
0
x sin πx =
f(x)dx > 0.
∫ x 2
0
f(t)dt.
Solution: Let F (x) = ∫ x
0
f(t)dt, as in the statement of the Fundamental Theorem of
Calculus (Part I). Then, applying the theorem, F ′ (x) = f(x). Now, the integral on the
right-hand side of the equation above is not F (x), but rather is F (x 2 ), for the upper limit of
integration is x 2 instead of simply x. Therefore, the derivative of the right-hand side is
d
dx [F (x2 )] = F ′ (x 2 d
) ·
dx [x2 ] = f(x 2 ) · [2x] = 2x · f(x 2 )