Math 112 (Calculus I) Final Exam Form A KEY Solutions, Exams of Calculus

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Math 112 (Calculus I)

Final Exam Form A KEY

Part I: Multiple Choice. Enter your answer on the scantron. Work will not be collected or reviewed.

Free response: Write your answer in the space provided. Answers not placed in this space will be ignored.

21. (8 points) Short answer. Two points each part. You do not need to show your work on this problem.

(a) Given ǫ > 0, and the statement

lim x→− 1

(− 2 x + 3) = 5,

find the largest δ > 0 so that the following statement is true:

If 0 < |x + 1| < δ, then | − 2 x + 3 − 5 | < ǫ.

Solution: ǫ/ 2

(b) Find lim x→∞

5 − 3 x^3 √ 81 x^6 − 16

Solution: − 1 / 3

(c) Evaluate the integral

(1 + x^2 ) dx.

Solution: x + x^3 /3 + C

(d) Let f (x) be the function whose graph is shown below. Use right hand sums with four rectangles to estimate

0

f (x)dx.

Solution: 11/2 = 5.

22. (5 points) Use the definition of the derivative to show:

If f (x) =

x − 1

, then f ′(x) = −

(x − 1)^2

26. (8 points) Suppose the area of a right triangle is 18 cm^2. Find the smallest possible length of its hypoteneuse. Solution: For a right triangle, let b denote the base and a the height. Then the area is given by A =

ab,

and the square of the hypoteneuse is given by:

h^2 = a^2 + b^2.

We know that A = 18 (cm^2 ), hence a = 36/b. Substitute this into our formula for h^2 :

h^2 =

(36)^2

b^2

• b^2.

Now, notice that h^2 and h will both have a minimum at the same values of a and b, so we can save ourselves a little algebra and minimize H(b) =

(36)^2

b^2

• b^2 rather than its square root (although using the square root will give the same answer). We find the derivative: H′(b) = − 2

(36)^2

b^3

• 2b

Set it equal to zero and solve for b:

(36)^2

b^3

• 2b = 0 ⇔ −2(36)^2 + 2b^4 = 0 ⇔ b^4 = (36)^2 ⇔ b = 6 (cm)

Use the second derivative test to check that this really gives a minimum:

H′′(b) = 6

(36)^2

b^4

• 2 > 0 when b = 6.

So b = 6 is a minimum, and H(6) =

(36)^2

• 36 = 72. Hence the shortest possible hypoteneuse is the square root of this, or 6

2 cm.

27. (8 points) The graph of a function f (x) is shown. Let g(x) =

∫ (^) x

0

f (t) dt, for 0 ≤ x ≤ 5.

(a) Evaluate g(0), g(2), and g(5). Solution: g(0) = 0, g(2) = 2. 5 , g(5) = − 3. 5 (b) Where is g(x) increasing on [0, 5]? Decreasing? Solution: Increasing on (0, 2), decreasing on (2, 5). (c) Sketch the graph of g(x) on the same axes of f (x), labeling all local maxima and minima. Solution: Graph should include the three points (0, 0), (2, 2 .5), and (5, − 3 .5). It should be increasing from (0, 0) to (2, 2 .5), then decreasing from (2, 2 .5) to (5, − 3 .5). The point (2, 2 .5) is a local maximum. There are no local minima.

28. (7 points) A pendulum swings with velocity v(t) = cos t. Find the total distance the pendulum

travels over 0 ≤ t ≤

5 π 2

Solution: The total distance traveled is the integral of | cos t| over [0, 5 π/2].

∫ (^5) π/ 2

0

| cos t|dt =

∫ (^) π/ 2

0

cos t dt −

∫ (^3) π/ 2

π/ 2

cos t dt +

∫ (^5) π/ 2

3 π/ 2

cos t dt

= sin t|π/ 0 2 − sin t|^3 π/π/ 22 + sin t|^53 π/π/^22 = 1 − 0 + 1 + 1 + 1 + 1 = 5

29. (6 points) Evaluate the integral.

x

2 + x dx

Solution: We use the substitution u = 2 + x. Then du = dx and x = u − 2. Then ∫ x

2 + x dx =

(u − 2)u^1 /^2 du =

(u^3 /^2 − 2 u^1 /^2 ) du

=

u^5 /^2 −

u^3 /^2 + C

=

(2 + x)^5 /^2 −

(2 + x)^3 /^2 + C.

END OF EXAM