Education.com
Try
Brainzy
Try
Plus

Definite Integrals Solutions to Practice Problems for AP Calculus

based on 5 ratings
By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these solutions can be found at: Definite Integrals Practice Problems for AP Calculus

Part A The use of a calculator is not allowed.

  1. Note that f (x )= cos xx2 is an even function. Thus you could have written and obtained the same result.

    Part B Calculators are allowed.

  2.  

 

  1.  
    1. See Figure 11.9-1.
    2. The function f has a relative minimum at x = – 5 and x =3, and f has a relative maximum at x = – 1 and x =7.

    3. See Figure 11.9-2.
    4. The function f is concave upward on intervals (–6, –3) and (1, 5).

    5. A change of concavity occurs at x = – 3, x =1, and x =5.
  2. Thus at each of the points at (1, 1) and (1, –5) the graph has a horizontal tangent.

    Enter [Solve] (9x 2 – 8x – 27=0, x ) and obtain x =3 or x = – 1.

    Thus at each of the points (3, –2) and (–1, –2), the graph has a vertical tangent. See Figure 11.9-3.

    Definite Integrals Solutions to Practice Problems

  3. (Calculator)
  4. Step 1. See Figure 11.9-4. Let P =x + y where P is the length of the pipe and x and y are as shown. The minimum value of P is the maximum length of the pipe to be able to turn in the corner. By similar triangles,

    Definite Integrals Solutions to Practice Problems

    Step 2.Find the minimum value of P. Enter Use the [Minimum] function of the calculator and obtain the minimum point (9.306, 22.388).

    Step 3.Verify with the First Derivative Test. Enter y 2=(y1(x ), x ) and observe. (See Figure 11.9-5.)

    Step 4.Check endpoints. The domain of x is (6,∞). Since x =9.306 is the only relative extremum, it is The absolute minimum. Thus the maximum length of the pipe is 22.388 feet.

Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed