2. Given the marginal cost, find the original cost equation. C ' ( x ) = 9 x 2 – 10 x + 7 ; fixed cost is $ 20. In algebra, we were told that what ever was done to one side of the equal sign we were to do to the other side. Therefore we will integrate both sides of the equation. ∫ C ' ( x ) dx = ∫ ( 9 x 2 – 10 x + 7 ) dx However we want the original cost function. You must understand what fixed cost means. Fixed cost is a cost encountered no matter what, even if nothing is being done (meaning x = 0 ). Integrating the left side simply takes us back to the original name C ( x ). Integrating the right side gives a general solution. C ( x ) = 3 x 3 – 5 x 2 + 7x 7x + C 20 = 3 ( 0 ) 3 – 5 ( 0 ) 2 + 7( 0 ) + C 20 = C Make certain to answer the question. C ( x ) = 3 x 3 – 5 x 2 + 7x 7x + 20 The dx’s and parentheses are necessary and must be written in the problem.

3. Do not make the mistake in believing that the constant C is equal to the fixed cost. This is only true because of the example chosen. If the marginal cost equation had been C ' ( x ) = 6 x 2 – 24 x + 11e x with fixed cost of $ 30 then ∫ C ' ( x ) dx = ∫ ( 6 x 2 – 24 x + 11 e x ) dx C ( x ) = 2 x 3 – 12 x 2 + 11 e x + C 30 = 2 ( 0 ) 3 – 12( 0 ) 2 + 11 e 0 + C 30 = 11 ( 1 ) + C 19 = C C ( x ) = 2 x 3 – 12 x 2 + 11 e x + 19 Look at problems on page 850. Remember any number, except zero, raised to the zero power is one.

4. Given the marginal cost, find the original cost equation. C ' ( x ) = 12 x 2 – 22 x + 7 ; three items cost $ 36 ∫ C ' ( x ) dx = ∫ ( 12 x 2 – 22 x + 7 ) dx C ( x ) = 4 x 3 – 11 x 2 + 7 x + C 36 = 4 ( 3 ) 3 – 11 ( 3 ) 2 + 7 ( 3 ) + C 36 = 4 ( 27 ) – 11 ( 9 ) + 7 ( 3 ) + C 36 = 108 – 99 + 21 + C 36 = 30 + C 6 = C C ( x ) = 4 x 3 – 11 x 2 + 5 x + 6 Do not make the mistake in believing that if three units cost $ 36 that one unit cost $ 12. The $ 12 would be an average cost not an actual cost. If you try it in the above example, you will not get C = 6. It is not necessary to show these intermediate steps on a test unless you just want to do it.

5. PROPERTIES OF DEFINITE INTEGRALS

6. To evaluate the definite integral. Write antiderivative twice, always separated by a minus sign, place the upper value in the first, the lower value in the second, and evaluate. Make certain when you key in the expression you do not use the [ ] symbols on your calculator; use the parentheses keys instead. If you decide to evaluate each term separately and remove the parentheses, make certain you remember that the minus sign affects each sign in the second group. You must also know and follow the rules of hierarchy of mathematics. The calculator knows and follows these rules without you having to remember. [ 2 ( ) 2 + 7 ( )] – [ 2 ( ) 2 + 7 ( = [ 2 ( 6 ) 2 + 7 ( 6 – [ 2 ( 3 ) 2 + 7 ( 3 = [ 2 ( 36 ) + 7 ( 6 )] – [ 2 ( 9 ) + 7 ( 3 = [ 72 + 42 ] – [ 18 + 21 ] = 72 + 42 – 18 – 21 = 114 – 39 = 75 Notice there is no + C written. It is not necessary for evaluating a definite integral because it would be eliminated in the final answer. FUNDAMENTAL THEOREM OF CALCULUS Let f be continuous on the interval [ a, b ] and let F be any antiderivative of f. Then

7. FINDING AREA To find the area bounded by f ( x ), x = a, x = b, and the x-axis, use the following steps. 1. Sketch the graph. 2. Find any x-intercepts of f ( x ) in [ a, b ]. These divide the total region into subgroups. 3. The definite integral will be positive for subregions above the x-axis and negative for subregions below the x-axis. Use separate integrals to find the (positive) areas of the subregions. 4. The total area is the sum of the areas of all the subregions. We will use absolute value symbols in finding the areas of subregions below the x-axis. The reason is that area can only be positive, never negative. If you fail to determine if there are negative subregions, the area will be incorrect.

8. Find the area for the interval 1 ≤ x ≤ 4 for f ( x ) = 15x 2 – 8x 8x – 63. (Hint: You must determine where the graph crosses the x-axis.) Step 1: Sketch the graph of the equation. On the test the sketch will be provided. Step 2: Find the x-intercept in the interval [ 1, 4 ]. This point must be found because it is the point that separates the negative subregion from the positive subregion.

9. To find the x-intercepts, set the equation equal to zero and solve by either factoring the equation or using the quadratic formula. Either method is acceptable. We will find this point using the quadratic formula a is the coefficient of x 2, b is the coefficient of x, and c is the constant. The sign is part of the value. Because the x-intercept on the far right is the only one needed, use only the + sign in the formula. There is no ± symbol on your calculator anyway. Keying this into this into the calculator and using the frac option that is found in the math menu, we obtain x = _ ( (–) ( (–) 8 ) + √ ( (–) 8) ^ 2 – 4 * 15 * (–) 63))/( 2 * 15 ) frac Note: (–) is the negative key at the bottom of the calculator. Do not confuse this with the – sign on the right side of the calculator. I am assuming you are using some type of TI 80 series calculator. f ( x ) = 15x 2 – 8x 8x – 63 = 0 a = 15, b = – 8, c = – 63

10. Step 3. So we will now separate our problem into the two subregions and because area must always be positive, we will take the absolute value of the negative subregion. Notice both expressions are exactly alike Next write each expression twice, separated by a minus sign, substituting the upper value in the first expression, and the lower value in the second. I suggest you work each expression separately and not try to get an immediate answer.

11. Notice these two are the same answer. Notice this answer is negative as expected. or 152.52 in decimal form. I prefer fraction form. Step 4. Study problems on page 882.