By Bob Miller
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Additional resources for Bob Miller's Calc for the Clueless: Calc I (Bob Miller's Clueless Series)
Substitute into T. Example 3— An open box with a square bottom is to be cut from a piece of cardboard 10 feet by 10 feet by cutting out the corners and folding the sides up. Find the dimensions that will result in the largest volume. The most difficult part of this problem is the picture, which is given above in three steps. The volume of a box is length times width times height. We reject 5 since it will give a volume of 0, a minimum. indicating a maximum. The length and width are each 10 - 2x = 10 - 2(5/3) = 20/3.
Now we are ready—not happy—but ready to start. 1. Divide the interval 3 < × < 6 into n equal parts. , and xn = 3 + n ∆x = 6. Solving for ∆x, ∆x = (6 - 3)/n = 3/n. 2. From before, the approximate sum is f(w1)∆x1 + f(w2)∆x2 + f(w3)∆x3 + ... + f(wn)∆xn. All of the ∆xi are equal to ∆x, and we will take the right end of each interval as the point where we will take the height. , and wn = xn. 3. Rewriting 2, we factor out the ∆x and get [f(x1) + f(x2) + f(x3) + ... + f(xn)∆x. 4. Now f(x) = x2 + 4x + 7.
X1 f(w1) represents the area of the first approximating rectangle. E. Do this for all the rectangles. We get The definite integral. In the above example, the definite integral represents the area. As in the case of derivatives, we would like the rules so that it would be unnecessary to do this process of approximating rectangles. Note If the velocity = v(t), represents the distance traveled from time a to time b. Rule 6 Rule 7 Rule 8 Rule 9 Rule 10 Rule 11 Rule 12 The fundamental theorem of integral calculus.
Bob Miller's Calc for the Clueless: Calc I (Bob Miller's Clueless Series) by Bob Miller