Page 35 - Calc for the Clueless

You must remember that the derivative of the tangent is the secant squared.

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Let u = tan 3x. du = 3 sec 3x dx. du/3 = sec 3x dx.
x = 0; tan (0) = 0. So u = 0.

x = π/12. tan (3π/12) = tan (π/4) = 1. So u = 1.

Finding the Area Under the Curve Using the Definition of the Definite Integral

One of the most laborious tasks is to find the area using the definition. Doing one of these problems will make
you forever grateful that there are some rules for antiderivatives, especially the fundamental theorem of
calculus.

Example 32—

Find dx using the definition of the definite integral.

Before we start, we need two facts:

These formulas for the sum of the first n positive integers and the sum of the squares of the first n positive
integers can be found in some precalculus books but are not too easily proved.

Now we are ready—not happy—but ready to start.

1. Divide the interval 3 < × < 6 into n equal parts. The left end 3 = x 0, x 1 = 3 + 1 ∆x, x 2 = 3 + 2 ∆x, x 3 = 3 + 3
∆x,..., and x n = 3 + n ∆x = 6. Solving for ∆x, ∆x = (6 - 3)/n = 3/n.

2. From before, the approximate sum is f(w 1)∆x 1 + f(w 2)∆x 2 + f(w 3)∆x 3 + ... + f(w n)∆x n. All of the ∆x i are equal
to ∆x, and we will take the right end of each interval as the point where we will take the height. Therefore w 1 =
x 1, w 2 = x 2,..., and w n = x n.

3. Rewriting 2, we factor out the ∆x and get [f(x 1) + f(x 2) + f(x 3) + ... + f(x n)∆x.

4. Now f(x) = x + 4x + 7.
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