Humongous Book Of Basic Math And Pre- Algebra Problems by W Michael Kelley
Are you struggling with math? Are you bored with dry and difficult study guides? Look no further than The Humongous Book of Basic Math and Pre-Algebra Problems! Unlike most math books, this one is designed to make math fun again. Written by W. Michael Kelley, who is anything but dry, this book is a breath of fresh air for those struggling with math concepts.
What sets this book apart from others is Kelley’s unique approach to problem-solving. Not only does he provide solved problems, but he also adds insightful notes along the margin. These notes include missing steps and simplified explanations, making even the most complex problems crystal clear. Suddenly, questions that used to baffle you become manageable.
One of the book’s greatest strengths is its ability to prepare you for those tricky problems that never seem to be covered in class but always show up on exams. Kelley’s expertise in math education shines through as he tackles these difficult and obscure problems head-on. With his guidance, you’ll be well-equipped to handle any challenge that comes your way.
In addition to the annotated notes within the book, Kelley’s impressive track record of helping students and math enthusiasts speaks for itself. His website, calculus-help.com, attracts thousands of students every month who rely on his expertise.
If you’re looking to conquer your math struggles and find joy in the subject again, The Humongous Book of Basic Math and Pre-Algebra Problems is the ultimate resource. Say goodbye to boring textbooks and hello to a fun and engaging way of learning math!
Calculating Integrals using Basic Substitution: Method described in Ch.18 Part 4
Welcome to Chapter 18, Part Four of our math series. In this post, we’ll be exploring Problem 18.44 and Problem 18.46, as well as evaluating the definite integral in Problem 18.47. Let’s dive in!
Problem 18.44 – Verifying the Anti-Derivative
In this problem, we need to verify the anti-derivative of the given function: ∫ tan(x) dx. To do this, we’ll use a basic substitution method. Let’s break it down step by step.
We start by rewriting the integrand in terms of u. We let u = cos(x), which means du = -sin(x) dx. Now, our integral becomes:
∫ -du/u
We can simplify this further to:
∫ -1/u du
The anti-derivative of -1/u (with respect to u) is -ln|u|. Therefore, the anti-derivative solution in terms of x is:
-ln|cos(x)| + c
And there you have it! Now, let’s move on to Problem 18.46.
Problem 18.46 – A New Integral to Solve
In this problem, we are given the integral: ∫ (3.14/6) tan(2x) dx. To solve this, we’ll make use of a substitution as well. Let’s walk through the steps.
First, we let u = 2x. This means du = 2 dx. We can rewrite the integral as:
(3.14/6) ∫ tan(u) (du/2)
Simplifying this further, we get:
(3.14/12) ∫ tan(u) du
The anti-derivative of tan(u) is -ln|cos(u)|. So, our final solution is:
(-3.14/12) ln|cos(2x)| + c
Excellent! Now, let’s move on to our last problem, Problem 18.47.
Problem 18.47 – Evaluating a Definite Integral
In this problem, we are tasked with evaluating the definite integral: ∫(5/13x)^(2x) dx. This requires a different approach called invertible substitution. Let’s see how it’s done.
We start by letting u = (13x)^(2x). This means du = 2(13x)^(2x) ln(13x) dx. We can rewrite the integral as:
∫ (u/2) du
Simplifying further:
(1/2) ∫ u du
The anti-derivative of u is (1/2)u^2. Now, let’s convert the limits of integration to terms of u.
When x = 1, u = (13(1))^2 = 169
When x = 5, u = (13(5))^2 = 4225
Now, our integral becomes:
(1/2) ∫ u du, with limits 169 to 4225
Using the fundamental theorem of calculus, we can calculate:
(1/2) [(4225)^2 – (169)^2]
After some calculation, the result is 7,094,550. And that wraps up Problem 18.47!
That’s it for these math problems! We hope this post helped you understand the concepts and techniques used in these integrals. Stay tuned for more math adventures in our next article!
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