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About Scotto
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Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

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I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

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I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

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You are here:  Experts > Teens > Homework/Study Tips > Calculus > Calculus I (Substitution Rule and Definite Integrals)

Calculus - Calculus I (Substitution Rule and Definite Integrals)


Expert: Scotto - 6/26/2008

Question
Please help! I'm not sure how to do this one and I must have it done by tomorrow (6-27-08)! I was able to do all of the indefinite integrals but this one is a definite and I am confused. Please give me step by step of how to do this.

Question: Evaluate the definite integral    ∫[0,4] ((x)/(sqrt(1+2x)))dx

(Note: The interval [0,4] follows the integral sign directly with the 4 being above the 0)

***I have attempted this problem and gotten (1/2)∫[1,9]  ((x)/(sqrt(u)))du by determining that u= 1+2x, du= 2dx, and du/2= sx as well as u=1 when x=0 and u=9 when x=4  (which are my new intervals). Please let me know if I was on the right track!!! I really appreciate it and hope you can help.  

Answer
Have you ever done ∫u dv = uv - ∫v du?

Let u=x, then du=dx.
Let dv = 1/(sqrt(1+2x)) dx, so v = sqrt((1+2x) since the half and the two cancel in the derivative.

The variables u and v are now both known, so all that has to be done is to integrate v.  The variable v is just (1+2x)^0.5.  Don't forget to worry about d(2x)/dx=2 when doing this.


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