Integration by parts, just need a small hand

In summary, the conversation discusses how to calculate definite integrals by first finding the antiderivatives and then evaluating them on a given interval. The individual terms for each integral are correctly calculated, but the final arithmetic is incorrect. The correct answer is 8/135, not 8/27 or 40/135.
  • #1
SYoungblood
64
1

Homework Statement


I'm going to cut from the initial part of the problem, which I am confident is good to go, and cut straight to the antiderivatives.

Homework Equations


All antiderivatives are to be integrated on the interval from 0 to π/18

(I1) = -1/9 cos 9x - (I2) (-2/27 * cos3(9x)) + (I3) (-1/45 * cos5(9x)

The Attempt at a Solution


For I1 -- (π)/18) -- -1/9 cos 9π/18 = -1/9 cos π/2 = 0
-1/9 cos 0 = -1/9
0 - (-1/9) = 1/9

ForI2 -- ( π/18) -- -2/27 cos 9π/18 = -2/27 cos π/2 = 0
-2/27 cos 0 = -2/27
0 - (-2/27) = 2/27

For I3 -- (π/18) -- -1/45 cos 9π/18 = -1/45 cos π/2 = 0
-1/45 cos 0 = -1/45
0- (-1/45) = 1/45

1/9 - 2/27 + 1/45 = 8/27, or 40/135 -- However, the answer in my text is 8/135.

Again, I am positive on my getting the antiderivatives right, I am sure the error is in the computing of the definite integrals. I thank you in advance for all help.

SY
 
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  • #2
SYoungblood said:

Homework Statement


I'm going to cut from the initial part of the problem, which I am confident is good to go, and cut straight to the antiderivatives.

Homework Equations


All antiderivatives are to be integrated on the interval from 0 to π/18

(I1) = -1/9 cos 9x - (I2) (-2/27 * cos3(9x)) + (I3) (-1/45 * cos5(9x)

The Attempt at a Solution


For I1 -- (π)/18) -- -1/9 cos 9π/18 = -1/9 cos π/2 = 0
-1/9 cos 0 = -1/9
0 - (-1/9) = 1/9

ForI2 -- ( π/18) -- -2/27 cos 9π/18 = -2/27 cos π/2 = 0
-2/27 cos 0 = -2/27
0 - (-2/27) = 2/27

For I3 -- (π/18) -- -1/45 cos 9π/18 = -1/45 cos π/2 = 0
-1/45 cos 0 = -1/45
0- (-1/45) = 1/45

1/9 - 2/27 + 1/45 = 8/27, or 40/135 -- However, the answer in my text is 8/135.

Again, I am positive on my getting the antiderivatives right, I am sure the error is in the computing of the definite integrals. I thank you in advance for all help.

SY

Your individual terms are correct and your almost-answer of ##1/9 - 2/27 + 1/45## is correct, but your arithmetic is wrong after that.
 
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  • #3
Ray Vickson said:
Your individual terms are correct and your almost-answer of ##1/9 - 2/27 + 1/45## is correct, but your arithmetic is wrong after that.
Got it, thank you very much for your help -- (15 - 10 + 3) / 135 = 8/ 135 -- just needed a push in the right direction.
 

FAQ: Integration by parts, just need a small hand

1. What is integration by parts?

Integration by parts is a calculus technique used to find the integral of a product of two functions by breaking it down into simpler parts.

2. When is integration by parts used?

Integration by parts is used when the integrand (the function being integrated) can be expressed as the product of two functions, and one of the functions is easier to integrate than the other.

3. How does integration by parts work?

Integration by parts involves using the formula ∫u dv = uv - ∫v du, where u and v are the two functions in the product and dv and du represent their derivatives. This formula is applied repeatedly until the integral can be easily evaluated.

4. What are the steps to use integration by parts?

The steps to use integration by parts are:

  1. Identify u and dv in the integrand
  2. Find the derivatives of u and v
  3. Apply the formula ∫u dv = uv - ∫v du
  4. Simplify the resulting integral
  5. Repeat the process until the integral can be easily evaluated

5. Are there any common mistakes to avoid when using integration by parts?

Yes, some common mistakes to avoid include: choosing the wrong u or dv, not simplifying the resulting integral, and not paying attention to the signs in the formula. It is important to double check your work and make sure all steps are correct.

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