Integration by Parts Evaluate the integral

In summary, integration by parts is a method used in calculus to evaluate integrals of products of functions. It should be used when the integral involves a product of two functions and there is no clear substitution that can be made. To use it, one must identify the "u" and "dv" functions and use the formula ∫u dv = uv - ∫v du. The purpose of integration by parts is to simplify the evaluation of integrals involving products of functions, allowing for the integration of more complex functions. It is particularly useful in engineering, physics, and other scientific fields.
  • #1
emmaerin
11
0

Homework Statement



Evaluate the integral. (Use C for the constant of integration.)

∫te ^ (-9t) dt

Homework Equations



∫udv = uv - ∫vdu

u=t dv= e ^ (-9t) dt
du=dt v=(-1/9) e ^(-9t)

The Attempt at a Solution



= -1/9 te^(-9t) - ∫-1/9 e ^(-9t) dt

Second Integral:
w=-9t
dw=-9dt
-81∫-1/9 * -81 e ^(-9t) dt
-81∫e^w * w
-81 * e^(-9t) +C

Final Answer:
= -1/9 te^(-9t) + 81 e ^(-9t) +C

This answer isn't right and I'm not sure where I'm going wrong, so any help would be appreciated. Thanks!
 
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  • #2
A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.
 
  • #3
emmaerin said:

Homework Statement



Evaluate the integral. (Use C for the constant of integration.)

∫te ^ (-9t) dt


Homework Equations



∫udv = uv - ∫vdu

u=t dv= (-1/9) e ^(-9t)
du=dt v=e ^ (-9t) dt
Where did that "-1/9" come from in "dv"?
The integral you are given is [itex]\int te^{-9t}dt[/itex]. Writing that as [itex]\int u dv[/itex], you could take u= t, [itex]dv= e^{-9t}[/itex].

Or you could write the integral as [itex]-9\int t(-(1/9)e^{-9t})dt[/itex] and then take [itex]dv= (-1/9)e^{-9t}dt[/itex] but you don't seem to have done that.

The Attempt at a Solution



= -1/9 te^(-9t) - ∫-1/9 e ^(-9t) dt

Second Integral:
w=-9t
dw=-9dt
-81∫-1/9 * -81 e ^(-9t) dt
-81∫e^w * w
-81 * e^(-9t) +C

Final Answer:
= -1/9 te^(-9t) + 81 e ^(-9t) +C

This answer isn't right and I'm not sure where I'm going wrong, so any help would be appreciated. Thanks!
 
  • #4
HallsofIvy said:
Where did that "-1/9" come from in "dv"?

You're completely right - I accidentally switched v and dv, thanks for pointing that out!
 
  • #5
Sir Beaver said:
A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.

So is the final answer -1/9 t e ^(-9t) + 1/81 e ^(-9t) + C ?
 
  • #6
emmaerin said:
So is the final answer -1/9 t e ^(-9t) + 1/81 e ^(-9t) + C ?


Differentiate it and see if you get your integrand.
 

1. What is integration by parts?

Integration by parts is a method used in calculus to evaluate integrals of products of functions. It involves using the product rule for differentiation to rewrite the integral in a different form, making it easier to solve.

2. When should integration by parts be used?

Integration by parts should be used when the integral involves a product of two functions and there is no clear substitution that can be made to simplify the integral. It is also useful when the integral involves logarithmic or inverse trigonometric functions.

3. How do you use integration by parts?

To use integration by parts, you must first identify which function will be the "u" and which will be the "dv" in the product rule. Then, use the formula: ∫u dv = uv - ∫v du, where u is the function being differentiated and v is the function being integrated. Repeat this process until the integral can be easily solved.

4. What is the formula for integration by parts?

The formula for integration by parts is: ∫u dv = uv - ∫v du, where u is the function being differentiated and v is the function being integrated. This formula is derived from the product rule for differentiation.

5. What is the purpose of integration by parts?

The purpose of integration by parts is to simplify the evaluation of integrals involving products of functions. It allows for the integration of more complex functions that cannot be solved using other methods, such as substitution. This method is particularly useful in solving integrals in engineering, physics, and other scientific fields.

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