Integration by parts of a function

In summary, the goal was to integrate the function c = 15te-.2t from t = 0 to t = 3. The integral was set up by taking out the constant 15 and setting u = t and dv = e-.2tdt. After following the integration by parts formula and evaluating the integral, the final answer in terms of the constant e was calculated to be 120.7130184. However, it was suggested to recalculate the answer without using a calculator and leaving it in terms of e.
  • #1
apiwowar
96
0
the function is c = 15te-.2t

the goal is to integrate it from t = 0 to t = 3

so to set up the integral i took out the 15 first so i got:

15 * integral from 0 to 3 of t*e-.2t

i set u = t and so du = dt
dv = e-.2tdt so v= -5e-.2t

so following the integration by parts formula i got:

15(-5*e-.2t + 5* integral of e-.2tdt

then after integrating the second part i got

15(-5e-.2t - 25e-.2t) evaluated from 0 to 3

when evaluated i got

(-123/4826181 - 205.8043635)-(-75-375) = 120.7130184

but that's wrong,

anyone have any hints or sees any mistakes that i made?

thanks
 
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  • #2
apiwowar said:
the function is c = 15te-.2t

the goal is to integrate it from t = 0 to t = 3

so to set up the integral i took out the 15 first so i got:

15 * integral from 0 to 3 of t*e-.2t

i set u = t and so du = dt
dv = e-.2tdt so v= -5e-.2t

so following the integration by parts formula i got:

15(-5*e-.2t + 5* integral of e-.2tdt
s

this should be

15(-5te-0.2t+5∫e-.2tdt
 
  • #3
your right, that's just a typo, i had the t there in my calculations
 
  • #4
it looks like you did it correct, is your answer supposed to be in terms of e?
 
  • #5
i don't think so, it just asks for the exact answer
 
  • #6
apiwowar said:
i don't think so, it just asks for the exact answer

Then you need to write it in terms of the constant e. So recalculate the answer, just not using a calculator.
 
  • #7
so just leave it as with the base e but calculate the exponents of e?
 
  • #8
apiwowar said:
so just leave it as with the base e but calculate the exponents of e?

Yes.
 
  • #9
ok, thanks alot
 

Related to Integration by parts of a function

1. What is integration by parts?

Integration by parts is a technique used in calculus to find the integral of a product of two functions. It is based on the product rule of differentiation, and is useful for solving integrals that cannot be solved by other methods.

2. How do you use integration by parts?

To use integration by parts, you need to identify the two functions in the integral and assign one as u and the other as dv. Then, use the formula ∫u dv = uv - ∫v du to find the integral. This technique is also known as the "LIATE" rule.

3. When should you use integration by parts?

Integration by parts is useful for integrals that involve products of functions, such as polynomials multiplied by trigonometric functions or exponential functions. It is also helpful for integrals that cannot be solved by substitution or other methods.

4. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are the two functions in the integral. This formula is derived from the product rule of differentiation.

5. What are the steps for using integration by parts?

The steps for using integration by parts are: 1) Identify the two functions in the integral and assign one as u and the other as dv. 2) Differentiate u and integrate dv. 3) Substitute the values into the formula ∫u dv = uv - ∫v du. 4) Simplify the resulting integral and solve for the answer. 5) If necessary, use integration by parts again or other methods to solve for the final answer.

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