Evaluate using Integration by parts

In summary, the integral from 1 to 5 of xg'(x)dx can be evaluated by using integration by parts to get the expression xg(x)-(the integral of g(x)dx from 1 to 5). Substituting in the given values, we get 37 as the result. However, this is incorrect as the integral of g(x)dx from 1 to 5 is already given as -9. So, we only need to substitute in the value of xg(x) from 1 to 5, which gives us 40-3=37. Then, we add the given value of -9 to get the final answer of 37+9=46.
  • #1
waealu
37
0

Homework Statement



If g(1)=3, g(5)=8 and the integral from 1 to 5 of g(x)dx=-9. Then, evaluate the integral from 1 to 5 of xg'(x)dx.

2. Homework Equations and attempt at solution

I used integration by parts to get =xg(x)-(integral of)g(x)dx from 1 to 5. Then substituting in, I get ((5)(8)+9)-((1)(3)+9)=37

Apparently that answer is incorrect. What am I missing? Thanks.
 
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  • #2
waealu said:

Homework Statement



If g(1)=3, g(5)=8 and the integral from 1 to 5 of g(x)dx=-9. Then, evaluate the integral from 1 to 5 of xg'(x)dx.

2. Homework Equations and attempt at solution

I used integration by parts to get =xg(x)-(integral of)g(x)dx from 1 to 5. Then substituting in, I get ((5)(8)+9)-((1)(3)+9)=37

Apparently that answer is incorrect. What am I missing? Thanks.

You didn't think of the integral in the right way. You are already provided the answer for the complete integral of g(x)dx. You don't need to apply it twice.

you should get xg(x) from 1 to 5 giving you 40-3=37. Then the integral will give you --9=+9

Final answer is 37+9=46

Note: [tex]\int_a^b u\; dv=[uv]_a^b-\int_a^b v\; du [/tex]
 

What is integration by parts?

Integration by parts is a technique used in calculus to evaluate integrals that involve products of functions. It is based on the product rule from differentiation, and involves breaking down an integral into two parts and applying a specific formula.

When should I use integration by parts?

Integration by parts should be used when the integral involves a product of two functions, and the integral cannot be easily evaluated by other means (such as substitution or using a table of integrals). It is also useful when the integral involves a function that becomes simpler after being differentiated, such as logarithmic or trigonometric functions.

How do I choose which part to differentiate and which part to integrate?

The general rule for choosing which part to differentiate and which part to integrate is to prioritize differentiating functions that become simpler after being differentiated (such as logarithmic or trigonometric functions), and integrating functions that become more complex after being integrated (such as polynomials or exponential functions).

What are the steps for using integration by parts?

The steps for using integration by parts are as follows:

  1. Identify the integral as a product of two functions.
  2. Choose which part to differentiate and which part to integrate.
  3. Apply the integration by parts formula: ∫u dv = uv - ∫v du
  4. Simplify the resulting integral and evaluate it using other integration techniques if necessary.

Are there any special cases or exceptions to using integration by parts?

Yes, there are a few special cases where integration by parts may not work or may require additional steps:

  • If the integral involves a product of two power functions (such as x^n and x^m), integration by parts may not work and a different approach may be needed.
  • If the integral is an indefinite integral (has no limits of integration), the resulting integral from integration by parts may require the use of integration by parts again.
  • If the integral involves trigonometric functions, a special trigonometric identity may be needed to simplify the resulting integral before it can be evaluated.

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