# How Does Substitution Impact Integration Limits?

• fashion_fever
In summary: So in summary, The Attempt at a Solution: There is no one definitive solution to the homework equations. Different students may come up with different solutions based on different methods and assumptions. However, one possible solution is to use substitution.
fashion_fever

## Homework Statement

evaluate:

higher limit of 36
lower limit of 0 (36+3x)^1/2 dx

## Homework Equations

i thought of using subsititution?

## The Attempt at a Solution

g(x)=36+3x
g'(x)=3

when x=0, u=36+3(0)=36
when x=36, u=36+3(36)=144

from lower limit of 36 to higher limit of 144

3(u)^1/2 du= 3(2/3)u^3/2 + C

substitute 36+3x back into u, i get: 2(36+3x)^3/2 + C

=[2(36+3(144))^3/2) + C ] - [ 2(36+3(36))^3/2] + C

= 2(468)^3/2 - 3456 + C

is this correct??

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Your final answer should just be a number, there should be no constant of integration. When you have -[ 2(36+3(36))^3/2] +C it should really be -[ 2(36+3(36))^3/2 + C] and the constants should cancel out

so it would just be: 2(468)^3/2 - 3456 ?

Checking it, you made a couple mistakes. First with your substitution. The derivative of $$2 (36+3x)^{3/2}$$ is $$9(36+3x)^{1/2}$$, which isn't right. Check your integration by substitution again and be careful with plugging in du for dx.

Also, you turned your answer back into x's, but then plugged in the upper and lower bounds for u.

um...i don't get it..i didn't find i wrote 9(36+3x)^1/2..so what should it be?
and for the x=144,x=36, i do substitute those in right?...

When you make a substitution to a new variable of integration, you can either a) fix your limits of integration by putting them through the function or b) leave them as "x=whatever" and then when you are done integrating, put back your original variable and evaluate. But not both, which is what you seem to have done here.

Let me see if I can make that clearer. Consider $\intop_1^2 (x^2-2x+1) dx$.
Now you can just evaluate this:
$$\frac{x^3}{3}-x^2+x \vert^1_2 = \frac{1}{3}$$.

But let's suppose you decide you want to do a substitution:
$u=x-1, du=dx$
$$\intop_1^2 (x^2-2x+1) dx = \intop^?_? u^2 du$$.

Now you have two choices, you can move the limits from x to u by plugging into your substitution:
$$\intop^1_0 u^2 du = \frac{u^3}{3} \vert^1_0 = \frac{1}{3}$$
or you can write:
$$\intop^{x=2}_{x=1} u^2 du = \frac{u^3}{3} \vert^{x=2}_{x=1} = \left[\frac{x^3}{3}-x^2+x-1 \vert^2_1 \right]= \frac{1}{3}$$

Sometimes it's more convenient to move your limits of integration to the new variable and sometimes it's not. But what you did was convert your limits of integration to the new variable u. Then when you were done integrating, you converted the expression to be evaluated back to x, but plugged in the "u" bounds.

## 1. What is integration by substitution?

Integration by substitution is a method used to solve integrals by substituting a new variable for the original variable in the integral. This new variable is chosen to simplify the integral and make it easier to solve.

## 2. How does integration by substitution work?

The first step in integration by substitution is to identify a new variable, usually denoted as u, and then rewrite the integral in terms of u. Next, the derivative of u with respect to the original variable is found and substituted into the integral. This allows for the original integral to be rewritten entirely in terms of u. The integral is then solved for u and the final answer is expressed in terms of the original variable.

## 3. When should I use integration by substitution?

Integration by substitution is most useful when the integrand (the function being integrated) contains a composite function, meaning a function within a function. It is also helpful when the integrand contains a function and its derivative or when the integrand can be simplified by substituting a new variable.

## 4. What are some common mistakes when using integration by substitution?

Some common mistakes when using integration by substitution include forgetting to include the differential of the new variable in the integral, not properly rewriting the integral in terms of the new variable, and not substituting back in the original variable at the end of the problem.

## 5. Can integration by substitution be used for all integrals?

No, integration by substitution can only be used for certain types of integrals, specifically those that contain a composite function or a function and its derivative. It may not work for integrals with more complex functions or those that cannot be simplified by substitution.

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