# Using integration to find limits

• waealu
In summary, the problem asks to integrate dx\(sqrt(abs(x-1))) from 0 to 2. After integrating and plugging in the bounds for the two limit equations, the answer is 0. However, this is incorrect as the absolute value must be taken into account. The correct solution is to split the integral into two parts, one for each interval of the absolute value, and then add them together. This results in the correct answer of 4.
waealu

## Homework Statement

The problem asks to integrate dx\(sqrt(abs(x-1))) from 0 to 2.
$\int(dx/(sqrt(abs(x-1))))$

## Homework Equations

I had the limit as b approaches 1 of the integral (from 0 to b) plus the limit as c approaches 1 of the integral (from c to 2).

## The Attempt at a Solution

After integrating, I got 2(sqrt(x-1)). Then I plugged in my bounds for the two limit equations and got (0-2)+(2-0)=0. However, I know the answer is 4. What am I missing? I have a feeling that I am forgetting to do something with the absolute value.

Last edited:
waealu said:

## Homework Statement

The problem asks to integrate dx\(sqrt(abs(x-1))) from 0 to 2.
$\int\frac{}{}dx$

## Homework Equations

I had the limit as b approaches 1 of the integral (from 0 to b) plus the limit as c approaches 1 of the integral (from c to 2).

## The Attempt at a Solution

After integrating, I got 2(sqrt(x-1)). Then I plugged in my bounds for the two limit equations and got (0-2)+(2-0)=0. However, I know the answer is 4. What am I missing? I have a feeling that I am forgetting to do something with the absolute value.

You are forgetting to do something with the absolute value. You can't just integrate pretending it isn't there. |x-1|=x-1 on the interval [1,2] and |x-1|=1-x on the interval [0,1]. Integrate over the two intervals separately and add them.

Er Dirk beat me to it.

Okay, thanks. I figured it out now.

## What is integration and how is it used to find limits?

Integration is a mathematical technique used to find the area under a curve. It involves breaking down a complex shape into smaller, simpler shapes and adding their areas together. To find limits using integration, we use the fundamental theorem of calculus to evaluate the integral at the desired limit point.

## Why is integration useful for finding limits?

Integration allows us to find precise and accurate values for limits, even for functions that are not continuous or differentiable. It also provides a more efficient method for finding limits compared to using algebraic manipulation or numerical methods.

## What are the steps for using integration to find limits?

The steps for using integration to find limits involve: 1) determining the function and the desired limit point, 2) setting up the integral by finding the antiderivative of the function, 3) evaluating the integral at the limit point, and 4) interpreting the result as the limit.

## Can integration be used to find limits for all types of functions?

Yes, integration can be used to find limits for all types of functions, including polynomial, rational, exponential, trigonometric, and logarithmic functions. However, the approach and techniques used may vary depending on the type of function.

## What are some common challenges when using integration to find limits?

Some common challenges when using integration to find limits include identifying the correct function, determining the appropriate limits of integration, and dealing with complex or undefined functions. It is important to carefully set up the integral and properly evaluate it to avoid errors in the final result.

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