- #1

PsychonautQQ

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how do i evaluate say a definite integral from [-3,3] of |x+1|? Any advice? I'm so confused.

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- Thread starter PsychonautQQ
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In summary, you shouldevaluate the integral from -3 to 3 of |x+1| by finding the area of two triangles and adding the two integrals together.

- #1

PsychonautQQ

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how do i evaluate say a definite integral from [-3,3] of |x+1|? Any advice? I'm so confused.

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- #2

Mark44

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Do this by getting rid of the absolute value. How is |x + 1| defined? If you're not sure of this, draw a graph of y = |x + 1|.PsychonautQQ said:how do i evaluate say a definite integral from [-3,3] of |x+1|? Any advice? I'm so confused.

- #3

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Split up the integral in two parts.

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HallsofIvy

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- #5

PsychonautQQ

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- #6

fishspawned

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Graph the function. Notice where it changes from one function to the other. Notice you are looking at two triangles

Now - consider what the functions are over each triangle. See if you cannot make an integral made up of two parts.

Remember that when you drop the abs value you are left with TWO cases.

- #7

PsychonautQQ

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Is this correct? integral(x)dx from 0 to 4 + integral(x)dx from 0 to 3?

- #8

fishspawned

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could you post your graph? I am not sure where you are deriving this equation.

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HallsofIvy

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No, that's not at all correct and I don't see how you could have got that from anything said here! Have you drawn the graph of this function as suggested? Have you at least determined where x+ 1= 0? Do you see why that is important? Do you see the two triangles I mentioned?PsychonautQQ said:Is this correct? integral(x)dx from 0 to 4 + integral(x)dx from 0 to 3?

- #10

PsychonautQQ

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So basically I'm saying the integral from -1 to -3 of x+1 is going to equal the integral of 0 to 2 of x, and the integral from -1 to 3 of x+1 is going to equal the integral of 0 to 4 of x.

- #11

HallsofIvy

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The way you say that is a little confusing! First you want to integratePsychonautQQ said:Okay so i drew the graph of |x+1. It looks like a big V that touches the x-axis at -1. To me and my nooby brain, it makes sense that the area would be equal to the integral of x(dx) from 0 to 2 plus the integral of x(dx) from 0 to 4.

This is because i am finding the total area under the graph from -3 to 3, the slope of the graph is 1, and it touches the x-axis at x = -1. Therefore, the total area under this V i have drawn will be equal to area of the graph of just x from 0 to 4 plus a graph of x from 0 to 2

So basically I'm saying the integral from -1 to -3 of x+1 is going to equal the integral of 0 to 2 of x, and the integral from -1 to 3 of x+1 is going to equal the integral of 0 to 4 of x.

I will point out again that this can be written as the sum of the areas of two triangles. Perhaps you can use that to check your result.

The integral of absolute value function is a mathematical concept that represents the area under the curve of an absolute value function. It is denoted by ∫|f(x)|dx and is calculated by finding the area between the curve and the x-axis.

The integral of absolute value function is different from other integrals because it takes into account the negative values of the function by considering the absolute value. This means that the integral of absolute value function is always positive or zero, unlike other integrals which can be negative.

The integral of absolute value function has various applications in mathematics, such as finding the distance traveled by an object, calculating the total displacement of a moving object, and determining the total change in a quantity over a given time interval.

The integral of absolute value function is commonly used in physics, engineering, and economics to solve problems related to distance, velocity, acceleration, and area under a curve. It is also used in statistics to calculate the mean absolute deviation of a data set.

Yes, the integral of absolute value function can be evaluated using traditional integration techniques such as the substitution method, integration by parts, and partial fraction decomposition. However, it may require a more complex approach depending on the specific function being integrated.

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