Definite Integral with Absolute Value.

In summary: The correct values are 33.342, 15.231, and 28.957, respectively. Therefore, in summary, there are errors in the values calculated for the integrals, resulting in the final answer being incorrect.
  • #1
m0gh
27
0
The problem is ∫x^2 - 3x - 5 with the lower limit being -4 and the upper limit 7.

I broke the integrals into three parts from [-4, -1.1926], [-1.1926, 4.1926], [4.1926, 7]

I did the integral and got (x^3)/3 - (3/2)x^2 - 5x

I subbed in the lower and upper limits and got 32.861 for [-4, -1.1926], 15.231 for [-1.1926, 4.1926], and finally 28.957 for [4.1926, 7].

I don't necessarily need a step by step solution ( though it would be greatly appreciated). I would really just like to know if you can spot which/where I am getting the wrong value(s).

EDIT: The final answer I keep getting is 76.691. The answer I get when I use a definite integral calculator is 83.2233
 
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  • #2
m0gh said:
The problem is ∫x^2 - 3x - 5 with the lower limit being -4 and the upper limit 7.
Is this your integral?
$$ \int_{-4}^7 |x^2 - 3x - 5|dx$$
m0gh said:
I broke the integrals into three parts from [-4, -1.1926], [-1.1926, 4.1926], [4.1926, 7]

I did the integral and got (x^3)/3 - (3/2)x^2 - 5x
The above is incorrect. You are ignoring the fact that there's an absolute value involved. The key idea is that |a| = a if a ≥ 0, but |a| = -a if a < 0.

Also, the approximate numbers you use aren't exact, so whatever answer you get will be off some.
m0gh said:
I subbed in the lower and upper limits and got 32.861 for [-4, -1.1926], 15.231 for [-1.1926, 4.1926], and finally 28.957 for [4.1926, 7].

I don't necessarily need a step by step solution ( though it would be greatly appreciated). I would really just like to know if you can spot which/where I am getting the wrong value(s).
It is against the rules in this forum to provide a complete answer, so a step-by-step solution isn't going to happen.
m0gh said:
EDIT: The final answer I keep getting is 76.691. The answer I get when I use a definite integral calculator is 83.2233
 
  • #3
The part you are saying is incorrect was set up by my professor. She put a negative sign in front of the integral for [-1.1926, 4.1926]
 
  • #4
m0gh said:
The part you are saying is incorrect was set up by my professor. She put a negative sign in front of the integral for [-1.1926, 4.1926]
Which you didn't mention.

Anyway, I get 83.2233 as well, so all I can say is that you have an error in one or more of your integrals. Also, as I mentioned already, you should be using the exact numbers for the limits of integration, rather than the decimal approximations. That is, you should be using ##3/2 - \sqrt{29}/2## and ##3/2 + \sqrt{29}/2##, although I suspect that the difference you're getting is caused by an error somewhere else.
 
  • #5
m0gh said:
The problem is ∫x^2 - 3x - 5 with the lower limit being -4 and the upper limit 7.

I broke the integrals into three parts from [-4, -1.1926], [-1.1926, 4.1926], [4.1926, 7]

I did the integral and got (x^3)/3 - (3/2)x^2 - 5x

I subbed in the lower and upper limits and got 32.861 for [-4, -1.1926], 15.231 for [-1.1926, 4.1926], and finally 28.957 for [4.1926, 7].

I don't necessarily need a step by step solution ( though it would be greatly appreciated). I would really just like to know if you can spot which/where I am getting the wrong value(s).

EDIT: The final answer I keep getting is 76.691. The answer I get when I use a definite integral calculator is 83.2233

The two in red are incorrect.
 

Related to Definite Integral with Absolute Value.

1. What is a definite integral with absolute value?

A definite integral with absolute value is a type of mathematical function that calculates the area under a curve, while also taking into account negative values. The absolute value ensures that the area is always positive, regardless of the direction of the curve.

2. How is a definite integral with absolute value different from a regular definite integral?

A regular definite integral calculates the area under a curve without taking into account negative values. This means that if the curve dips below the x-axis, the area will be subtracted from the total. In contrast, a definite integral with absolute value will always calculate the total area as positive, regardless of the direction of the curve.

3. What is the purpose of using a definite integral with absolute value?

The purpose of using a definite integral with absolute value is to accurately calculate the total area under a curve, even when the curve has negative values. This is particularly useful in real-world applications, where negative values may represent important data points.

4. How is a definite integral with absolute value calculated?

To calculate a definite integral with absolute value, the absolute value is first applied to the function being integrated. The resulting function is then integrated using standard integration techniques, such as the Fundamental Theorem of Calculus or integration by substitution.

5. What are some practical applications of definite integrals with absolute value?

Definite integrals with absolute value have many practical applications in fields such as physics, engineering, and economics. For example, they can be used to calculate the displacement of an object, the total work done by a force, or the area under a demand curve in economics.

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