Does anyone know why the answer is not 45, and instead its 175/3? (Integrating to find the area under a curve)

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In summary, the speaker integrated y using the upper limit of 5 and lower limit of 2 in order to get the result of 45/3, but the correct answer should have been 175/3. This discrepancy may be due to a typo or calculation error by the person who set the question. Additionally, using the notation (1/3)x^3 instead of 1/3x^3 can help avoid confusion.
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i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
 
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homeworkhelpls said:
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i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
Both Wolfram Alpha and I agree with you.
 
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Me too!
 
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The integral would evaluate to 175/3 if the lower limit were x=-2. I suspect a silly typo or calculation slip by the question setter.
 
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homeworkhelpls said:
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i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
Just a notation tip. 1/3x^3 can be misread as 1/(3x^3) placing the x^3 in the denominator. To be precise, we can instead write (1/3)x^3 to ensure x^3 is in the numerator and not mistakenly placed in the denominator.
 

1. Why is the answer not 45?

The answer is not 45 because when integrating to find the area under a curve, you are essentially finding the sum of infinitely small rectangles under the curve. Therefore, the answer will be a decimal or fraction, rather than a whole number like 45.

2. How do you integrate to find the area under a curve?

To integrate and find the area under a curve, you must first find the antiderivative of the function. Then, you can use the fundamental theorem of calculus to evaluate the integral at the given limits of integration.

3. Why is the answer in the form of 175/3?

The answer is in the form of 175/3 because it is the exact value of the integral. In some cases, the integral may not have a simplified decimal or whole number answer, so it is left in fraction form.

4. Can you explain the concept of integration in simpler terms?

Integration is essentially finding the area under a curve, which can be thought of as the sum of infinitely small rectangles. It is used to solve problems involving rates of change and accumulation.

5. What are some real-world applications of integrating to find the area under a curve?

Integrating to find the area under a curve has many real-world applications, such as calculating the distance traveled by an object given its velocity function, finding the total revenue of a business over a certain time period, and determining the amount of medication in a patient's bloodstream over time.

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