Problems computing area with integrals.

In summary, the conversation discusses finding the total area enclosed by two given functions using integrals. The speaker initially sets up the integral with the lower limit at 0 and the upper limit at 5, but is told that the functions actually intersect at three points instead of two. After following some advice and finding the third point of intersection, the speaker is able to solve the problem and get the correct answer of 10.416.
  • #1
haydn
27
0

Homework Statement



Find the total area enclosed by the graphs of

y=9x^2–x^3+x
y=x^2+16x


Homework Equations



No real equations, just using integrals

The Attempt at a Solution



I graph the functions and find they intersect at 0 and 5, and that y=x^2+16x seems to be the upper function while y=9x^2-x^3+x is the lower function.

I set up an integral with the lower limit 0 and the upper limit 5, and x^2+16x - (9x^2-x^3+x) dx inside the integral.

I solve by simplifying and using the fundamental theorem of calculus and get 10.416. The homework website I'm using is telling me this is wrong...

Thank you.
 
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  • #2
They intersect at three points, not two. One is a cubic equation. Can you find them?
 
Last edited:
  • #3
I've checked it which sage and I get 10.416... I say forget what the website says.
 
  • #4
Nevermind. Follow Dick's advice.
 
  • #5
Dick said:
They intersect at three points, not two. One is a cubic equation. Can you find them?

Found it and got the correct answer. Thanks a bunch!
 

1. What is the basic concept behind computing area with integrals?

The basic concept behind computing area with integrals is using the fundamental theorem of calculus, which states that the area under a curve can be calculated as the integral of that curve's function. This means that the area can be found by evaluating the definite integral of the function over a specific interval.

2. How do you set up the integral to compute the area under a curve?

To set up the integral, you must first identify the function that represents the curve and determine the interval over which you want to calculate the area. Then, you must take the integral of the function over that interval, using the limits of integration as the endpoints. This will give you the area under the curve.

3. Can integrals be used to compute the area of irregular shapes?

Yes, integrals can be used to compute the area of irregular shapes. This is because the integral is able to handle curves and changing rates of change, making it a powerful tool for calculating areas that may not have a simple geometric formula.

4. Are there any limitations to using integrals to compute area?

One limitation of using integrals to compute area is that it can only be used for continuous functions. If the function has any discontinuities or sharp edges, the integral may not accurately capture the area under the curve. Additionally, integrals can be difficult to solve for complex functions, which may make it challenging to compute the area.

5. How can computing area with integrals be applied in real-world situations?

Integrals are commonly used in physics and engineering to calculate the area under curves that represent velocity, acceleration, and other physical quantities. They can also be used in economics to find the area under demand or supply curves. In general, integrals are a powerful tool for solving real-world problems that involve finding the area under a curve.

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