Evaluate the definite integral:

In summary, the homework statement is to evaluate the definite integral ∫(x101-√(9-x^2))dx from -3 to 3. Homework equations state that this problem can be done without anti-differentiation, but hints are needed on how to do it. Integrating x101 and sqrt(9-x^2) separately yields the same answer as finding the area under the curve for sqrt(9-x^2), confirming that the two functions have the same magnitude, but opposite sign.
  • #1
1question
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Homework Statement



Evaluate the definite integral ∫(x101-√(9-x^2))dx from -3 to 3.
Hint: This problem can be done without anti-differentiation.

Homework Equations


The Attempt at a Solution



I am stuck. I tried to do it with with anti-differentiation and it didn't work/very computation-intensive.
Hints on how to do it without anti-differentiation, please?
 
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  • #2
1question said:

Homework Statement



Evaluate the definite integral ∫(x101-√(9-x^2))dx from -3 to 3.
Hint: This problem can be done without anti-differentiation.


Homework Equations





The Attempt at a Solution



I am stuck. I tried to do it with with anti-differentiation and it didn't work/very computation-intensive.
Hints on how to do it without anti-differentiation, please?

Think about what the graphs of x^101 and sqrt(9-x^2) look like.
 
  • #3
1question said:
I am stuck. I tried to do it with with anti-differentiation and it didn't work/very computation-intensive.

Actually, using indefinite integrals and then applying limits of integration is a pretty good way of solving it. Integral of x101 is easy to calculate and it gets simple to integrate the other term when you substitute x = 3sinθ.
 
  • #4
Dick said:
Think about what the graphs of x^101 and sqrt(9-x^2) look like.

Well, I could use the fact that the - and + parts have identical areas (and thus find 1 and multiply by 2 to get the total area?)
 
  • #5
Sunil Simha said:
Actually, using indefinite integrals and then applying limits of integration is a pretty good way of solving it. Integral of x101 is easy to calculate and it gets simple to integrate the other term when you substitute x = 3sinθ.

No, it's not hard to do the integration. But the point here is to find an even easier shortcut.
 
  • #6
1question said:
Well, I could use the fact that the - and + parts have identical areas (and thus find 1 and multiply by 2 to get the total area?)

Treat the two function x^101 and sqrt(9-x^2) differently. Start with x^101. What do you say about the relation between the + part and the - part in this case? Are they really the same? For sqrt(9-x^2) sketch a graph. It might be a common geometric shape that you know the area of.
 
  • #7
Dick said:
Treat the two function x^101 and sqrt(9-x^2) differently. Start with x^101. What do you say about the relation between the + part and the - part in this case? Are they really the same? For sqrt(9-x^2) sketch a graph. It might be a common geometric shape that you know the area of.

They have the same magnitude, but opposite sign.

Well, I know that sqrt(9-x^2) is a semi-circle at x=+/- 3 and height = 3.
 
  • #8
1question said:
They have the same magnitude, but opposite sign.

Well, I know that sqrt(9-x^2) is a semi-circle at x=+/- 3 and height = 3.

Good! You've basically got it. If you add "same magnitude, but opposite sign" what do you get? And the integral is the same as the area under the curve, right? So if you know a formula for the area of a semicircle, you know the integral.
 
  • #9
Dick said:
Good! You've basically got it. If you add "same magnitude, but opposite sign" what do you get? And the integral is the same as the area under the curve, right? So if you know a formula for the area of a semicircle, you know the integral.

0?

Oh...

I got it (confirmed with Wolfram) - first term = 0, and the second = πr^2/2 (b/c semi-circle), and so the answer is:

= ...
= (0-πr^2/2)
= 0-π(3)^2/2
= -9π/2

Thank you!
 

FAQ: Evaluate the definite integral:

What is a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve. It is represented by the symbol ∫ and has two limits, a lower and an upper bound, which determine the range of integration.

How do you evaluate a definite integral?

To evaluate a definite integral, you must first identify the function and the limits of integration. Then, you can use various techniques such as substitution, integration by parts, or the fundamental theorem of calculus to find the area under the curve.

What is the purpose of evaluating a definite integral?

The purpose of evaluating a definite integral is to find the area under a curve, which can represent physical quantities such as displacement, velocity, or acceleration in real-world applications. It can also be used to solve optimization problems and calculate volumes and areas of irregular shapes.

What are the common methods used to evaluate a definite integral?

Some of the common methods used to evaluate a definite integral include the substitution method, integration by parts, trigonometric substitution, and the use of special formulas such as the power rule or the logarithmic rule. The choice of method depends on the complexity of the integral and the functions involved.

Are there any limitations to evaluating a definite integral?

Yes, there are limitations to evaluating a definite integral. Some integrals may not have closed-form solutions and require numerical methods to approximate the value. Additionally, improper integrals, which have infinite limits or infinite discontinuities, may not be possible to evaluate using traditional methods.

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