To use the Mean Value Theorem for Integrals (MVTI) in solve problems, follow these simple steps:
The Mean Value Theorem for Integrals tells us that if a function ( f ) is continuous on the closed interval ([a, b]), there is at least one point ( c ) in the interval ( (a, b) ) where:
[ f(c) = \frac{1}{b - a} \int_a^b f(x) , dx ]
This means the function has an average value over the interval, and at least at one point in that interval, it matches that average.
Before using this theorem, make sure the function meets some key conditions:
Continuity: Check that ( f ) is continuous on ([a, b]). If ( f ) is a polynomial or a trigonometric function, then it is continuous over any interval.
Closed Interval: Be sure you are working with a closed interval ([a, b]).
Now, calculate the definite integral of the function from ( a ) to ( b ). You might need to:
For example, for the function ( f(x) = x^2 ) over the interval ([1, 3]), we find the integral:
[ \int_1^3 x^2 , dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} ]
After calculating the integral, find the average value of the function over the interval:
[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) , dx ]
Using our previous example:
[ \text{Average value} = \frac{1}{3 - 1} \cdot \frac{26}{3} = \frac{26}{6} = \frac{13}{3} ]
Now that you have the average value, you need to find a specific value of ( c ) in ( (a, b) ) where ( f(c) ) equals the average value.
Set up the equation:
[ f(c) = \text{Average value} ]
In our example with ( f(x) = x^2 ), we set it equal to:
[ c^2 = \frac{13}{3} ]
Taking the square root gives:
[ c = \sqrt{\frac{13}{3}} \approx 2.08 ]
Finally, check if the value of ( c ) is within the interval ( (a, b) ). Here, since ( 2.08 ) is between ( 1 ) and ( 3 ), the theorem works!
Here’s a quick recap of the steps to apply the Mean Value Theorem for Integrals:
By following these steps, you can successfully apply MVTI to different calculus problems involving the area under curves.
To use the Mean Value Theorem for Integrals (MVTI) in solve problems, follow these simple steps:
The Mean Value Theorem for Integrals tells us that if a function ( f ) is continuous on the closed interval ([a, b]), there is at least one point ( c ) in the interval ( (a, b) ) where:
[ f(c) = \frac{1}{b - a} \int_a^b f(x) , dx ]
This means the function has an average value over the interval, and at least at one point in that interval, it matches that average.
Before using this theorem, make sure the function meets some key conditions:
Continuity: Check that ( f ) is continuous on ([a, b]). If ( f ) is a polynomial or a trigonometric function, then it is continuous over any interval.
Closed Interval: Be sure you are working with a closed interval ([a, b]).
Now, calculate the definite integral of the function from ( a ) to ( b ). You might need to:
For example, for the function ( f(x) = x^2 ) over the interval ([1, 3]), we find the integral:
[ \int_1^3 x^2 , dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} ]
After calculating the integral, find the average value of the function over the interval:
[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) , dx ]
Using our previous example:
[ \text{Average value} = \frac{1}{3 - 1} \cdot \frac{26}{3} = \frac{26}{6} = \frac{13}{3} ]
Now that you have the average value, you need to find a specific value of ( c ) in ( (a, b) ) where ( f(c) ) equals the average value.
Set up the equation:
[ f(c) = \text{Average value} ]
In our example with ( f(x) = x^2 ), we set it equal to:
[ c^2 = \frac{13}{3} ]
Taking the square root gives:
[ c = \sqrt{\frac{13}{3}} \approx 2.08 ]
Finally, check if the value of ( c ) is within the interval ( (a, b) ). Here, since ( 2.08 ) is between ( 1 ) and ( 3 ), the theorem works!
Here’s a quick recap of the steps to apply the Mean Value Theorem for Integrals:
By following these steps, you can successfully apply MVTI to different calculus problems involving the area under curves.