# Find the Average Value of the Function g(t)=t/( square root of 5+t^2) , [2,5] ,
To find the average value of a function, the function should be continuous on the closed interval . To find whether is continuous on or not, find the domain of .
Set the radicand in greater than or equal to to find where the expression is defined.
Solve for .
Subtract from both sides of the inequality.
Since the left side has an even power, it is always positive for all real numbers.
All real numbers
All real numbers
Set the denominator in equal to to find where the expression is undefined.
Solve for .
To remove the radical on the left side of the equation, square both sides of the equation.
Simplify each side of the equation.
Multiply the exponents in .
Apply the power rule and multiply exponents, .
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Simplify.
Raising to any positive power yields .
Solve for .
Subtract from both sides of the equation.
Take the square root of both sides of the equation to eliminate the exponent on the left side.
The complete solution is the result of both the positive and negative portions of the solution.
Simplify the right side of the equation.
Rewrite as .
Rewrite as .
Rewrite as .
The complete solution is the result of both the positive and negative portions of the solution.
First, use the positive value of the to find the first solution.
Next, use the negative value of the to find the second solution.
The complete solution is the result of both the positive and negative portions of the solution.
The domain is all real numbers.
Interval Notation:
Set-Builder Notation:
Interval Notation:
Set-Builder Notation:
is continuous on .
is continuous
The average value of function over the interval is defined as .
Substitute the actual values into the formula for the average value of a function.
Let . Then , so . Rewrite using and .
Let . Find .
Differentiate .
By the Sum Rule, the derivative of with respect to is .
Since is constant with respect to , the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
Substitute the lower limit in for in .
Simplify.
Raise to the power of .
Substitute the upper limit in for in .
Simplify.
Raise to the power of .
The values found for and will be used to evaluate the definite integral.
Rewrite the problem using , , and the new limits of integration.
Simplify.
Multiply and .
Move to the left of .
Since is constant with respect to , move out of the integral.
Apply basic rules of exponents.
Use to rewrite as .
Move out of the denominator by raising it to the power.
Multiply the exponents in .
Apply the power rule and multiply exponents, .
Combine and .
Move the negative in front of the fraction.
By the Power Rule, the integral of with respect to is .
Substitute and simplify.
Evaluate at and at .
Simplify.
Rewrite as .
Apply the power rule and multiply exponents, .
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Evaluate the exponent.
Multiply by .
Evaluate.
Apply the distributive property.
Cancel the common factor of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
Cancel the common factor of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
Subtract from .
Apply the distributive property.
Combine and .
Cancel the common factor of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
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