Calculus Examples

Find the Average Value of the Derivative y=25-x^2 , [-5,5]
y=25-x2y=25x2 , [-5,5][5,5]
Step 1
Write y=25-x2y=25x2 as a function.
f(x)=25-x2f(x)=25x2
Step 2
Find the derivative of f(x)=25-x2f(x)=25x2.
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Step 2.1
Find the first derivative.
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Step 2.1.1
Differentiate.
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Step 2.1.1.1
By the Sum Rule, the derivative of 25-x225x2 with respect to xx is ddx[25]+ddx[-x2]ddx[25]+ddx[x2].
ddx[25]+ddx[-x2]ddx[25]+ddx[x2]
Step 2.1.1.2
Since 2525 is constant with respect to xx, the derivative of 2525 with respect to xx is 00.
0+ddx[-x2]0+ddx[x2]
0+ddx[-x2]0+ddx[x2]
Step 2.1.2
Evaluate ddx[-x2]ddx[x2].
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Step 2.1.2.1
Since -11 is constant with respect to xx, the derivative of -x2x2 with respect to xx is -ddx[x2]ddx[x2].
0-ddx[x2]0ddx[x2]
Step 2.1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=2n=2.
0-(2x)0(2x)
Step 2.1.2.3
Multiply 22 by -11.
0-2x02x
0-2x02x
Step 2.1.3
Subtract 2x2x from 00.
f(x)=-2x
f(x)=-2x
Step 2.2
The first derivative of f(x) with respect to x is -2x.
-2x
-2x
Step 3
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Interval Notation:
(-,)
Set-Builder Notation:
{x|x}
Step 4
f(x) is continuous on [-5,5].
f(x) is continuous
Step 5
The average value of function f over the interval [a,b] is defined as A(x)=1b-abaf(x)dx.
A(x)=1b-abaf(x)dx
Step 6
Substitute the actual values into the formula for the average value of a function.
A(x)=15+5(5-5-2xdx)
Step 7
Since -2 is constant with respect to x, move -2 out of the integral.
A(x)=15+5(-25-5xdx)
Step 8
By the Power Rule, the integral of x with respect to x is 12x2.
A(x)=15+5(-2(12x2]5-5))
Step 9
Simplify the answer.
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Step 9.1
Combine 12 and x2.
A(x)=15+5(-2(x22]5-5))
Step 9.2
Substitute and simplify.
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Step 9.2.1
Evaluate x22 at 5 and at -5.
A(x)=15+5(-2((522)-(-5)22))
Step 9.2.2
Simplify.
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Step 9.2.2.1
Raise 5 to the power of 2.
A(x)=15+5(-2(252-(-5)22))
Step 9.2.2.2
Raise -5 to the power of 2.
A(x)=15+5(-2(252-252))
Step 9.2.2.3
Combine the numerators over the common denominator.
A(x)=15+5(-225-252)
Step 9.2.2.4
Subtract 25 from 25.
A(x)=15+5(-2(02))
Step 9.2.2.5
Cancel the common factor of 0 and 2.
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Step 9.2.2.5.1
Factor 2 out of 0.
A(x)=15+5(-22(0)2)
Step 9.2.2.5.2
Cancel the common factors.
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Step 9.2.2.5.2.1
Factor 2 out of 2.
A(x)=15+5(-22021)
Step 9.2.2.5.2.2
Cancel the common factor.
A(x)=15+5(-22021)
Step 9.2.2.5.2.3
Rewrite the expression.
A(x)=15+5(-2(01))
Step 9.2.2.5.2.4
Divide 0 by 1.
A(x)=15+5(-20)
A(x)=15+5(-20)
A(x)=15+5(-20)
Step 9.2.2.6
Multiply -2 by 0.
A(x)=15+5(0)
A(x)=15+5(0)
A(x)=15+5(0)
A(x)=15+5(0)
Step 10
Add 5 and 5.
A(x)=1100
Step 11
Multiply 110 by 0.
A(x)=0
Step 12
 [x2  12  π  xdx ]