Calculus Examples

y=x-2y=x2 , (2,7)(2,7)
Step 1
The Root Mean Square (RMS) of a function ff over a specified interval [a,b][a,b] is the square root of the arithmetic mean (average) of the squares of the original values.
frms=1b-abaf(x)2dxfrms=1babaf(x)2dx
Step 2
Substitute the actual values into the formula for the root mean square of a function.
frms=17-2(72(x-2)2dx)frms= 172(72(x2)2dx)
Step 3
Evaluate the integral.
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Step 3.1
Let u=x-2u=x2. Then du=dxdu=dx. Rewrite using uu and dduu.
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Step 3.1.1
Let u=x-2u=x2. Find dudxdudx.
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Step 3.1.1.1
Differentiate x-2x2.
ddx[x-2]ddx[x2]
Step 3.1.1.2
By the Sum Rule, the derivative of x-2x2 with respect to xx is ddx[x]+ddx[-2]ddx[x]+ddx[2].
ddx[x]+ddx[-2]ddx[x]+ddx[2]
Step 3.1.1.3
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn1 where n=1n=1.
1+ddx[-2]1+ddx[2]
Step 3.1.1.4
Since -22 is constant with respect to xx, the derivative of -22 with respect to xx is 00.
1+01+0
Step 3.1.1.5
Add 11 and 00.
11
11
Step 3.1.2
Substitute the lower limit in for xx in u=x-2u=x2.
ulower=2-2ulower=22
Step 3.1.3
Subtract 22 from 22.
ulower=0ulower=0
Step 3.1.4
Substitute the upper limit in for xx in u=x-2u=x2.
uupper=7-2uupper=72
Step 3.1.5
Subtract 22 from 77.
uupper=5uupper=5
Step 3.1.6
The values found for ulowerulower and uupperuupper will be used to evaluate the definite integral.
ulower=0ulower=0
uupper=5uupper=5
Step 3.1.7
Rewrite the problem using uu, dudu, and the new limits of integration.
50u2du50u2du
50u2du50u2du
Step 3.2
By the Power Rule, the integral of u2u2 with respect to uu is 13u313u3.
13u3]5013u3]50
Step 3.3
Substitute and simplify.
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Step 3.3.1
Evaluate 13u313u3 at 55 and at 00.
(1353)-1303(1353)1303
Step 3.3.2
Simplify.
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Step 3.3.2.1
Raise 55 to the power of 33.
13125-1303131251303
Step 3.3.2.2
Combine 1313 and 125125.
1253-130312531303
Step 3.3.2.3
Raising 00 to any positive power yields 00.
1253-1301253130
Step 3.3.2.4
Multiply 00 by -11.
1253+0(13)1253+0(13)
Step 3.3.2.5
Multiply 00 by 1313.
1253+01253+0
Step 3.3.2.6
Add 12531253 and 00.
12531253
12531253
12531253
12531253
Step 4
Simplify the root mean square formula.
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Step 4.1
Multiply 17-2172 by 12531253.
frms=125(7-2)3frms=125(72)3
Step 4.2
Subtract 22 from 77.
frms=12553frms=12553
Step 4.3
Reduce the expression 1255312553 by cancelling the common factors.
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Step 4.3.1
Factor 55 out of 125125.
frms=52553frms=52553
Step 4.3.2
Factor 55 out of 5353.
frms=5255(3)frms=5255(3)
Step 4.3.3
Cancel the common factor.
frms=52553
Step 4.3.4
Rewrite the expression.
frms=253
frms=253
Step 4.4
Rewrite 253 as 253.
frms=253
Step 4.5
Simplify the numerator.
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Step 4.5.1
Rewrite 25 as 52.
frms=523
Step 4.5.2
Pull terms out from under the radical, assuming positive real numbers.
frms=53
frms=53
Step 4.6
Multiply 53 by 33.
frms=5333
Step 4.7
Combine and simplify the denominator.
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Step 4.7.1
Multiply 53 by 33.
frms=5333
Step 4.7.2
Raise 3 to the power of 1.
frms=5333
Step 4.7.3
Raise 3 to the power of 1.
frms=5333
Step 4.7.4
Use the power rule aman=am+n to combine exponents.
frms=5331+1
Step 4.7.5
Add 1 and 1.
frms=5332
Step 4.7.6
Rewrite 32 as 3.
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Step 4.7.6.1
Use nax=axn to rewrite 3 as 312.
frms=53(312)2
Step 4.7.6.2
Apply the power rule and multiply exponents, (am)n=amn.
frms=533122
Step 4.7.6.3
Combine 12 and 2.
frms=53322
Step 4.7.6.4
Cancel the common factor of 2.
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Step 4.7.6.4.1
Cancel the common factor.
frms=53322
Step 4.7.6.4.2
Rewrite the expression.
frms=533
frms=533
Step 4.7.6.5
Evaluate the exponent.
frms=533
frms=533
frms=533
frms=533
Step 5
The result can be shown in multiple forms.
Exact Form:
frms=533
Decimal Form:
frms=2.88675134
Step 6
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