Calculus Examples
y=x-2y=x−2 , (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-a⋅∫baf(x)2dxfrms=√1b−a⋅∫baf(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=
⎷17−2⋅(∫72(x−2)2dx)
Step 3
Step 3.1
Let u=x-2u=x−2. Then du=dxdu=dx. Rewrite using uu and dduu.
Step 3.1.1
Let u=x-2u=x−2. Find dudxdudx.
Step 3.1.1.1
Differentiate x-2x−2.
ddx[x-2]ddx[x−2]
Step 3.1.1.2
By the Sum Rule, the derivative of x-2x−2 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-1nxn−1 where n=1n=1.
1+ddx[-2]1+ddx[−2]
Step 3.1.1.4
Since -2−2 is constant with respect to xx, the derivative of -2−2 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=x−2.
ulower=2-2ulower=2−2
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=x−2.
uupper=7-2uupper=7−2
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.
∫50u2du∫50u2du
∫50u2du∫50u2du
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.
Step 3.3.1
Evaluate 13u313u3 at 55 and at 00.
(13⋅53)-13⋅03(13⋅53)−13⋅03
Step 3.3.2
Simplify.
Step 3.3.2.1
Raise 55 to the power of 33.
13⋅125-13⋅0313⋅125−13⋅03
Step 3.3.2.2
Combine 1313 and 125125.
1253-13⋅031253−13⋅03
Step 3.3.2.3
Raising 00 to any positive power yields 00.
1253-13⋅01253−13⋅0
Step 3.3.2.4
Multiply 00 by -1−1.
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
Step 4.1
Multiply 17-217−2 by 12531253.
frms=√125(7-2)⋅3frms=√125(7−2)⋅3
Step 4.2
Subtract 22 from 77.
frms=√1255⋅3frms=√1255⋅3
Step 4.3
Reduce the expression 1255⋅31255⋅3 by cancelling the common factors.
Step 4.3.1
Factor 55 out of 125125.
frms=√5⋅255⋅3frms=√5⋅255⋅3
Step 4.3.2
Factor 55 out of 5⋅35⋅3.
frms=√5⋅255(3)frms=√5⋅255(3)
Step 4.3.3
Cancel the common factor.
frms=√5⋅255⋅3
Step 4.3.4
Rewrite the expression.
frms=√253
frms=√253
Step 4.4
Rewrite √253 as √25√3.
frms=√25√3
Step 4.5
Simplify the numerator.
Step 4.5.1
Rewrite 25 as 52.
frms=√52√3
Step 4.5.2
Pull terms out from under the radical, assuming positive real numbers.
frms=5√3
frms=5√3
Step 4.6
Multiply 5√3 by √3√3.
frms=5√3⋅√3√3
Step 4.7
Combine and simplify the denominator.
Step 4.7.1
Multiply 5√3 by √3√3.
frms=5√3√3√3
Step 4.7.2
Raise √3 to the power of 1.
frms=5√3√3√3
Step 4.7.3
Raise √3 to the power of 1.
frms=5√3√3√3
Step 4.7.4
Use the power rule aman=am+n to combine exponents.
frms=5√3√31+1
Step 4.7.5
Add 1 and 1.
frms=5√3√32
Step 4.7.6
Rewrite √32 as 3.
Step 4.7.6.1
Use n√ax=axn to rewrite √3 as 312.
frms=5√3(312)2
Step 4.7.6.2
Apply the power rule and multiply exponents, (am)n=amn.
frms=5√3312⋅2
Step 4.7.6.3
Combine 12 and 2.
frms=5√3322
Step 4.7.6.4
Cancel the common factor of 2.
Step 4.7.6.4.1
Cancel the common factor.
frms=5√3322
Step 4.7.6.4.2
Rewrite the expression.
frms=5√33
frms=5√33
Step 4.7.6.5
Evaluate the exponent.
frms=5√33
frms=5√33
frms=5√33
frms=5√33
Step 5
The result can be shown in multiple forms.
Exact Form:
frms=5√33
Decimal Form:
frms=2.88675134…
Step 6