Enter a problem...
Calculus Examples
f(x)=x2+2xf(x)=x2+2x , 0<x<60<x<6
Step 1
Step 1.1
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∈ℝ}{x|x∈R}
Step 1.2
f(x)f(x) is continuous on [0,6][0,6].
The function is continuous.
The function is continuous.
Step 2
Step 2.1
Find the derivative.
Step 2.1.1
Find the first derivative.
Step 2.1.1.1
Differentiate.
Step 2.1.1.1.1
By the Sum Rule, the derivative of x2+2xx2+2x with respect to xx is ddx[x2]+ddx[2x]ddx[x2]+ddx[2x].
ddx[x2]+ddx[2x]ddx[x2]+ddx[2x]
Step 2.1.1.1.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=2n=2.
2x+ddx[2x]2x+ddx[2x]
2x+ddx[2x]2x+ddx[2x]
Step 2.1.1.2
Evaluate ddx[2x]ddx[2x].
Step 2.1.1.2.1
Since 22 is constant with respect to xx, the derivative of 2x2x with respect to xx is 2ddx[x]2ddx[x].
2x+2ddx[x]2x+2ddx[x]
Step 2.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
2x+2⋅12x+2⋅1
Step 2.1.1.2.3
Multiply 22 by 11.
f′(x)=2x+2f'(x)=2x+2
f′(x)=2x+2f'(x)=2x+2
f′(x)=2x+2f'(x)=2x+2
Step 2.1.2
The first derivative of f(x)f(x) with respect to xx is 2x+22x+2.
2x+22x+2
2x+22x+2
Step 2.2
Find if the derivative is continuous on [0,6][0,6].
Step 2.2.1
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∈ℝ}{x|x∈R}
Step 2.2.2
f′(x)f'(x) is continuous on [0,6][0,6].
The function is continuous.
The function is continuous.
Step 2.3
The function is differentiable on [0,6][0,6] because the derivative is continuous on [0,6][0,6].
The function is differentiable.
The function is differentiable.
Step 3
For arc length to be guaranteed, the function and its derivative must both be continuous on the closed interval [0,6][0,6].
The function and its derivative are continuous on the closed interval [0,6][0,6].
Step 4
Step 4.1
Differentiate.
Step 4.1.1
By the Sum Rule, the derivative of x2+2xx2+2x with respect to xx is ddx[x2]+ddx[2x]ddx[x2]+ddx[2x].
ddx[x2]+ddx[2x]ddx[x2]+ddx[2x]
Step 4.1.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=2n=2.
2x+ddx[2x]2x+ddx[2x]
2x+ddx[2x]2x+ddx[2x]
Step 4.2
Evaluate ddx[2x]ddx[2x].
Step 4.2.1
Since 22 is constant with respect to xx, the derivative of 2x2x with respect to xx is 2ddx[x]2ddx[x].
2x+2ddx[x]2x+2ddx[x]
Step 4.2.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
2x+2⋅12x+2⋅1
Step 4.2.3
Multiply 22 by 11.
2x+22x+2
2x+22x+2
2x+22x+2
Step 5
To find the arc length of a function, use the formula L=∫ba√1+(f′(x))2dxL=∫ba√1+(f'(x))2dx.
∫60√1+(2x+2)2dx∫60√1+(2x+2)2dx
Step 6
Step 6.1
Complete the square.
Step 6.1.1
Use the form ax2+bx+cax2+bx+c, to find the values of aa, bb, and cc.
a=4a=4
b=8b=8
c=5c=5
Step 6.1.2
Consider the vertex form of a parabola.
a(x+d)2+ea(x+d)2+e
Step 6.1.3
Find the value of dd using the formula d=b2ad=b2a.
Step 6.1.3.1
Substitute the values of aa and bb into the formula d=b2ad=b2a.
d=82⋅4d=82⋅4
Step 6.1.3.2
Simplify the right side.
Step 6.1.3.2.1
Cancel the common factor of 88 and 22.
Step 6.1.3.2.1.1
Factor 22 out of 88.
d=2⋅42⋅4d=2⋅42⋅4
Step 6.1.3.2.1.2
Cancel the common factors.
Step 6.1.3.2.1.2.1
Factor 22 out of 2⋅42⋅4.
d=2⋅42(4)d=2⋅42(4)
Step 6.1.3.2.1.2.2
Cancel the common factor.
d=2⋅42⋅4d=2⋅42⋅4
Step 6.1.3.2.1.2.3
Rewrite the expression.
d=44d=44
d=44d=44
d=44d=44
Step 6.1.3.2.2
Cancel the common factor of 44.
Step 6.1.3.2.2.1
Cancel the common factor.
d=44d=44
Step 6.1.3.2.2.2
Rewrite the expression.
d=1d=1
d=1d=1
d=1d=1
d=1d=1
Step 6.1.4
Find the value of ee using the formula e=c-b24ae=c−b24a.
Step 6.1.4.1
Substitute the values of cc, bb and aa into the formula e=c-b24ae=c−b24a.
e=5-824⋅4e=5−824⋅4
Step 6.1.4.2
Simplify the right side.
Step 6.1.4.2.1
Simplify each term.
Step 6.1.4.2.1.1
Raise 88 to the power of 22.
e=5-644⋅4e=5−644⋅4
Step 6.1.4.2.1.2
Multiply 44 by 44.
e=5-6416e=5−6416
Step 6.1.4.2.1.3
Divide 6464 by 1616.
e=5-1⋅4e=5−1⋅4
Step 6.1.4.2.1.4
Multiply -1−1 by 44.
e=5-4e=5−4
e=5-4e=5−4
Step 6.1.4.2.2
Subtract 44 from 55.
e=1e=1
e=1e=1
e=1e=1
Step 6.1.5
Substitute the values of aa, dd, and ee into the vertex form 4(x+1)2+14(x+1)2+1.
∫60√4(x+1)2+1dx∫60√4(x+1)2+1dx
∫60√4(x+1)2+1dx∫60√4(x+1)2+1dx
Step 6.2
Let u=x+1u=x+1. Then du=dxdu=dx. Rewrite using uu and dduu.
Step 6.2.1
Let u=x+1u=x+1. Find dudxdudx.
Step 6.2.1.1
Differentiate x+1x+1.
ddx[x+1]ddx[x+1]
Step 6.2.1.2
By the Sum Rule, the derivative of x+1x+1 with respect to xx is ddx[x]+ddx[1]ddx[x]+ddx[1].
ddx[x]+ddx[1]ddx[x]+ddx[1]
Step 6.2.1.3
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
1+ddx[1]1+ddx[1]
Step 6.2.1.4
Since 11 is constant with respect to xx, the derivative of 11 with respect to xx is 00.
1+01+0
Step 6.2.1.5
Add 11 and 00.
11
11
Step 6.2.2
Substitute the lower limit in for xx in u=x+1u=x+1.
ulower=0+1ulower=0+1
Step 6.2.3
Add 00 and 11.
ulower=1ulower=1
Step 6.2.4
Substitute the upper limit in for xx in u=x+1u=x+1.
uupper=6+1uupper=6+1
Step 6.2.5
Add 66 and 11.
uupper=7uupper=7
Step 6.2.6
The values found for ulowerulower and uupperuupper will be used to evaluate the definite integral.
ulower=1ulower=1
uupper=7uupper=7
Step 6.2.7
Rewrite the problem using uu, dudu, and the new limits of integration.
∫71√4u2+1du∫71√4u2+1du
∫71√4u2+1du∫71√4u2+1du
Step 6.3
Let u=12tan(t)u=12tan(t), where -π2≤t≤π2−π2≤t≤π2. Then du=sec2(t)2dtdu=sec2(t)2dt. Note that since -π2≤t≤π2−π2≤t≤π2, sec2(t)2sec2(t)2 is positive.
∫1.499488861.10714871√4(12tan(t))2+1sec2(t)2dt∫1.499488861.10714871√4(12tan(t))2+1sec2(t)2dt
Step 6.4
Simplify terms.
Step 6.4.1
Simplify √4(12tan(t))2+1√4(12tan(t))2+1.
Step 6.4.1.1
Simplify each term.
Step 6.4.1.1.1
Combine 1212 and tan(t)tan(t).
∫1.499488861.10714871√4(tan(t)2)2+1sec2(t)2dt∫1.499488861.10714871√4(tan(t)2)2+1sec2(t)2dt
Step 6.4.1.1.2
Apply the product rule to tan(t)2tan(t)2.
∫1.499488861.10714871√4tan2(t)22+1sec2(t)2dt∫1.499488861.10714871√4tan2(t)22+1sec2(t)2dt
Step 6.4.1.1.3
Raise 22 to the power of 22.
∫1.499488861.10714871√4tan2(t)4+1sec2(t)2dt∫1.499488861.10714871√4tan2(t)4+1sec2(t)2dt
Step 6.4.1.1.4
Cancel the common factor of 44.
Step 6.4.1.1.4.1
Cancel the common factor.
∫1.499488861.10714871√4tan2(t)4+1sec2(t)2dt∫1.499488861.10714871√4tan2(t)4+1sec2(t)2dt
Step 6.4.1.1.4.2
Rewrite the expression.
∫1.499488861.10714871√tan2(t)+1sec2(t)2dt∫1.499488861.10714871√tan2(t)+1sec2(t)2dt
∫1.499488861.10714871√tan2(t)+1sec2(t)2dt∫1.499488861.10714871√tan2(t)+1sec2(t)2dt
∫1.499488861.10714871√tan2(t)+1sec2(t)2dt∫1.499488861.10714871√tan2(t)+1sec2(t)2dt
Step 6.4.1.2
Apply pythagorean identity.
∫1.499488861.10714871√sec2(t)sec2(t)2dt∫1.499488861.10714871√sec2(t)sec2(t)2dt
Step 6.4.1.3
Pull terms out from under the radical, assuming positive real numbers.
∫1.499488861.10714871sec(t)sec2(t)2dt∫1.499488861.10714871sec(t)sec2(t)2dt
∫1.499488861.10714871sec(t)sec2(t)2dt∫1.499488861.10714871sec(t)sec2(t)2dt
Step 6.4.2
Simplify.
Step 6.4.2.1
Combine sec(t)sec(t) and sec2(t)2sec2(t)2.
∫1.499488861.10714871sec(t)sec2(t)2dt∫1.499488861.10714871sec(t)sec2(t)2dt
Step 6.4.2.2
Multiply sec(t)sec(t) by sec2(t)sec2(t) by adding the exponents.
Step 6.4.2.2.1
Multiply sec(t)sec(t) by sec2(t)sec2(t).
Step 6.4.2.2.1.1
Raise sec(t)sec(t) to the power of 11.
∫1.499488861.10714871sec1(t)sec2(t)2dt∫1.499488861.10714871sec1(t)sec2(t)2dt
Step 6.4.2.2.1.2
Use the power rule aman=am+naman=am+n to combine exponents.
∫1.499488861.10714871sec(t)1+22dt∫1.499488861.10714871sec(t)1+22dt
∫1.499488861.10714871sec(t)1+22dt∫1.499488861.10714871sec(t)1+22dt
Step 6.4.2.2.2
Add 11 and 22.
∫1.499488861.10714871sec3(t)2dt∫1.499488861.10714871sec3(t)2dt
∫1.499488861.10714871sec3(t)2dt∫1.499488861.10714871sec3(t)2dt
∫1.499488861.10714871sec3(t)2dt∫1.499488861.10714871sec3(t)2dt
∫1.499488861.10714871sec3(t)2dt∫1.499488861.10714871sec3(t)2dt
Step 6.5
Since 1212 is constant with respect to tt, move 1212 out of the integral.
12∫1.499488861.10714871sec3(t)dt12∫1.499488861.10714871sec3(t)dt
Step 6.6
Apply the reduction formula.
12(tan(t)sec(t)2]1.499488861.10714871+12∫1.499488861.10714871sec(t)dt)12(tan(t)sec(t)2]1.499488861.10714871+12∫1.499488861.10714871sec(t)dt)
Step 6.7
The integral of sec(t)sec(t) with respect to tt is ln(|sec(t)+tan(t)|)ln(|sec(t)+tan(t)|).
12(tan(t)sec(t)2]1.499488861.10714871+12ln(|sec(t)+tan(t)|)]1.499488861.10714871)12(tan(t)sec(t)2]1.499488861.10714871+12ln(|sec(t)+tan(t)|)]1.499488861.10714871)
Step 6.8
Simplify.
Step 6.8.1
Combine 1212 and ln(|sec(t)+tan(t)|)]1.499488861.10714871ln(|sec(t)+tan(t)|)]1.499488861.10714871.
12(tan(t)sec(t)2]1.499488861.10714871+ln(|sec(t)+tan(t)|)]1.499488861.107148712)
Step 6.8.2
To write tan(t)sec(t)2]1.499488861.10714871 as a fraction with a common denominator, multiply by 22.
12(tan(t)sec(t)2]1.499488861.10714871⋅22+ln(|sec(t)+tan(t)|)]1.499488861.107148712)
Step 6.8.3
Combine tan(t)sec(t)2]1.499488861.10714871 and 22.
12(tan(t)sec(t)2]1.499488861.10714871⋅22+ln(|sec(t)+tan(t)|)]1.499488861.107148712)
Step 6.8.4
Combine the numerators over the common denominator.
12⋅tan(t)sec(t)2]1.499488861.10714871⋅2+ln(|sec(t)+tan(t)|)]1.499488861.107148712
Step 6.8.5
Move 2 to the left of tan(t)sec(t)2]1.499488861.10714871.
12⋅2⋅(tan(t)sec(t)2]1.499488861.10714871)+ln(|sec(t)+tan(t)|)]1.499488861.107148712
Step 6.8.6
Multiply 12 by 2(tan(t)sec(t)2]1.499488861.10714871)+ln(|sec(t)+tan(t)|)]1.499488861.107148712.
2(tan(t)sec(t)2]1.499488861.10714871)+ln(|sec(t)+tan(t)|)]1.499488861.107148712⋅2
Step 6.8.7
Multiply 2 by 2.
2(tan(t)sec(t)2]1.499488861.10714871)+ln(|sec(t)+tan(t)|)]1.499488861.107148714
2(tan(t)sec(t)2]1.499488861.10714871)+ln(|sec(t)+tan(t)|)]1.499488861.107148714
Step 6.9
Substitute and simplify.
Step 6.9.1
Evaluate tan(t)sec(t)2 at 1.49948886 and at 1.10714871.
2((tan(1.49948886)sec(1.49948886)2)-tan(1.10714871)sec(1.10714871)2)+ln(|sec(t)+tan(t)|)]1.499488861.107148714
Step 6.9.2
Evaluate ln(|sec(t)+tan(t)|) at 1.49948886 and at 1.10714871.
2((tan(1.49948886)sec(1.49948886)2)-tan(1.10714871)sec(1.10714871)2)+(ln(|sec(1.49948886)+tan(1.49948886)|))-ln(|sec(1.10714871)+tan(1.10714871)|)4
Step 6.9.3
Remove unnecessary parentheses.
2(tan(1.49948886)sec(1.49948886)2-tan(1.10714871)sec(1.10714871)2)+ln(|sec(1.49948886)+tan(1.49948886)|)-ln(|sec(1.10714871)+tan(1.10714871)|)4
2(tan(1.49948886)sec(1.49948886)2-tan(1.10714871)sec(1.10714871)2)+ln(|sec(1.49948886)+tan(1.49948886)|)-ln(|sec(1.10714871)+tan(1.10714871)|)4
Step 6.10
Use the quotient property of logarithms, logb(x)-logb(y)=logb(xy).
2(tan(1.49948886)sec(1.49948886)2-tan(1.10714871)sec(1.10714871)2)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11
Simplify.
Step 6.11.1
Simplify the numerator.
Step 6.11.1.1
Evaluate tan(1.10714871).
2(tan(1.49948886)sec(1.49948886)2-2sec(1.10714871)2)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.1.2
Evaluate sec(1.10714871).
2(tan(1.49948886)sec(1.49948886)2-2⋅2.236067972)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
2(tan(1.49948886)sec(1.49948886)2-2⋅2.236067972)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.2
Multiply 2 by 2.23606797.
2(tan(1.49948886)sec(1.49948886)2-4.472135952)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.3
Divide 4.47213595 by 2.
2(tan(1.49948886)sec(1.49948886)2-1⋅2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.4
Multiply -1 by 2.23606797.
2(tan(1.49948886)sec(1.49948886)2-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.5
Simplify each term.
Step 6.11.5.1
Simplify the numerator.
Step 6.11.5.1.1
Evaluate tan(1.49948886).
2(14sec(1.49948886)2-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.5.1.2
Evaluate sec(1.49948886).
2(14⋅14.035668842-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
2(14⋅14.035668842-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.5.2
Multiply 14 by 14.03566884.
2(196.499363862-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.5.3
Divide 196.49936386 by 2.
2(98.24968193-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
2(98.24968193-2.23606797)+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.6
Subtract 2.23606797 from 98.24968193.
2⋅96.01361395+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.7
Multiply 2 by 96.01361395.
192.02722791+ln(|sec(1.49948886)+tan(1.49948886)||sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.8
sec(1.49948886)+tan(1.49948886) is approximately 28.03566884 which is positive so remove the absolute value
192.02722791+ln(sec(1.49948886)+tan(1.49948886)|sec(1.10714871)+tan(1.10714871)|)4
Step 6.11.9
sec(1.10714871)+tan(1.10714871) is approximately 4.23606797 which is positive so remove the absolute value
192.02722791+ln(sec(1.49948886)+tan(1.49948886)sec(1.10714871)+tan(1.10714871))4
192.02722791+ln(sec(1.49948886)+tan(1.49948886)sec(1.10714871)+tan(1.10714871))4
192.02722791+ln(sec(1.49948886)+tan(1.49948886)sec(1.10714871)+tan(1.10714871))4
Step 7
The result can be shown in multiple forms.
Exact Form:
192.02722791+ln(sec(1.49948886)+tan(1.49948886)sec(1.10714871)+tan(1.10714871))4
Decimal Form:
48.47926750…
Step 8
