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ƒ()
ƒ'()

n =
Given ƒ(x) = 2x3 - 4x2 - 22x + 24dx
Determine the integral ∫ƒ(x)
Go through and integrate each term

Integrate term 1

ƒ(x) = 2x3

Use the power rule

∫ƒ(x) of the expression axn
ax(n + 1)
n + 1

= 2, n = 3
and x is the variable we integrate
∫ƒ(x)  =  2x(3 + 1)
  3 + 1

∫ƒ(x)  =  2x4
  4

Simplify our fraction.
Divide top and bottom by 2
∫ƒ(x)  =  x4
  2

Integrate term 2

ƒ(x) = -4x2

Use the power rule

∫ƒ(x) of the expression axn
ax(n + 1)
n + 1

= -4, n = 2
and x is the variable we integrate
∫ƒ(x)  =  -4x(2 + 1)
  2 + 1

∫ƒ(x)  =  -4x3
  3

Integrate term 3

ƒ(x) = -22x

Use the power rule

∫ƒ(x) of the expression axn
ax(n + 1)
n + 1

= -22, n = 1
and x is the variable we integrate
∫ƒ(x)  =  -22x(1 + 1)
  1 + 1

∫ƒ(x)  =  -22x2
  2

Simplify our fraction.
Divide top and bottom by 2
∫ƒ(x) = -11x2

Integrate term 4

ƒ(x) = 24

Use the power rule

∫ƒ(x) of the expression axn
ax(n + 1)
n + 1

= 24, n = 0
and x is the variable we integrate
∫ƒ(x)  =  24x(0 + 1)
  0 + 1

∫ƒ(x) = 24x

Collecting all of our integrated terms we get:

∫ƒ(x) = x4/2 - 4x3/3 - 11x2 + 24x

Evaluate ∫ƒ(x) on the interval [0,1]

The value of the integral over an interval is ∫ƒ(1) - ∫ƒ(0)

Evaluate ∫ƒ(1)

∫ƒ(1) = (1)4/2 - 4(1)3/3 - 11(1)2 + 24(1)
∫ƒ(1) = (1)/2 - 4(1)/3 - 11(1) + 24(1)
∫ƒ(1) = 0.5 - 1.3333333333333 - 11 + 24
∫ƒ(1) = 12.166666666667

Evaluate ∫ƒ(0)

∫ƒ(0) = (0)4/2 - 4(0)3/3 - 11(0)2 + 24(0)
∫ƒ(0) = (0)/2 - 4(0)/3 - 11(0) + 24(0)
∫ƒ(0) = 0 - 0 - 0 - 0
∫ƒ(0) = 0

Determine our answer

∫ƒ(x) on the interval [0,1] = ∫ƒ(1) - ∫ƒ(0)
∫ƒ(x) on the interval [0,1] = 12.166666666667 - 0

Answer
Success!
∫ƒ(x) on the interval [0,1] = 12.166666666667

↓Steps Explained:↓

Final Answer

∫ƒ(x) on the interval [0,1] = 12.166666666667
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