PROBLEM
(4) Find the area of the region bounded by f(x)=x4/3 and g(x)=2x1/3
STEP 1
1. The functions f(x)=x4/3 and g(x)=2x1/3 are well-defined for x≥0.
2. The area between two curves f(x) and g(x) over an interval [a,b] is given by the integral of the absolute difference of the functions over that interval.
3. The points of intersection between f(x) and g(x) need to be determined to establish the limits of integration.
STEP 2
1. Find the points of intersection of f(x) and g(x).
2. Set up the integral to find the area between the curves.
3. Evaluate the integral to find the area.
STEP 3
Set the functions equal to each other to find the points of intersection:
x4/3=2x1/3
STEP 4
Rewrite the equation in a more manageable form:
x4/3=2x1/3 Divide both sides by x1/3 (assuming x=0):
x=2
STEP 5
Check if x=0 is also a solution:
04/3=2⋅01/3⇒0=0 So, the points of intersection are x=0 and x=2.
STEP 6
Set up the integral to find the area between the curves from x=0 to x=2:
Area=∫02∣f(x)−g(x)∣dx Since f(x)≤g(x) on [0,2]:
Area=∫02(2x1/3−x4/3)dx
STEP 7
Evaluate the integral:
∫02(2x1/3−x4/3)dx Break it into two separate integrals:
∫022x1/3dx−∫02x4/3dx
STEP 8
Evaluate the first integral:
∫022x1/3dx Using the power rule for integration:
∫xndx=n+1xn+1+C For 2x1/3:
2∫x1/3dx=2[4/3x4/3]02=2[43x4/3]02=23[x4/3]02
STEP 9
Evaluate at the bounds:
23[24/3−04/3]=23⋅24/3
STEP 10
Evaluate the second integral:
∫02x4/3dx Using the power rule for integration:
[7/3x7/3]02=[73x7/3]02
STEP 11
Evaluate at the bounds:
73[27/3−07/3]=73⋅27/3
STEP 12
Combine the results from the two integrals:
Area=23⋅24/3−73⋅27/3 Factor out the common term 3⋅24/3:
Area=3⋅24/3(21−72)
STEP 13
Simplify the fraction inside the parenthesis:
21−72=147−144=143
SOLUTION
Combine the terms to get the final area:
Area=3⋅24/3⋅143=149⋅24/3 Therefore, the area of the region bounded by f(x)=x4/3 and g(x)=2x1/3 is:
Area=149⋅24/3
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