Math / Calculus

Question(4) Find the area of the region bounded by $f(x)=x^{4 / 3}$ and $g(x)=2 x^{1 / 3}$

Studdy Solution

STEP 1

1. The functions $f(x) = x^{4/3}$ and $g(x) = 2x^{1/3}$ are well-defined for $x \geq 0$ .
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:
$x^{4/3} = 2x^{1/3}$

STEP 4

Rewrite the equation in a more manageable form:
$x^{4/3} = 2x^{1/3}$ Divide both sides by $x^{1/3}$ (assuming $x \neq 0$ ):
$x = 2$

STEP 5

Check if $x = 0$ is also a solution:
$0^{4/3} = 2 \cdot 0^{1/3} \Rightarrow 0 = 0$
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 $x = 2$ :
$\text{Area} = \int_{0}^{2} |f(x) - g(x)| \, dx$ Since $f(x) \leq g(x)$ on $[0, 2]$ :
$\text{Area} = \int_{0}^{2} (2x^{1/3} - x^{4/3}) \, dx$

STEP 7

Evaluate the integral:
$\int_{0}^{2} (2x^{1/3} - x^{4/3}) \, dx$ Break it into two separate integrals:
$\int_{0}^{2} 2x^{1/3} \, dx - \int_{0}^{2} x^{4/3} \, dx$

STEP 8

Evaluate the first integral:
$\int_{0}^{2} 2x^{1/3} \, dx$ Using the power rule for integration:
$\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$ For $2x^{1/3}$ :
$2 \int x^{1/3} \, dx = 2 \left[ \frac{x^{4/3}}{4/3} \right]_{0}^{2} = 2 \left[ \frac{3}{4} x^{4/3} \right]_{0}^{2} = \frac{3}{2} \left[ x^{4/3} \right]_{0}^{2}$

STEP 9

Evaluate at the bounds:
$\frac{3}{2} \left[ 2^{4/3} - 0^{4/3} \right] = \frac{3}{2} \cdot 2^{4/3}$

STEP 10

Evaluate the second integral:
$\int_{0}^{2} x^{4/3} \, dx$ Using the power rule for integration:
$\left[ \frac{x^{7/3}}{7/3} \right]_{0}^{2} = \left[ \frac{3}{7} x^{7/3} \right]_{0}^{2}$

STEP 11

Evaluate at the bounds:
$\frac{3}{7} \left[ 2^{7/3} - 0^{7/3} \right] = \frac{3}{7} \cdot 2^{7/3}$

STEP 12

Combine the results from the two integrals:
$\text{Area} = \frac{3}{2} \cdot 2^{4/3} - \frac{3}{7} \cdot 2^{7/3}$ Factor out the common term $3 \cdot 2^{4/3}$ :
$\text{Area} = 3 \cdot 2^{4/3} \left( \frac{1}{2} - \frac{2}{7} \right)$

STEP 13

Simplify the fraction inside the parenthesis:
$\frac{1}{2} - \frac{2}{7} = \frac{7}{14} - \frac{4}{14} = \frac{3}{14}$

STEP 14

Combine the terms to get the final area:
$\text{Area} = 3 \cdot 2^{4/3} \cdot \frac{3}{14} = \frac{9}{14} \cdot 2^{4/3}$
Therefore, the area of the region bounded by $f(x)=x^{4/3}$ and $g(x)=2x^{1/3}$ is:
$\text{Area} = \frac{9}{14} \cdot 2^{4/3}$

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