Math / Calculus

QuestionFind the total area of the shaded regions.
The area is $\square$ . (Simplify your answer.)

Studdy Solution

STEP 1

1. The curves given are $y = 4x$ and $y = x^4$ .
2. The shaded region is between these curves from $x = -3$ to $x = 3$ .
3. The area between the curves is calculated by integrating the difference of the functions over the given interval.

STEP 2

1. Determine the points of intersection of the curves.
2. Set up the integral for the area between the curves.
3. Evaluate the integral to find the total area of the shaded regions.

STEP 3

Find the points of intersection by setting the equations equal:
$4x = x^4$
Rearrange the equation:
$x^4 - 4x = 0$
Factor the equation:
$x(x^3 - 4) = 0$
Solve for $x$ :
$x = 0 \quad \text{or} \quad x^3 = 4$
The real solutions are:
$x = 0 \quad \text{and} \quad x = \sqrt[3]{4}$

STEP 4

Set up the integral for the area between the curves from $x = -3$ to $x = 3$ :
The area $A$ between the curves is given by:
$A = \int_{-3}^{3} (4x - x^4) \, dx$

STEP 5

Evaluate the integral:
$A = \int_{-3}^{3} (4x - x^4) \, dx$
Calculate the antiderivative:
$\int (4x - x^4) \, dx = 2x^2 - \frac{x^5}{5} + C$
Evaluate from $-3$ to $3$ :
$A = \left[ 2x^2 - \frac{x^5}{5} \right]_{-3}^{3}$
Calculate the definite integral:
$A = \left( 2(3)^2 - \frac{(3)^5}{5} \right) - \left( 2(-3)^2 - \frac{(-3)^5}{5} \right)$
$A = \left( 18 - \frac{243}{5} \right) - \left( 18 + \frac{243}{5} \right)$
$A = \left( 18 - 48.6 \right) - \left( 18 + 48.6 \right)$
$A = -30.6 - 66.6$
$A = -97.2$
Since area cannot be negative, take the absolute value:
$A = 97.2$
The total area of the shaded regions is:
$\boxed{97.2}$

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