Set up the definite integral.

Question

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To calculate the magnitude of the area bounded by the graph $y = -x^3 - 9x^2 + 90$ , the $x$ -axis, and the lines $x = -4$ and $x = 2$ , we need to evaluate the definite integral of the function from $x = -4$ to $x = 2$ .

From the graph, we can observe that the function $y = -x^3 - 9x^2 + 90$ is above the $x$ -axis for the entire interval $[-4, 2]$ . Therefore, the area can be calculated directly by integrating the function.

Set up the definite integral.
$A = \int_{-4}^{2} (-x^3 - 9x^2 + 90)\,dx$
Find the antiderivative of the function.
The power rule for integration states that $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ .
$\int (-x^3 - 9x^2 + 90)\,dx = -\frac{x^{3+1}}{3+1} - 9\frac{x^{2+1}}{2+1} + 90x + C$ $= -\frac{x^4}{4} - 9\frac{x^3}{3} + 90x + C$ $= -\frac{x^4}{4} - 3x^3 + 90x + C$
Evaluate the definite integral using the Fundamental Theorem of Calculus.
$A = \left[-\frac{x^4}{4} - 3x^3 + 90x\right]_{-4}^{2}$
First, evaluate the antiderivative at the upper limit $x=2$ :
$F(2) = -\frac{(2)^4}{4} - 3(2)^3 + 90(2)$ $F(2) = -\frac{16}{4} - 3(8) + 180$ $F(2) = -4 - 24 + 180$ $F(2) = 152$
Next, evaluate the antiderivative at the lower limit $x=-4$ :
$F(-4) = -\frac{(-4)^4}{4} - 3(-4)^3 + 90(-4)$ $F(-4) = -\frac{256}{4} - 3(-64) - 360$ $F(-4) = -64 + 192 - 360$ $F(-4) = 128 - 360$ $F(-4) = -232$
Subtract the value at the lower limit from the value at the upper limit.
$A = F(2) - F(-4)$ $A = 152 - (-232)$ $A = 152 + 232$ $A = 384$
The magnitude of the window area is $384 \text{ square units}$ .

Set up the definite integral.

Answer

Step-by-step solution

Set up the definite integral.

Find the antiderivative of the function.

Evaluate the definite integral using the Fundamental Theorem of Calculus.

Subtract the value at the lower limit from the value at the upper limit.

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