find the area bounded by the graphs of the indicated equations over the given interval.
y = -5x; y = 0; -2 ≤ x ≤ 3
the area is square units.
(type an integer or decimal rounded to three decimal places as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Determine where $y = - 5x$ is above or below $y = 0$
Set $-5x=0$, we get $x = 0$.
For $x\in[-2,0]$, $y=-5x\geqslant0$ (since when $x\in[-2,0]$, $-5x$ is non - negative).
For $x\in[0,3]$, $y=-5x\leqslant0$.

## Step2: Calculate the area using definite integrals
The area $A=\int_{-2}^{0}(- 5x)dx+\int_{0}^{3}-(-5x)dx$
First integral: $\int_{-2}^{0}(-5x)dx=-5\times\frac{x^{2}}{2}\big|_{-2}^0$
$$
\begin{align*}
-5\times\frac{x^{2}}{2}\big|_{-2}^0&=(-\frac{5}{2}(0)^{2})-(-\frac{5}{2}(-2)^{2})\
&=0 - (-\frac{5}{2}\times4)\
&=10
\end{align*}
$$
Second integral: $\int_{0}^{3}5xdx = 5\times\frac{x^{2}}{2}\big|_{0}^3$
$$
\begin{align*}
5\times\frac{x^{2}}{2}\big|_{0}^3&=\frac{5}{2}(3)^{2}-\frac{5}{2}(0)^{2}\
&=\frac{45}{2}-0\
& = 22.5
\end{align*}
$$

## Step3: Sum the two areas
$A = 10+22.5$

# Answer:
$32.5$

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QUESTION IMAGE

Question

Explanation:

Step1: Determine where \(y = - 5x\) is above or below \(y = 0\)

Step2: Calculate the area using definite integrals

Step3: Sum the two areas

Answer: