Question Statement

Find the area of the region bounded by the graphs of $y=x$, $y=-2x$ and $x=3$.

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Background and Explanation

This problem involves finding the area between two curves using definite integration. To solve this, you need to identify where the curves intersect to determine the limits of integration, then integrate the vertical distance between the upper and lower functions over that interval.

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Solution

First, identify the bounding curves: $y=x\text{(line 1)}$ $y=-2x\text{(line 2)}$

Step 1: Find the intersection point

To locate where these lines meet, set the equations equal to each other:

$\begin{array}{rl}x& =-2x\\ 3x& =0\\ x& =0\end{array}$

The curves intersect at $x=0$ (the origin). The vertical line $x=3$ provides the right boundary, so our integration interval is $[0,3]$.

Step 2: Determine the upper and lower functions

On the interval $[0,3]$, we check which function has greater values. For any $x>0$:

• $y=x$ produces positive values
• $y=-2x$ produces negative values

Therefore, $f(x)=x$ is the upper function and $g(x)=-2x$ is the lower function.

Step 3: Set up the definite integral

The area between two curves is given by ${\int }_a^b[f(x)-g(x)]dx$. Substituting our functions:

$\begin{array}{rl}\text{Area}& ={\int }_0^3(f(x)-g(x))dx\\ & ={\int }_0^3(x-(-2x))dx\\ & ={\int }_0^3(x+2x)dx\\ & ={\int }_0^{3}3xdx\end{array}$

Step 4: Evaluate the integral

Factor out the constant and apply the power rule:

$\begin{array}{rl}\text{Area}& =3{\int }_0^{3}xdx\\ & =3{\left|\frac{x^2}{2}\right|}_0^3\\ & =\frac{3}{2}\left((3{)}^2-0\right)\\ & =\frac{3}{2}(9)\\ & =\frac{27}{2}\text{ sq. unit}\end{array}$

Thus, the area of the bounded region is $\boxed{\frac{27}{2}}$ square units (or $13.5$ square units).

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Key Formulas or Methods Used

• Area between two curves: $\text{Area}={\int }_a^b[f(x)-g(x)]dx$, where $f(x)\ge g(x)$ on $[a,b]$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ (for $n\ne -1$)
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is the antiderivative of $f$

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Summary of Steps

1. Find intersection: Set $x=-2x$ to find the curves meet at $x=0$
2. Establish limits: Region spans from $x=0$ to $x=3$
3. Identify upper/lower: On $[0,3]$, $y=x$ is above $y=-2x$
4. Set up integral: ${\int }_0^3[x-(-2x)]dx={\int }_0^{3}3xdx$
5. Evaluate: $3\cdot \frac{x^2}{2}{|}_0^3=\frac{3}{2}(9)=\frac{27}{2}$