Question

Sketch the following regions (if a figure is not given) and then find the area. The region bounded by \(y=1-|x|\) and \(2 y-x+1=0\)

Short answer

The area of the region bounded by the given equations is \(\frac{1}{3}\).

Step-by-step solution

Sketch the Regions and Find Intersection Points

First, let's sketch the equation \(y=1-|x|\). As per the definition of an absolute value, we split the graph into two portions with the intersection point at x=0. For \(x\geq 0\), we have \(y=1-x\), and for \(x<0\), we have \(y=1+x\). Next, let's sketch the line equation \(2y-x+1=0\), which can be rewritten as \(y=\frac{1}{2}x+\frac{1}{2}\). Now, to find the intersection points, let's solve the two equations: 1. For \(x\geq 0\), equate \(1-x=\frac{1}{2}x+\frac{1}{2}\): \begin{align*} 1 - x &= \frac{1}{2} x + \frac{1}{2} \\ \frac{3}{2} x &= \frac{1}{2} \\ x &= \frac{1}{3} \end{align*} So, when \(x=\frac{1}{3}\); \(y=1-\frac{1}{3}=\frac{2}{3}\). 2. For \(x<0\), equate \(1+x=\frac{1}{2}x+\frac{1}{2}\): \begin{align*} 1 + x &= \frac{1}{2} x + \frac{1}{2} \\ \frac{1}{2} x &= \frac{1}{2} \\ x &= -1 \end{align*} So, when \(x=-1\); \(y=1+(-1)=0\). Now, we have the intersection points A and B (-1, 0) and (\(\frac{1}{3}\), \(\frac{2}{3}\)). After sketching the graphs and the regions, we can find the area between them.

Setting up the Integral

The area between the two curves can be found using the formula: \(A=\int_a^b (f(x)-g(x))\:dx\). To find the integral limits, we will use the intersection points we found earlier. For the region on the left half (\(x\leq 0\)), we use the endpoints -1 and 0. The equation for the line is the upper bound and the equation of the curve for \(x\leq0\), \(1+x\), is the lower bound: $$A_1 = \int_{-1}^0 \left(\frac{1}{2}x+\frac{1}{2}-(1+x)\right)dx$$ For the region on the right half (\(x\geq 0\)), we use the endpoints 0 and \(\frac{1}{3}\). The equation of the curve for \(x\geq0\), \(1-x\), is the upper bound and the equation of the line is the lower bound: $$A_2 = \int_0^{\frac{1}{3}} \left((1-x)-\left(\frac{1}{2}x+\frac{1}{2}\right)\right)dx$$

Calculate the Integrals

Now, let's evaluate the integrals. For \(A_1\): \begin{align*} A_1 &= \int_{-1}^0 \left(\frac{1}{2}x +\frac{1}{2} -(1+x)\right)dx \\ &= \int_{-1}^0 (-\frac{1}{2}x -\frac{1}{2})dx \\ &= \left[\frac{1}{4}x^2 - \frac{1}{2}x\right]_{-1}^0 \\ &= \left(0 - 0\right) - \left(\frac{1}{4} - \frac{1}{2}\right) \\ &= \frac{1}{4} \end{align*} For \(A_2\): \begin{align*} A_2 &= \int_0^{\frac{1}{3}} \left(1-x-\left(\frac{1}{2}x+\frac{1}{2}\right)\right)dx \\ &= \int_0^{\frac{1}{3}} \left(\frac{1}{2}-\frac{3}{2}x\right)dx \\ &= \left[\frac{1}{2}x -\frac{3}{4}x^2\right]_0^{\frac{1}{3}} \\ &= \left(\frac{1}{6} - \frac{1}{12}\right) - (0-0) \\ &= \frac{1}{12} \end{align*}

Compute the Total Area

Finally, let's compute the total area by summing the areas of the two regions, \(A_1\) and \(A_2\): $$A = A_1 + A_2 = \frac{1}{4} + \frac{1}{12} = \frac{1}{3}$$ So, the area of the region bounded by the given equations is \(\boxed{\frac{1}{3}}\).

Key concepts

Area between curves

When we talk about the area between curves, we are looking at the space that is enclosed by different function graphs. Imagine two curves on a coordinate plane. The area we want is the region between them. To compute this area, we usually find the integral of the top function minus the bottom function over a specific interval.

• The 'top' function is the one that is higher or "above" on the graph in the given interval.
• The 'bottom' function is the one that sits below in this region.

This process helps us determine the exact size of the space confined between two different curves.

Integral calculation

Integral calculation is essential when finding areas under or between curves. Integration is like an advanced form of adding up slices to find a total. It's how we sum things continuously.

• To set up the integral, ensure you've determined which curve is on the top and which is on the bottom within the interval defined by the intersection points.
• Use the formula \( A = \int_a^b (f(x) - g(x)) \, dx \), where \( f(x) \) is the top curve and \( g(x) \) is the bottom curve.
• Replace \( a \) and \( b \) with the x-values from where your curves intersect.

This helps us compute areas more accurately.

Absolute value function

An absolute value function, like \( y = |x| \), creates a V-shaped graph. This function's main trait is that it turns all negative inputs to positive, which is why the graph has symmetry.
In absolute value, breaking the function into two separate equations helps when dealing with problems involving it:

• For \( x \geq 0 \), use the given function directly as it is.
• For \( x < 0 \), replace \( |x| \) with \( -x \), effectively "flipping" the negative x-values.

This understanding is crucial when finding intersections or graphing specific portions of such functions.

Intersection points of graphs

Intersection points are where two graphs meet on a coordinate plane. These points determine the interval over which we find the area between curves.

• To find these points, set the equations equal to each other. Solve for \( x \) to find where the curves cross.
• Once \( x \) is found, plug it back into one of the original equations to find the corresponding \( y \) value.
• For example, if the curves are \( y = 1 - |x| \) and \( y = \frac{1}{2}x + \frac{1}{2} \), solve \( 1-|x| = \frac{1}{2}x + \frac{1}{2} \).

These calculations are the first step in many area problems, letting us know the bounds for our integrals.