Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$y=\sin (\pi x / 2) \quad$$ and $$\quad y=x$$

Short answer

The area between the curves \(y=\sin (\pi x / 2)\) and \(y=x\) over the interval [0,1] is the definite integral \(\int_{0}^{1} (x - \sin(\pi x / 2)) dx\).

Step-by-step solution

Find Points of Intersection

First, you need to find the x-coordinates at which the two functions intersect. This can be done by setting \(y=\sin (\pi x / 2)\) equal to \(y=x\) and solving for x. Solving the equation \(\sin(\pi x / 2) =x \) we get two solutions: x=0 and x=1.

Set Up the Integral for the Area

Now that we have the points of intersection, we set up the integral to compute the area. The area A between the two curves is given by the formula \(A= \int_{a}^{b} |f(x)-g(x)| dx\), where f(x) and g(x) are the two functions, and [a,b] is the interval from the smallest to largest x-values of the points of intersection. In case one function is not always larger than the other, the interval of integration [a,b] would be divided into sub intervals such that in each sub interval, one function is always larger than the other.

Perform the Integration

To calculate the area, we integrate from the smallest x-value of the intersection points (x=0) to the largest x-value (x=1). Taking \(f(x)=\sin (\pi x / 2)\) and \(g(x)=x\), the integral becomes \(\int_{0}^{1} |\sin(\pi x / 2) - x| dx\). However, since \(x \geq \sin (\pi x / 2)\) on the interval [0,1], the absolute value can be dropped, resulting to \(\int_{0}^{1} (x - \sin(\pi x / 2)) dx\). Finally, perform the integral to find the area between the curves.

Key concepts

Integration

Imagine that you are designing a beautiful patch of garden, and you've got two different paths, one curving gently and one straight as an arrow, that border the space where you want the flowers to bloom. To plan your planting, you need to know the exact area you have to work with. This is akin to finding the area between two curves in mathematics, a task commonly tackled using integration.

Integration is like a super-powered version of addition. It's used to sum up infinitely many infinitesimally small pieces to find the whole. When dealing with areas under curves, integration lets us add up these pieces along the x-axis (horizontally) to get the total area under the curve. It's a core tool for mathematicians, scientists, and engineers, and comes in especially handy in calculating areas bounded by curves, not just straight lines like the rectangles and triangles you might be more used to.

When using integration to find the area between two curves, we're essentially taking the integral of the vertical 'slices' between the curves over the interval where they 'sandwich' the space between them. In practice, you set up an integral where one curve is subtracted from the other, and this difference gives us the 'height' of each infinitesimal slice of area. By integrating this difference across the length of the interval, we can find the total area sandwiched between the curves.

Functions Intersections

To create a stunning landscape feature, your garden's paths must intersect at just the right spots—too close, and there's not enough room for flowers; too far apart, and the garden looks bare. In calculus, before we can figure out the area between curves, we need to know where exactly these 'garden paths' meet.

Intersections of functions are the points where two or more graphs cross. These points are crucial because they define the boundaries of the area we're interested in. To find these meeting points, we set the functions equal to each other and solve for the points of intersection. These solutions will set the limits for our integral since they mark where we start and stop adding up the slices of area.

For the functions at hand, we're given the sinuous \(y = \frac{\textrm{sin}( \textrm{\( \pi \)} x)}{2}\) and the straight path \(y = x\). By equating them, we get the intersection points as x-values, which in this case are x=0 and x=1. These x-values are akin to 'edge stones' in our path that mark where the curvature of the sine wave and the steady incline of the line meet to carve out the space for our garden—the enclosed area we're after.

Definite Integrals

Once the layout of the pathways is determined, it's time for the satisfying part—planting the flowers, or in our mathematical venture, calculating the area with a definite integral.

Definite integrals have limits—these are the starting and ending points of our interval, much like the opening and closing hours of the garden. If we return to our garden paths analogy, think of the definite integral as planting flowers along the path in a straight line from the start of the garden to the end within the interval.

Specifically, a definite integral computes the accumulation of quantities, like adding up soil volume slice by slice along a garden bed. When we apply this to finding the area between the curves of \(y = \frac{\textrm{sin}( \textrm{\( \pi \)} x)}{2}\) and \(y = x\), we set up our definite integral from 0 to 1—since we already determined that's where our paths cross. The result of this definite integral gives us not only the total area but also confirms that we've planted our garden precisely within the bounds of our intersecting paths. It reflects the culmination of our efforts in creating a beautiful space between.