Question

Sketch the region whose area is represented by the definite integral. Then use a geometric formula to evaluate the integral. $$ \int_{0}^{4} \frac{x}{2} d x $$

Short answer

The value of the integral \( \int_{0}^{4} \frac{x}{2} dx \) is equivalent to the area of the sketched region, which is 4.

Step-by-step solution

Identify the Function

The given integral is \( \int_{0}^{4} \frac{x}{2} dx \). The function to be integrated, \( f(x) = \frac{x}{2} \), is a linear function. This can be interpreted as a line with a slope of 0.5, which intersects the y-axis at 0.

Sketch the Function

Draw the line for \( f(x) = \frac{x}{2} \) on a coordinate axis. Extend the line from x = 0 to x = 4. The graph will be a straight line from the origin (0,0) to the point (4,2).

Identify the Geometric Shape

The graph of the function from 0 to 4 under the x-axis forms a right-angled triangle, with its hypotenuse as the line \( f(x) = \frac{x}{2} \), base on the x-axis from 0 to 4 and height on the y-axis from 0 to 2.

Calculate the Area

For a right-angled Triangle, the formula to calculate the area is \( \frac{1}{2} * base * height \). By applying this formula, the area of the triangle becomes \( \frac{1}{2} * 4 * 2 = 4 \). Thus, the value of the integral \( \int_{0}^{4} \frac{x}{2} dx \) is 4.

Key concepts

Geometric Interpretation of Integrals

Understanding the geometric interpretation of integrals is like discovering a secret language where algebra meets geometry. When you calculate a definite integral, like \( \[ \int_{0}^{4} \frac{x}{2} dx \] \), you're actually finding the area between the graph of the function and the x-axis over a certain interval. Imagine drawing a picture where the x-axis is your floor and the function \( f(x) = \frac{x}{2} \) is a wall leaning against it. From \( x = 0 \) to \( x = 4 \), the wall gradually grows taller. \
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The area of the shape formed by the 'wall', the 'floor', and the 'vertical lines' at either end is what the integral measures. For the given function, this shape turns out to be a right-angled triangle. In other words, even though integrals might seem abstract, they have a very real and visual interpretation as the space enclosed by curves and lines on a graph.

Area Calculation Using Integrals

To find out how much 'paint' you would need to cover a certain shape on the floor, you'd use area calculation techniques. In mathematics, calculating the area under a curve using integrals follows a similar principle. With the integral \( \[ \int_{0}^{4} \frac{x}{2} dx \] \), we can employ a simple geometry formula since the graph of \( f(x) = \frac{x}{2} \) forms a right-angled triangle for this interval. \
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The formula for the area of a triangle is \( \frac{1}{2} \times base \times height \), which tells us exactly how much 'paint' (or area) we need. Plugging in the base, which is the interval from 0 to 4, and the height, which is the function's value at \( x = 4 \), we get the same answer as evaluating the integral – in this case, 4 square units. So, calculating the area with integrals is a powerful tool for visualizing and solving geometry problems in calculus.

Linear Functions and Integrals

When dealing with linear functions, such as \( f(x) = \frac{x}{2} \), integrals simplify to finding the area of basic geometric shapes. A linear function's graph is a straight line, which makes the integral's geometric interpretation pretty straightforward—either triangles or trapezoids are formed depending on where the line crosses the x-axis. \
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For our function, which is bisected by the origin, the area under the curve from \( x = 0 \) to \( x = 4 \) is just the area of a right-angled triangle. This simplicity is unique to linear functions and is an elegant example of how certain types of algebraic expressions correspond to particular geometric shapes. In terms of integrals, this means we can often bypass the complex integral calculus and use geometry instead!