Problem 25 - Essential Calculus

Question

In Exercises 21-30, sketch the region whose area is given by the definite integral. Then use a geometric formula to evaluate the integral \((a>0, r>0)\) $$ \int_{0}^{2}(2 x+5) d x $$

Step-by-Step Solution

Answer

The area under the curve is 14 square units.

Step 1: Identify the geometric shape

Since the function \(2x+5\) is a straight line, it will form a trapezoidal shape between the limits of 0 and 2 when plotted.

Step 2: Calculate the area of the trapezoid

The area of a trapezoid is represented by the formula: \(Area = 0.5 * (base1 + base2) * height\). Using this formula, the bases are the value of the function at the limits 0 and 2, and height would be the difference between the limits. Calculate the function values as: \(base1 = f(0) = 2*0+5 = 5\), \(base2 = f(2) = 2*2+5 = 9\) and the height is \(height = 2-0 = 2\). Calculate the area using these values.

Step 3: Final Calculation

Substitute the calculated values into the trapezoidal area formula given in Step 2: \(Area = 0.5 * (base1 + base2) * height = 0.5 * (5+9) * 2\). Solve for the Area.

Key Concepts

Trapezoidal Area

When we talk about the trapezoidal area in relation to definite integrals, we're looking at an incredible way of interpreting integration through simple geometry. Let's break this down a little further. Imagine you have a function defining a straight line, represented as \(2x+5\) in this case.

When you plot this line between the limits 0 and 2, you form a shape under the line bound by the x-axis, which is a trapezoid. The unique aspect of a trapezoid is it has two parallel sides, known as bases. In our problem, the lengths of these bases are determined by the function values at the limits \(x = 0\) and \(x = 2\). We go on to calculate the bases as 5 and 9.

• Base 1 is the function value at \(x = 0\), which is 5.
• Base 2 is the function value at \(x = 2\), which is 9.

The distance between these two points, which are at \(x = 0\) and \(x = 2\), gives us our height of 2. The area of the trapezoid thus becomes:\[\text{Area} = 0.5 \times (\text{Base 1} + \text{Base 2}) \times \text{Height} = 0.5 \times (5 + 9) \times 2 = 14\] This is how the trapezoidal area helps in evaluating the definite integral geometrically.

Function Graph

The graph of a function is like the picture of its behavior; it tells us how the function values change with different inputs. For the function \(f(x) = 2x + 5\), plugging in various values of \(x\) will give us corresponding \(y\) values. Here's how it works:

The graph of \(2x+5\) is a straight line, which shows a steady increase. This is because of the coefficient of \(x\), which is 2 in this formula, indicating the slope of the line. A positive slope means the function is increasing.

• At \(x = 0\), the value is 5, marked as one point on the graph.
• At \(x = 2\), the value rises to 9.

The function graph crosses the y-axis at the point \(y = 5\), which is known as the y-intercept. The straight line between these points represents how the function changes smoothly and linearly between them, creating our trapezoid when you mark from 0 to 2 on the x-axis.

Geometric Interpretation

Geometry provides a visual approach to understanding problems, making it a handy tool in calculus. The geometric interpretation of a definite integral involves viewing the integral as an area under a curve. Instead of solely relying on calculations, you can actually see what the integral represents, which is the accumulated sum represented by this area.

In our case, by translating the integral \(\int_{0}^{2}(2x + 5)dx\) into the area of a trapezoid, we not only solve it numerically but also visually understand the problem. A definite integral sums up infinite small vertical slices under a curve, and geometry helps combine these slices into recognizable shapes like triangles or trapezoids.

• This approach demystifies the concept of integration, making it easier to grasp.
• It also simplifies calculations by applying familiar geometric formulas.

Practicing with geometric interpretations enhances intuition and ensures you get the concepts more deeply, alongside honing your skills in solving integrals.