Question

Find the abscissas of the points of intersection of the graph of the functions \(\mathrm{F}_{1}(\mathrm{x})=\int_{3}^{\mathrm{x}}(2 \mathrm{t}-5) \mathrm{dt}, \mathrm{F}_{2}(\mathrm{x})=\mathrm{x}^{2}-5 \mathrm{x}+6\)

Short answer

Answer: The functions \(F_1(x)\) and \(F_2(x)\) are the same, and they intersect at all points along their curve. There isn't a unique answer to the problem as there are infinitely many points of intersection with all x values in the domain of the functions being the abscissas of the points of intersection.

Step-by-step solution

Evaluate the Integral in \(F_1(x)\)

First, let's find the explicit function for \(F_1(x)\) by evaluating the integral: $$ F_1(x) = \int_{3}^{x} (2t - 5) dt $$ To find the antiderivative of \((2t - 5)\), we can use the power rule where we increase the power of t by 1 and divide by the new power: $$ \int (2t - 5) dt = t^2 - 5t + C $$ Now, we can apply the fundamental theorem of Calculus which states that: $$ F_1(x) = t^2 - 5t \Big| _3^x $$ Substituting the limits of integration, we get: $$ F_1(x) = (x^2 - 5x) - (3^2 - 5(3)) $$ Simplifying the expression, we have: $$ F_1(x) = x^2 - 5x + 6 $$

Set \(F_1(x)\) Equal to \(F_2(x)\) to Find the Points of Intersection

Now that we have the explicit functions for \(F_1(x) = x^2 - 5x + 6\) and \(F_2(x) = x^2 - 5x + 6\), we can set them equal to each other to find the x-values at which they intersect: $$ x^2 - 5x + 6 = x^2 - 5x + 6 $$ Notice that both sides of the equation are identical. This means that the graphs of \(F_1(x)\) and \(F_2(x)\) are the same, and they intersect at all points along their curve. Therefore, there isn't a unique answer to the question since they have infinitely many points of intersection.

Verify the Solution

Graphing the functions \(F_1(x) = x^2 - 5x + 6\) and \(F_2(x) = x^2 - 5x + 6\) confirms that they are the same function. As a result, they have infinitely many points of intersection, and thus, there are infinitely many abscissas of the points of intersection. In conclusion, since \(F_1(x)\) and \(F_2(x)\) are the same function, they intersect at all points along their curve, and there isn't a unique answer to the problem. The abscissas of the points of intersection are all x values in the domain of the functions.

Key concepts

Definite Integral

The definite integral represents the accumulated sum, or 'net area', under a curve when bounded by limits on a graph. Imagine a function described by an equation like \(y = f(x)\), which forms a curve on the Cartesian plane. A definite integral takes this curve and calculates the total area between the curve itself, the x-axis, and the vertical lines at the boundaries, called the limits of integration.

How to Compute a Definite Integral

For a function \(f(t)\), the definite integral from \(a\) to \(b\) is denoted as \(\int_a^b f(t) \, dt\). Finding the value of this integral involves determining an antiderivative, often denoted by \(F(t)\), and evaluating it at each boundary limit. This is formulaically represented as \(F(b) - F(a)\) after applying the fundamental theorem of calculus.

When we apply this to an exercise involving finding the intersection of functions, we evaluate the integral within the given boundaries, which provides us with a specific function value that we can use to compare with other functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of antiderivatives and definite integrals, signifying a deep connection between differentiation and integration. This theorem consists of two parts: the first relates the derivative to the integral, while the second allows for the evaluation of the definite integral.

First Part: Connection with Derivatives

The first part states that if function \(f\) is continuous over an interval and \(F\) is an antiderivative of \(f\) on that interval, then the derivative of \(F\) is \(f\). In other words, \(F'(x) = f(x)\).

Second Part: Evaluation of Definite Integrals

The second part of the theorem is particularly useful for calculating the area under a curve. It says that the definite integral of \(f\) from \(a\) to \(b\) is given by \(F(b) - F(a)\), where \(F\) is any antiderivative of \(f\). This enables us to compute integrals by finding antiderivatives rather than summing infinite rectangles under the curve, greatly simplifying the process.

Graph of Function

The graph of a function is a visual representation of all the possible ordered pairs \((x, f(x))\) that satisfy the function equation. When plotted on a Cartesian plane, the x-coordinates (abscissas) and y-coordinates (ordinates) form a curve or a line that embodies the function's behavior.

Understanding Intersections

When we consider the intersection of functions, such as in the textbook exercise, we're looking for points where two function graphs share the same coordinates. This is typically found by setting the functions equal to each other and solving for x, which gives the abscissas where the graphs meet.

Graphing both functions can confirm whether our algebraic solutions are correct. In the exercise provided, since our evaluation led us to identical functions, the graphs are superimposed, indicating an infinite number of intersections along the same curve.

Antiderivative

The antiderivative, also known as the indefinite integral, of a function \(f(x)\) is another function, often represented by \(F(x)\), whose derivative is the original function: \(F'(x) = f(x)\). The process of finding an antiderivative is called integration.

Finding an Antiderivative

For a function in the form \(2t - 5\), for example, we can utilize commonly known rules such as the power rule for integration to determine the antiderivative. The function \(F(x) = x^2 - 5x + C\), where \(C\) represents the constant of integration, is the antiderivative of \(2t - 5\).

Finding the antiderivative is crucial in solving definite integrals. It allows us to evaluate the integral by substituting the limits and computing the difference, as seen in the provided exercise. The understanding of antiderivatives is essential while solving such problems and in grasping the broader concept of integration.