Question

If \(\mathrm{f}(1)=12, \mathrm{f}^{\prime}\) is continuous, and \(\int_{1}^{4} \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=17\), what is the value of \(\mathrm{f}(4)\) ?

Short answer

Answer: The value of f(4) is 29.

Step-by-step solution

Use the integral and fundamental theorem of calculus

According to the fundamental theorem of calculus, the integral of the derivative of a function on an interval gives us the value of the original function in that interval. Therefore, we have: \(\int_{1}^{4} f'(x) dx = f(4) - f(1)\)

Use the given value of f(1)

We are given that f(1) = 12. Now, we can substitute this value back into the equation above: \(\int_{1}^{4} f'(x) dx = f(4) - 12\)

Use the given value of the integral

The given exercise also provides the value for the integral of f'(x) from 1 to 4, which is 17. Substituting this value into the equation, we get: \(17 = f(4) - 12\)

Solve for f(4)

Finally, we just need to solve the equation above for f(4): \(f(4) = 17 + 12\) \(f(4) = 29\) So the value of f(4) is 29.

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size or value, such as areas under curves, from known rates of change or slopes, which are represented by derivatives. It is fundamentally about adding up an infinite number of infinitesimally small quantities, a process known as integration.

To visualize this, imagine a graph with a curve representing a function. The area under this curve between two points can be thought of as a sum of very thin rectangles underneath. Integral calculus tells us how to calculate this area accurately, even though there are infinitely many rectangles to consider. In the context of the original exercise, integral calculus helps us relate the derivative of a function to the original function through the area under its derivative's curve.

Derivative of a Function

The derivative of a function represents the rate at which the function's value is changing at any given point. In simpler terms, it's like looking at a car's speed at a specific moment during a road trip. Just as speed can tell us much about the car's journey, derivatives provide insights into the behavior of functions.

If we were to graph a function, the derivative at any point is the slope of the tangent line to the curve at that point. This is essential for understanding how functions grow or decrease, and has broad applications in fields ranging from physics to economics. In our exercise, the continuous derivative, or the slope of f(x), is integrated to uncover values of the original function f(x).

Definite Integrals

Definite integrals are like precise measurements; they give us the exact value of the area under a function's curve between two points on the x-axis. To calculate a definite integral, you need to know the function you're integrating as well as the starting and ending points, which are called the limits of integration.

For example, the expression \(\int_{a}^{b} f(x) dx\) represents the integral of f(x) with respect to x, from the start point a to the end point b. This produces a number that gives the net area between the function and the x-axis, considering both the 'heights' and 'depths' in cases where the function dips below the axis. In our solution, the definite integral from 1 to 4 results in the net change in the function f(x) from f(1) to f(4), which is a key to unlocking the value of f(4).

Solving Integrals

Solving integrals is a bit like solving a puzzle. You take a function and figure out what shape its graph makes, then calculate the area of that shape. This process can involve several techniques, such as substitution, integration by parts, and even numerical methods when an integral can't be solved analytically.

When we solve definite integrals, like in the given exercise, we are essentially finding the sum of all the change happening in the function between two points. It encapsulates the total cumulative effect the function has as it moves from the starting to the ending value. The solution of the exercise requires applying the Fundamental Theorem of Calculus, which bridges derivatives and integrals, to solve for the value of the function at a particular point. This theorem significantly simplifies the process of solving integrals.