Problem 79 - Biocalculus Calculus for the Life Sciences

Question

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

Step-by-Step Solution

Answer

\( f(4) = 29 \).

Step 1: Understanding the problem

We need to find the value of the function \( f(x) \) at \( x=4 \). We are given \( f(1)=12 \), that the derivative \( f'(x) \) is continuous, and the integral \( \int_{1}^{4} f'(x) \ dx = 17 \).

Step 2: Using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that \( F(b) - F(a) = \int_{a}^{b} f'(x) \ dx \). This relates the antiderivative (\( f(x) \)) to its derivative (\( f'(x) \)). Here we set \( a = 1 \) and \( b = 4 \).

Step 3: Applying the values to the theorem

According to the theorem, \( f(4) - f(1) = \int_{1}^{4} f'(x) \ dx \). We substitute the known values: \( f(1) = 12 \) and \( \int_{1}^{4} f'(x) \ dx = 17 \).

Step 4: Solving for \( f(4) \)

The equation \( f(4) - 12 = 17 \) needs to be solved. Add 12 to both sides to isolate \( f(4) \).

Step 5: Calculating \( f(4) \)

Add 12 to 17: \( f(4) = 17 + 12 = 29 \).

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with integrals, which can be understood as the opposite of derivatives. In simple terms, integrals help us find the accumulation of quantities. Imagine coloring in an area under a curve on a graph; that's essentially what integration does.
It allows us to calculate the total work done, total distance traveled, or the area under curves.

• The definite integral provides the aggregate value between two points. In the original exercise, this was represented as \( \int_{1}^{4} f'(x) \, dx = 17 \).
• An integral helps in finding total quantities by adding up infinitely small pieces of data, often over an interval.

One key part of integral calculus is the Fundamental Theorem of Calculus. This powerful tool links the definite integral with antiderivatives, simplifying the process of evaluating integrals considerably. Using it, we deduce that the change in a function over an interval \([a, b]\) is given by the integral of its derivative over that interval. This is the theorem we used to solve the original problem.

Antiderivative

An antiderivative, or primitive, of a function is a function whose derivative is the original function. If we differentiate the antiderivative, we return to the original function. It's essentially working backwards from taking a derivative.

• For example, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F'(x) = f(x) \).
• In the context of integrals, antiderivatives help in reversing differentiation. Thus, they play a crucial role in solving integrals.

The exercise used this idea, predicting that the integral \( \int_{a}^{b} f'(x) \, dx = F(b) - F(a) \) relates directly to changes in the antiderivative function \( f(x) \). This allows for the calculation of values like \( f(4) \) efficiently, without directly summing areas under curves.

Continuous Function

A continuous function is a function without interruptions in its graph; it can be drawn without lifting a pen. Such a property might seem trivial, but it's a necessary condition for many calculus concepts.
In calculus, for a function to be integrable or differentiable over an interval, continuity is a must.

• Continuity ensures smooth transitions; no jumps or holes exist in the interval considered.
• When a function's derivative is continuous, as in the exercise, it confirms that changes between values like \( f(1) \) and \( f(4) \) are smooth and calculable using the Fundamental Theorem of Calculus.

Encountering a continuous function guarantees that the integral calculus methods will apply safely, facilitating the precise calculation of changes in functions over specified intervals with tools like derivatives and integrals.