Problem 37 - Biocalculus Calculus for the Life Sciences

Question

Suppose that \(f(1)=2, f(4)=7, f^{\prime}(1)=5, f^{\prime}(4)=3,\) and \(f^{\prime \prime}\) is continuous. Find the value of \(\int_{1}^{4} x f^{\prime \prime}(x) d x\)

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Understand the Problem

We need to calculate the value of \(\int_{1}^{4} x f''(x) \, dx\) given some values and derivatives of the function \(f(x)\). The problem involves integration by parts.

Step 2: Set Up Integration by Parts Formula

Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). We choose \(u = x\) and \(dv = f''(x) \, dx\) so that \(du = dx\) and \(v = f'(x)\).

Step 3: Apply Integration by Parts

Using the integration by parts setup: \[\int_{1}^{4} x f''(x) \, dx = \left[ x f'(x) \right]_{1}^{4} - \int_{1}^{4} f'(x) \, dx\]

Step 4: Evaluate Terms at Bounds

Calculate \(\left[ x f'(x) \right]_{1}^{4}\):- For \(x = 4\), \(4 \cdot f'(4) = 4 \cdot 3 = 12\)- For \(x = 1\), \(1 \cdot f'(1) = 1 \cdot 5 = 5\)So, \(\left[ x f'(x) \right]_{1}^{4} = 12 - 5 = 7\).

Step 5: Find the Remaining Integral

Find \(\int_{1}^{4} f'(x) \, dx\) which is equal to \(f(4) - f(1)\). By substituting the given values \(f(4) = 7\) and \(f(1) = 2\), we get \(7 - 2 = 5\).

Step 6: Calculate the Final Result

Substitute the values obtained in Steps 4 and 5 into the formula from Step 3:\[ 7 - 5 = 2 \]Thus, the value of \(\int_{1}^{4} x f''(x) \, dx\) is 2.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique in calculus, often used when dealing with complex integrals. It is essentially a tool that helps us integrate the product of two functions. The formula we use is \[ \int u \, dv = uv - \int v \, du \]where we pick specific parts of the integrand as \(u\) and \(dv\). The choice of \(u\) and \(dv\) is crucial in simplifying the integral. In our case, the integral involves \(x\) and \(f''(x)\), so we chose \(u = x\) and \(dv = f''(x) \, dx\). From these, we find \(du = dx\) and \(v = f'(x)\).

• Choose \(u\) where differentiating simplifies the expression.
• Pick \(dv\) where integrating doesn’t complicate the process unnecessarily.

Once set up, the formula helps us break down the original integral into more manageable pieces.

Calculus Problem Solving

Solving calculus problems often involves a mix of integration techniques and differentiation rules, and sometimes, creative thinking is necessary. In our exercise, recognizing the need for integration by parts came from understanding the composition of the integral \(\int x f''(x) \, dx\).
Here are a few tips for tackling such problems:

• Analyze the problem carefully: identify the type of integral and clues indicating a suitable technique.
• State the known values and check any provided derivative information, as they might be needed later.
• Break the problem into smaller parts: Find calculations that can be completed first and then use them to simplify later steps.

This structured approach ensures you don't miss out on using valuable given information, as seen with the function values and derivatives provided here.

Integral Bounds Evaluation

Evaluating an integral, especially a definite integral, requires understanding of how bounds affect the calculation. For our problem, evaluating bounds was performed when calculating terms like \( \left[ x f'(x) \right]_{1}^{4} \)
This involves:

• Substituting the upper and lower bounds into the antiderivative, taking care to keep track of signs and coefficients.
• In our example, \(x\) values of 4 and 1 were respectively used, and our function \(f'(x)\) was calculated at these points.
• "Subtract the lower bound evaluation from the upper bound evaluation" ensures we get the accurate net result between the two points.

Such steps help us correctly compute the contributions each boundary value brings to the integral value.

Derivatives and Continuity

Understanding derivatives and their continuity is key in many calculus problems. Here, we're dealing with a second derivative \(f''(x)\), which is continuous, simplifying the complexity of our task.
Continuity ensures that we can apply fundamental calculus theorems without worrying about unexpected behavior at any point within our interval.

• The function values and their derivatives are consistent across the bounds, making our problem solvable with standard techniques.
• Such continuity allows precise evaluations during the integration by parts process and when computing expressions like \(f(4) - f(1)\).
• Recognizing that derivatives must satisfy continuity gives confidence in proceeding with integrated evaluations and even estimating values based on derivative data.

Hence, the seamless behavior of \(f''(x)\) across the interval \[ [1, 4] \] allows successful application of the integration methods and ensures correctness of solution steps.