Problem 3 - Active Calculus

Question

Find the average value of \(f(x)=4 x+4\) over [4,9] average value \(=\) _________________

Step-by-Step Solution

Answer

The average value is 30.

Step 1 - Understand the Formula for the Average Value of a Function

The average value of a continuous function \(f(x)\) over the interval \([a, b]\) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \] For this exercise, we need to use this formula to find the average value of \(f(x) = 4x + 4\) over the interval \([4, 9]\).

Step 2 - Set Up the Integral

Set up the integral for the given function over the interval \([4, 9]\): \[ \frac{1}{9-4} \int_4^9 (4x + 4) \, dx \] Calculate the length of the interval which is \(9 - 4 = 5\). Hence, the expression becomes: \[ \frac{1}{5} \int_4^9 (4x + 4) \, dx \]

Step 3 - Compute the Integral

To solve the integral \int_4^9 (4x + 4) \, dx\, find the antiderivative of \(4x + 4\): \[ \frac{1}{5} \[ \frac{4x^2}{2} + 4x \]_4^9 \] This simplifies to: \[ \frac{1}{5} \[ 2x^2 + 4x \]_4^9 \]

Step 4 - Evaluate the Antiderivative at the Bounds

Substitute the upper and lower bounds into the antiderivative: \[ \frac{1}{5} \left[ (2(9)^2 + 4(9)) - (2(4)^2 + 4(4)) \right] \] Calculate the values: \[ \frac{1}{5} \left[ (2 \cdot 81 + 36) - (2 \cdot 16 + 16) \right] \] This simplifies to: \[ \frac{1}{5} \left[ 162 + 36 - 32 - 16 \right] \] Which further simplifies to: \[ \frac{1}{5} \left[ 162 + 36 - 48 \right] = \frac{1}{5} \[ 150 \] = 30 \]

Step 5 - Write the Final Answer

The average value of the function \(f(x) = 4x + 4\) over the interval \([4,9]\) is 30.

Key Concepts

integral calculus

Integral calculus is a crucial part of mathematics. It deals with the concept of integration, which combines small parts to determine a whole. Just like summing up individual numbers gives a total, integration sums up tiny slices of area under a curve.To interpret an integral, think of it as finding the accumulated value. When applied to a function, it helps find areas, volumes, central points, and many useful properties. In our exercise, we used an integral to find the average value of a function over an interval.

antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. It reverses differentiation, returning to the original function before it was differentiated. Mathematically, if you have a function \(f(x)\), an antiderivative is a function \(F(x)\) such that \(F'(x) = f(x)\).In our exercise, we found the antiderivative of \(4x + 4\), which is \(2x^2 + 4x\). Once you have the antiderivative, you can evaluate it at the upper and lower bounds of the interval to compute the definite integral.

definite integral

A definite integral calculates the net area under a function's curve over a specified interval. It also determines the accumulation of quantities, like distance over time or total area. The formula for a definite integral from \(a\) to \(b\) is \( \int_a^b f(x) \, dx \).In our problem, we calculated the definite integral of \(4x + 4\) over \([4, 9]\). First, we found its antiderivative, then evaluated it at \(x = 9\) and \(x = 4\). Subtracting the lower value from the upper value gave us the definite integral, which was crucial in finding the average value of the function.

continuous function

A continuous function means there are no breaks, jumps, or holes in its graph. You can draw it without lifting your pencil from the paper. This property is crucial for applying integral calculus because it ensures the function behaves predictably over the interval.In our exercise, \(f(x) = 4x + 4\) is a continuous function. This continuity allowed us to confidently apply the integral calculus techniques to find the average value over the interval \([4, 9]\). A continuous function ensures smooth, uninterrupted calculations.