Question

Estimate the change in $$ f(x)=\int_{0}^{x} \frac{e^{-t}}{t^{2}+0.51} d t $$ if \(x\) changes from \(0.7\) to \(0.71\).

Short answer

The change in \( f(x) \) is approximately 0.004966.

Step-by-step solution

- Define the Function and Derivative

The given function is \[ f(x) = \int_{0}^{x} \frac{e^{-t}}{t^{2}+0.51} dt \l. To estimate the change in the function, we first find the derivative of \( f(x) \), which is \[ f'(x) = \frac{e^{-x}}{x^{2}+0.51}. \] \]

- Evaluate the Derivative at Initial Point

Evaluate the derivative \( f'(x) \) at the initial point \( x = 0.7 \): \[ f'(0.7) = \frac{e^{-0.7}}{0.7^{2} + 0.51} \]. Substitute the values to get \[ f'(0.7) = \frac{e^{-0.7}}{0.49 + 0.51} = \frac{e^{-0.7}}{1}. \] Using a calculator, \[ f'(0.7) \approx 0.4966. \]

- Estimate the Change Using Linear Approximation

The change in \( f(x) \) when \( x \) changes from 0.7 to 0.71 can be approximated using a linear approximation: \[ \Delta f \approx f'(0.7) \cdot \Delta x. \] Here, \Delta x = 0.71 - 0.7 = 0.01, so \[ \Delta f \approx 0.4966 \cdot 0.01 = 0.004966. \]

Key concepts

Understanding Integration

Integration is a fundamental concept in calculus that allows us to find the accumulated quantity, such as area under a curve. Essentially, when you integrate a function, you're summing up infinitesimal parts to get a whole.
In our example, we integrated the function \(f(x) = \int_{0}^{x} \frac{e^{-t}}{t^{2}+0.51} dt\).
This integration helps us find the total value based on the function from the lower limit 0 to the upper limit \(x\). The result is a function \(f(x)\) representing the cumulative effect up to \(x\). Integration is often used in physics to find quantities like distance from velocity, or in economics to find total cost from marginal cost.

Importance of Derivatives

Derivatives measure how a function changes as its input changes. It's the 'rate of change' or 'slope' of the function. We often denote this as \(f'(x)\).
In our exercise, we needed the derivative of \(f(x)\) to estimate the change. The derivative tells us how quickly \(f(x)\) changes, which is useful for predicting future behavior.
For our function \(f(x)\), we found the derivative to be \(f'(x) = \frac{e^{-x}}{x^{2} + 0.51}\). Evaluating this at \(x = 0.7\) gave us \(f'(0.7) \approx 0.4966\), showing how fast \(f(x)\) changes around 0.7.

Linear Approximation

Linear approximation is a handy method in calculus to estimate the value of a function near a point using its derivative. Think of it as the tangent line to the curve at a certain point.
The formula for linear approximation is \( \Delta f \approx f'(x) \cdot \Delta x \)
In our example, we estimate the change in \(f(x)\) from \(x = 0.7\) to \(x = 0.71\). We already calculated \(f'(0.7) \approx 0.4966\). With \(\Delta x = 0.01\), the estimated change \( \Delta f\) is \(0.4966 \cdot 0.01 = 0.004966\).
This tells us that for a small change in \(x\), the resulting change in \(f(x)\) is approximately 0.004966.

Estimating Changes

Change estimation using calculus is very powerful. By understanding the relationship between a function and its derivative, we can predict how the function behaves for small changes in \(x\).
This is highly practical in many fields:

• In physics, predicting how position changes with time knowing the velocity function.
• In economics, forecasting cost increases given marginal cost.
• In engineering, estimating stress and strain in materials.

Our example exercise showed how a small change in \(x\) from 0.7 to 0.71 impacts \(f(x)\) using the derivative \(f'(x)\) and linear approximation. The change in \(f(x)\) was effectively estimated as 0.004966.