Problem 10 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x}\) \(y=\int_{1}^{x} t e^{-t^{2}} d t\)

Step-by-Step Solution

Answer

Using the Fundamental Theorem of Calculus, Part 1, we find \( \frac{d y}{d x} = x e^{-x^2} \).

Step 1: Understand the Problem

We need to find the derivative of the integral function given with respect to the upper limit of integration. The function provided is \( y = \int_{1}^{x} t e^{-t^2} dt \). Here, \( x \) is the variable for which we are differentiating.

Step 2: Apply the Fundamental Theorem of Calculus, Part 1

According to the Fundamental Theorem of Calculus, Part 1, if \( F(x) \) is defined as an integral function \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative of \( F(x) \) with respect to \( x \) is \( f(x) \).

Key Concepts

Understand the Concept of Derivative

The concept of a derivative is fundamental in calculus. It measures how a function changes as its input changes. More formally, the derivative represents the rate of change of the function with respect to a variable. In simpler terms, it tells us the slope of the function at any given point.

For example, if you have a function that describes the position of an object over time, the derivative of that function will give you the velocity of that object at any time. It is the "instantaneous" rate of change.

To find the derivative of a function, we typically use various rules of differentiation, like the power rule, product rule, or chain rule. In the problem given, we find the derivative of an integral which involves applying specific rules from calculus, such as the Fundamental Theorem of Calculus.

Integral Functions and Their Connections

Integral functions are the opposite of derivative functions in calculus. They represent the cumulative sum, or accumulation, of a function over a specific interval.

For instance, in an integral \[ y = \int_{a}^{b} f(t) \, dt \], the function \( f(t) \) is accumulated from \( a \) to \( b \), which geometrically corresponds to the area under the curve \( f(t) \) from \( a \) to \( b \).

In our original exercise, the function \( y = \int_{1}^{x} t e^{-t^2} \, dt \) accumulates the function \( t e^{-t^2} \) as \( t \) goes from 1 to \( x \). Integral functions, therefore, help us to capture the behavior of a quantity as it changes over a specific domain. The application of integral functions is vast, ranging from physics to economics, where they enable the calculation of things like total distance traveled or total revenue generated over time.

Differentiation via the Fundamental Theorem of Calculus

Differentiation is another core component of calculus. It is the process through which we find a derivative. While the general differentiation involves rules and manipulation, the key insight to the given problem is the Fundamental Theorem of Calculus (FTC).

The FTC connects integration and differentiation in a profound way: it states that if you have an integral function defined as \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is simply \( f(x) \). This tells us that differentiation can directly "undo" integration in this context, providing the function inside the integral as the result of the derivative.

In our original exercise, this means that the derivative of \( y = \int_{1}^{x} t e^{-t^2} \, dt \) with respect to \( x \) simplifies directly to the expression \( t e^{-t^2} \) evaluated at \( t = x \), which is \( x e^{-x^2} \). This theorem thus builds an intuitive bridge between the two operations and showcases the elegant symmetry of calculus.