Problem 6 - Advanced Engineering Mathematics

Question

Evaluate the given partial integral. $$ \int_{x^{3}}^{x} e^{2 y / x} d y $$

Step-by-Step Solution

Answer

The derivative of the integral is \( e^2 - 3x^2 \cdot e^{2x^2} \).

Step 1: Identify the integral function

The given function is a definite integral of the form\[\int_{a(x)}^{b(x)} f(y, x) \, dy,\]where \( f(y, x) = e^{2y/x} \), \( a(x) = x^3 \), and \( b(x) = x \).

Step 2: Differentiate using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus for derivatives of integrals, for an integral\[\int_{a(x)}^{b(x)} f(y, x) \, dy,\]the derivative with respect to \( x \) is\[ f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x).\]

Step 3: Compute \( f(b(x), x) \cdot b'(x) \)

Substitute \( b(x) = x \) into the function: \( f(b(x), x) = e^{2x/x} = e^2 \).Then, differentiate \( b(x) \): \( b'(x) = 1 \).So, \( f(b(x), x) \cdot b'(x) = e^2 \cdot 1 = e^2 \).

Step 4: Compute \( f(a(x), x) \cdot a'(x) \)

Now, substitute \( a(x) = x^3 \) into the function: \( f(a(x), x) = e^{2x^3/x} = e^{2x^2} \).Then, differentiate \( a(x) \): \( a'(x) = 3x^2 \).So, \( f(a(x), x) \cdot a'(x) = e^{2x^2} \cdot 3x^2 \).

Step 5: Combine results to find the derivative of the integral

Putting it all together:\[\frac{d}{dx} \int_{x^3}^{x} e^{2y/x} \, dy = e^2 - 3x^2 \cdot e^{2x^2}.\]

Key Concepts

Definite Integral

The concept of a definite integral is foundational in calculus. A definite integral represents the accumulation of quantities, often interpreted as the area under a curve for a given interval. In the function \[ \int_{x^3}^{x} e^{2y/x} \, dy \]the definite integral refers to the computation of the area between the curve \( e^{2y/x} \) and the y-axis, over the interval from \( x^3 \) to \( x \).
This process calculates the net area, meaning it accounts for areas above the x-axis as positive and below as negative. To perform a definite integral correctly, we consider the boundaries—known as the limits of integration—that define where the calculation starts and ends on the curve. In our problem, these limits are not constant but rather functions of \( x \): \( a(x) = x^3 \) and \( b(x) = x \).
Ultimately, evaluating a definite integral with variable limits involves finding the antiderivative of the integration variable and then applying the Fundamental Theorem of Calculus, allowing us to find the exact area or accumulated value in a variety of contexts, from physics to statistics.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function changes as its input changes. In the context of the Fundamental Theorem of Calculus, differentiation connects to integration through the concept of reversing accumulation. When dealing with variable limits of integration as in\[ \int_{x^3}^{x} e^{2y/x} \, dy, \]differentiation helps us determine how the output of the integral changes as the input \( x \) changes; this involves a different procedure from standard derivatives because both the function and its limits depend on \( x \).
The Fundamental Theorem of Calculus states that if you differentiate an integral with variable limits, \[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(y, x) \, dy, \]you can use the chain rule to find the derivative: this means evaluating the integrand at both limits and multiplying each by the derivative of the respective limit. This principle links the derivative to the integral by showing how an integral's value responds to changes with respect to its upper and lower bounds, thus bridging the concepts of accumulation and rate of change.

Variable Limits of Integration

Variable limits of integration occur when the bounds of a definite integral are expressed as functions of a variable rather than fixed numbers. In our example integral:\[ \int_{x^3}^{x} e^{2y/x} \, dy \]the limits \( a(x) = x^3 \) and \( b(x) = x \) depend on \( x \), making this an integral with variable limits. This situation introduces additional complexity when applying the Fundamental Theorem of Calculus since both limits must be tackled during differentiation.
When evaluating such an integral, as you differentiate with respect to \( x \), each limit of integration contributes an additional term to the derivative.

• Upper limit \( b(x) \) contributes \( f(b(x), x) \cdot b'(x) \).
• Lower limit \( a(x) \) contributes \(-f(a(x), x) \, a'(x) \).

This approach accounts for how the integral's area changes as the limits change with respect to \( x \). Handling variable limits extends the range of applications for definite integrals, allowing for more dynamic modeling in physics and engineering scenarios.