91Ó°ÊÓ

When dealing with functions and graphing, one important concept is understanding the area under a curve. For a function plotted on a coordinate geometry plane, calculating the area under a curve between two points on the x-axis involves integration. Specifically, the definite integral is used for this purpose.
To compute the area under the curve for the function \( y = x e^{x^2} \) from \( x=1 \) to \( x=3 \), we set up the definite integral \( \int_{1}^{3} x e^{x^2} \, dx \). This integral captures the accumulation of small pieces of area, from the starting point \( x=1 \) to the ending point \( x=3 \).

• The integral provides a net area, considering regions above or below the x-axis.
• If all areas are above the x-axis, this is simply the total area.
• If any part dips below the x-axis, it is subtracted.

Thus, the integral \( \int_{1}^{3} x e^{x^2} \, dx \) gives the exact area enclosed by the curve and the x-axis over the specified interval.

Substitution is a powerful technique for evaluating integrals, especially when dealing with complex expressions. In our exercise, the function \( y = x e^{x^2} \) poses a challenge because of its mixed terms \( x \) and \( e^{x^2} \).
To simplify this process, we use a substitution method, where we choose \( u = x^2 \), transforming the integral into a more manageable form.

• We differentiate \( u = x^2 \) to get \( du = 2x \, dx \), meaning \( x \, dx = \frac{1}{2} du \).
• Substitute into the integral to change variables: \( \int x e^{x^2} \, dx = \int e^u \cdot \frac{1}{2} \, du \).
• Change the integration limits: when \( x = 1 \), \( u = 1 \), and when \( x = 3 \), \( u = 9 \).

This conversion simplifies our integral, making it easier to handle as \( \int e^u \, du \), which integrates straightforwardly.

Exponential functions, such as \( e^x \), are prominent in calculus due to their unique properties. They grow faster than polynomial or linear functions and are crucial in modeling growth processes.
In our integral, \( e^{x^2} \) represents an exponential function with a quadratic exponent, demonstrating a steeper increase.

• The fundamental derivative and integral of \( e^x \) simplify to the function itself, \( \int e^u \, du = e^u + C \), which highlights how they change in predictable patterns.
• These functions are useful for calculating continuous growth or decay processes in natural sciences and finances.

Thus, understanding \( e^x \) is critical, as we used \( \int e^u \, du \) to evaluate part of our integral, obtaining \( \frac{1}{2} [e^u]_{1}^{9} \), which then allowed us to find the area under the curve efficiently."