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An antiderivative is essentially a function that reverses the process of differentiation. When we have a function, like the one in the example, and we want to find its antiderivative, we're looking for a new function that gives us back the original when differentiated. In our problem, the function to integrate is \( f(x) = \frac{1}{\sqrt{x}} \), which we can rewrite as \( x^{-\frac{1}{2}} \). The antiderivative of \( x^{-\frac{1}{2}} \) is \( 2x^{\frac{1}{2}} + C \), where \( C \) is a constant that we can usually ignore in definite integrals because it cancels out during evaluation.

• Antiderivatives are sometimes called indefinite integrals.
• Finding an antiderivative allows us to calculate areas or solve differential equations.

Antiderivatives play a crucial role when solving for areas under curves because they allow us to use the fundamental theorem of calculus.

The fundamental theorem of calculus bridges the gap between differentiation and integration, showing a deep connection. It states that if a function \( f \) is continuous over an interval \([a, b]\), and \( F \) is an antiderivative of \( f \) on that interval, then:\[\int_a^b f(x) \, dx = F(b) - F(a)\]This theorem allows us to evaluate definite integrals easily. In our example, to find the area under the curve from \( x = 4 \) to \( x = 9 \), we calculate:

• Find the antiderivative: \( F(x) = 2x^{\frac{1}{2}} \)
• Evaluate: \( A = F(9) - F(4) = 6 - 4 = 2 \)

The fundamental theorem makes it possible to compute definite integrals without finding the limit of a sum, simplifying the process significantly.

Finding the area under a curve is a common application of definite integrals in calculus. It's like finding the "accumulated" value of the function between two points. The area under the curve of \( f(x) = \frac{1}{\sqrt{x}} \) from \( x = 4 \) to \( x = 9 \) is solved by setting up and evaluating the definite integral \( \int_4^9 \frac{1}{\sqrt{x}} \, dx \).Some key points include:

• The area calculated using a definite integral is net area. It considers if the curve is above or below the x-axis.
• In this problem, we find a positive area because the curve is above the x-axis between the specified \( x \)-values.

Calculating this "under-the-curve" area helps understand both geometric and real-world interpretations, like finding the total distance traveled over time.

Sketching the curve of a function helps visualize the solution to an integral problem. For \( f(x) = \frac{1}{\sqrt{x}} \), this involves drawing the graph of the function over the specified interval \( x = 4 \) to \( x = 9 \). The function \( \frac{1}{\sqrt{x}} \) decreases as \( x \) increases because as the denominator gets larger, the overall fraction value decreases.Here are some tips:

• Start by drawing the axis and marking important points like \( x = 4 \) and \( x = 9 \).
• Plot calculated points to ensure the curve's accuracy, like checking \( f(x = 4) \approx 0.5 \) and \( f(x = 9) \approx 0.33 \).
• Shade the region under the curve within the relevant bounds, which represents the area calculated by the definite integral.

Sketching aids in understanding the context and confirming that the calculated area logically fits the problem setting.