91Ó°ÊÓ

The antiderivative is essentially the reverse process of differentiation, also called the indefinite integral. When you compute the antiderivative of a function, you're looking for another function whose derivative matches the original.
In the given exercise, we're finding the antiderivative of the function \(8 \sqrt{x}\). To simplify, \(\sqrt{x}\) means \(x\) raised to the power of \(\frac{1}{2}\).
The power of \(x\) is crucial in using the power rule. Hence, \(8 \sqrt{x}\) equals \(8x^{1/2}\).
When solving for the antiderivative, you'll first transform \(8x^{1/2}\) considering the power rule, which allows us to reverse-engineer differentiation. This step is like asking, "What function differentiates to \(8 \sqrt{x}\)?"

The Power Rule is one of the simplest methods for finding antiderivatives of polynomial expressions.
When you have a variable \(x\) raised to a power \(n\), the power rule states:

• Add 1 to the power: \(n+1\)
• Divide by the new power: \(\frac{1}{n+1}\)
• Don't forget the coefficient! Multiply everything by any existing front coefficient of the original term.

In this exercise, \(n=\frac{1}{2}\), so you add 1 to get \(\frac{3}{2}\). Dividing by \(\frac{3}{2}\) and simplifying gives \(\frac{2}{3}x^{3/2}\) before multiplying by 8 gives \(\frac{16}{3}x^{3/2}\).
This function, \(\frac{16}{3}x^{3/2}\), is your antiderivative.

The definite integral calculates the net area under the curve of a function across a specific interval. This interval is defined by its limits or bounds: lower and upper bounds.
In this exercise, the bounds are from 1 to 9. Applying these bounds involves evaluating the antiderivative at these points.
To do this:

• Substitute the upper bound (9) into the antiderivative, giving you \(\frac{16}{3} \cdot 9^{3/2}\)
• Do likewise for the lower bound (1): \(\frac{16}{3} \cdot 1^{3/2}\)
• Subtract the result of the lower bound from the upper bound result.

This subtraction gives the net area or total accumulated value between \(x=1\) and \(x=9\). Finally, simplify the expression to find the solution to the definite integral, which in this case was calculated as 128.