91Ó°ÊÓ

Definite integrals help us find the area under a curve within certain limits. Think of the definite integral as a tool that calculates the total accumulation of quantities.
In our exercise, the function we deal with is given by \(y = 2x - x^2 \). The area under this curve, from \(x = 1\) to \(x = 2\), is what we want to discover.
By using definite integrals, we establish the boundaries, or limits, indicated by the values of \(x\). This means instead of getting a general formula with a constant (\(C\)), the computation includes these limits to provide a specific numerical answer.
Remember that the definite integral gives us a significant value by calculating \(F(b) - F(a)\), where \(a\) and \(b\) are the limits. This value corresponds to the exact area under the curve between these two points.

Integrating polynomial functions can be quite straightforward. Polynomial functions are those with terms like \(ax^n\). Here, we have a simple polynomial function: \(2x - x^2\).
To integrate a polynomial function, apply the power rule: increase the exponent by one and divide by the new exponent.

• The integral of \(2x\) becomes \(x^2\).
• The integral of \(-x^2\) becomes \(-\frac{x^3}{3}\).

Upon integrating \(y = 2x - x^2\), we obtain \(F(x) = x^2 - \frac{x^3}{3} + C\).
The constant \(C\) typically vanishes when evaluating a definite integral, as it cancels out when subtracting \(F(a)\) from \(F(b)\). This is why in definite integration, \(C\) is not needed for the final result.

Limits of integration are crucial in narrowing down the area you want from a function. It is like setting the boundaries for what part of the curve to consider.
When you perform a definite integral, you need these limits to pinpoint the starting and ending x-values. For our example, the limits are \(x = 1\) and \(x = 2\).
Applying these limits means you need to evaluate the antiderivative, or integrated function, at these points. Calculate \(F(x)\) at these limits and subtract the results to find the definite integral.

• Evaluate the function at the upper limit: \(F(2) = 2^2 - \frac{2^3}{3}\).
• Evaluate the function at the lower limit: \(F(1) = 1^2 - \frac{1^3}{3}\).
• Subtract these results: \(F(2) - F(1) = \frac{4}{3} - \frac{1}{3} = \frac{1}{3}\).

This operation provides the final area under the curve from \(x=1\) to \(x=2\).