91Ó°ÊÓ

When working with integrals, especially double integrals, understanding antiderivatives is crucial. An antiderivative is a function whose derivative is the given function. Essentially, it's the reverse process of differentiation. In the original exercise, our goal with antiderivatives is to simplify the function inside the integral before evaluating it over a certain range.
For instance, when we have the expression \(xy\), integrating with respect to \(y\) involves finding an antiderivative for \(xy\). This results in \(\frac{1}{2}xy^2\). It's important to note the power rule here, which helps in finding antiderivatives: in general, \(x^n\) becomes \(\frac{x^{n+1}}{n+1}\). The concept of antiderivatives also plays a crucial role when integrating with respect to \(x\), like turning \(4x^3\) into \(\frac{4}{4}x^4\) and \(x^5\) into \(\frac{1}{6}x^6\).
Antiderivatives simplify expressions for further analysis and are the building block for evaluating definite integrals over specified limits.

When dealing with integrals, deciding the boundaries or limits of integration is key to solving the problem accurately. In the exercise, we have two limits - one for \(y\) (from \(x^2\) to \(2x\)) and another for \(x\) (from \(0\) to \(2\)). These limits define the region of integration on the Cartesian plane.

• The inner integral \(\int_{x^2}^{2x} x y d y\) means we are holding \(x\) constant and integrating with respect to \(y\)
• The outer integral \(\int_{0}^{2} \cdots dx\) specifies that after computing the inner integral, the resulting expression is integrated over \(x\).

After finding the antiderivative in terms of \(y\), we substitute the \(y\) limits into our expression. This is seen when we transition from \(\frac{1}{2}x y^2\) to \(\frac{1}{2}x (2x)^2-\frac{1}{2}x (x^2)^2\).
Correctly substituting these limits is essential for arriving at the correct answer.

Once the antiderivatives are found and limits of integration are substituted, the final step is evaluating the integral. This process includes computing the values at the upper and lower limits and subtracting them.
In our solved exercise, the antiderivative of the expression \(4x^3-x^5\) was evaluated from \(0\) to \(2\). This involves substituting \(2\) into our antiderivative \(\frac{4}{4}x^4-\frac{1}{6}x^6\), resulting in \(16 - \frac{64}{3}\). Then we substitute \(0\) which gives \(0\). Subtracting these results gives \(\frac{16}{3}\), which is the value of the double integral.
This subtraction cancels out terms that arise from the lower limit, ensuring that the solution reflects the accumulated value from the defined starting point to the end.