91Ó°ÊÓ

When tackling a complex integral, the substitution method can simplify the process. Essentially, it involves changing variables to make integration more manageable. For our given function \(\int_{-2}^{0} x^{2} e^{x^{3} / 2} d x\), we cannot integrate directly. Instead, we use substitution: set \(u = x^{3}/2\).
This alters the variable of integration from \(x\) to \(u\). The derivative of \(u\), \(du = \frac{3}{2} x^2 dx\), helps us express \(dx\) in terms of \(du\). Solving for \(dx\), we get \(dx = \frac{2}{3x^2} du\). Replacing these in the integral gives a simpler form: \(\int e^u \frac{2 du}{3x^2}\). After substitution, any expression of \(x\) should ideally cancel out. Here, replacing \(dx\) transforms our complex integral to \(\frac{2}{3} \int e^u du\). This is a core step in solving integrals using substitution, providing a pathway to evaluate more challenging integrals.

The Fundamental Theorem of Calculus bridges the gap between derivatives and integrals, giving us a tool to evaluate definite integrals. After simplifying the integral using substitution, we found \(\frac{2}{3} \int e^u du\), which is keyed for the theorem.
The theorem consists of two main parts. The first part ensures an antiderivative exists for a continuous function on a closed interval. The second part specifies that if \(F\) is the antiderivative of \(f\), then:

• The definite integral of \(f\) from \(a\) to \(b\) is \(F(b) - F(a)\).

Applying this, once we get our antiderivative \(\frac{2}{3} e^u\), the theorem allows us to evaluate it at the new limits. This effectively calculates the original definite integral, providing a connection between indefinite and definite integrals. In a nutshell, the theorem simplifies what would otherwise be tedious area calculations under curves.

In calculus, finding an antiderivative, or an indefinite integral, is crucial to solve definite integrals. With our substituted integral \(\frac{2}{3} \int e^u du\), identifying the antiderivative is straightforward.
Since the derivative of \(e^u\) is \(e^u\) itself, its antiderivative follows the same pattern. Thus, the antiderivative of \(\frac{2}{3} \int e^u du\) is \(\frac{2}{3} e^u + C\), where \(C\) is the constant of integration.
In the context of definite integrals, we often omit \(C\) because its value cancels out when computing the definite integral over a specific interval. Recognizing the antiderivative quickly transforms the integral into a problem of finding the values at specific points, thanks to the substitution method earlier. Therefore, a strong grasp of antiderivatives lays the groundwork for integrating complex functions and evaluating definite integrals.

Limits of integration offer the boundaries within which we evaluate a definite integral. In our scenario, original limits were given as \(-2\) to \(0\). Substitution transformed these \(x\) limits to \(u\) limits because of the change of variable from \(x^{3}/2\) to \(u\).

• For the lower limit: Substitute \(x = -2\) into \(u = x^{3}/2\) to get \(u_{-2} = (-2)^{3}/2 = -4\).
• For the upper limit: Substitute \(x = 0\) into \(u = x^{3}/2\) to get \(u_0 = 0\).

Once these limits are set, you evaluate the antiderivative \(\frac{2}{3} e^u\) at these points, specifically substituting \(u_0\) and \(u_{-2}\). The value \(\frac{2}{3}(e^0 - e^{-4})\) gives the area between the curve and the \(x\)-axis from \(x = -2\) to \(0\). Understanding how to adjust and use limits robustly ensures the completion of the definite integral evaluation accurately.