91Ó°ÊÓ

When we talk about definite integrals, we are referring to the area under a curve within a specific interval. The notation you might see often looks like this: \(\text{\int_{a}^{b} f(x) dx}\). Here, \(a\) and \(b\) are the bounds, and \(f(x)\) is our function. For instance, in our exercise: \(\text{\int_{-1}^{0} (x^{3} + x^{2}) dx + \int_{0}^{1} (x^{3} + x^{2}) dx}\), the bounds are -1 to 0 and 0 to 1.
We combined these into one integral, \(\int_{-1}^{1} (x^{3} + x^{2}) dx\), covering the entire range from -1 to 1. This simplification helps us in reducing computation steps and makes solving easier.
Definite integrals are useful because they provide us with exact areas and, therefore, precise calculations. They also have broad applications, from physics to engineering.

The antiderivative is the reverse process of differentiation. It is essentially a function whose derivative gives back the original function. So if \(F(x)\) is an antiderivative of \(f(x)\), then \(F'(x) = f(x)\).
In our example, we needed to find the antiderivative of the function \(x^{3} + x^{2}\). We did this separately for each term. The antiderivative of \(x^{3}\) is \(\frac{x^4}{4}\) since \(d/dx (\frac{x^4}{4}) = x^{3}\). Similarly, for \(x^{2}\), the antiderivative is \(\frac{x^3}{3}\) because \(d/dx (\frac{x^3}{3}) = x^{2}\). Therefore, our antiderivative becomes \(\frac{x^4}{4} + \frac{x^3}{3}\).
This antiderivative allows us to then evaluate the integral over the given bounds.

One of the most important tools in calculus, the Fundamental Theorem of Calculus links antiderivatives with definite integrals. It states that if \(F(x)\) is an antiderivative of \(f(x)\) on the interval \[a, b \], then:
\(\text{\int_{a}^{b} f(x) dx = F(b) - F(a)}\).
In simpler terms, you evaluate the antiderivative at the upper limit, subtract the evaluation at the lower limit, and this gives the value of the definite integral.
Applying this to our example, we evaluated our antiderivative \(F(x) = \frac{x^4}{4} + \frac{x^3}{3}\) at the bounds 1 and -1:
\(F(1) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}\)
\(F(-1) = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12}\).
Subtracting these, we get \(F(1) - F(-1) = \frac{7}{12} - (-\frac{1}{12}) = \frac{8}{12} = \frac{2}{3}\). This gives us our final answer: \(\frac{2}{3}\).