91Ó°ÊÓ

The absolute value function, denoted \(|x|\), is a fundamental mathematical concept. It transforms negative input values into their positive counterparts. Basically, \(|x|\) returns \(x\) if \(x \geq 0\) and \(-x\) if \(x < 0\). This means that every value you put into the function comes out non-negative. Absolute value functions are critical in various algebraic and calculus applications where the magnitude of a number is more important than its sign. For instance, in the integral \(\[\begin{equation} \begin{split} \int_{-3}^{4}\big|x^3\big| d x \end{split} \end{equation}\]\), \(x^3\) can return negative results for inputs less than zero. By applying the absolute value function, these negative results are transformed into positive values, ensuring that the integrand \(\big|x^3\big|\) remains non-negative across the entire interval.

Breaking down integrals is a strategy used to simplify complex integrals. This is especially useful with absolute value functions, where the function’s behavior changes at certain points. In our example, the integral \(\[\begin{equation} \begin{split} \int_{-3}^{4}\big|x^3\big| d x \end{split} \end{equation}\]\) can be divided at \(x = 0\) since the absolute value function’s behavior changes here. This results in two simpler integrals:

• From \(-3\) to \(0\)
• From \(0\) to \(4\)

By splitting the integral at \(x = 0\), we can handle the integrals separately, simplifying the overall process. This technique not only makes calculations more manageable but also helps in understanding the behavior of the function over different intervals.

Integral evaluations involve integrating functions within defined limits. In our case, the integrals to be evaluated are on two intervals:

• \(\[\begin{equation} \begin{split} \int_{-3}^{0}-x^3 dx \end{split} \end{equation}\]\), representing the interval where \(x \textless 0\) and so \(\big|x^3\big| = -x^3\)
• \(\[\begin{equation} \begin{split} \int_{0}^{4} x^3 d x \end{split} \end{equation}\]\) representing the interval where \(x \geq 0\) and so \(\big|x^3\big| = x^3\)

Calculations are as follows:

• For \( \begin{aligned} \int_{-3}^{0} -x^3 dx = \bigg[ \frac{-x^4}{4} \bigg]_{-3}^{0} = 0 - \left( \frac{-81}{4} \right) = -\left(\frac{81}{4}\right) = -20.25 \end{aligned} \)
• For \(\begin{aligned} \int_{0}^{4} x^3 dx = \bigg[ \frac{x^4}{4} \bigg]_{0}^{4} = 64 \end{aligned} \)

Finally, add the absolute values of the results to compute the total area under the curve: \(\begin{aligned} -\bigg(-20.25\bigg) + 64 = 20.25 + 64 = 84.25 \end{aligned} \).