91Ó°ÊÓ

Integral calculus is a fundamental part of calculus that deals with integrals, which are essentially the opposite of derivatives. Integrals allow us to find the area under a curve, the accumulated quantity, and are at the heart of solving many physics and engineering problems. In terms of notation, an integral is represented by the elongated 'S' symbol:

• The expression inside the integral sign, called the integrand, is the function we want to integrate.
• The variable of integration, indicated usually as 'dx' for integration with respect to x, specifies how integration is carried out.

Understanding integral calculus involves:

• Definite integrals, which calculate the exact area under a curve between two limits.
• Indefinite integrals, which represent a family of functions and include the constant of integration, C.

In the problem given, the expression to be integrated \(\int \frac{3x \, dx}{\sqrt[3]{10-x^{2}}}\)involves evaluating the area or accumulated quantity as defined by the integrand. The complication here is due to the cube root and binomial in the denominator, which requires additional techniques to integrate successfully.

The substitution method, also known as u-substitution, is a powerful technique in integral calculus often used to simplify the integration process. When faced with a complex integrand, we can substitute part of the expression with a new variable, 'u', to make the integral easier to evaluate. Here’s how substitution works:

• Identify a part of the integrand to simplify with substitution: Choose something that will simplify the expression when replaced. In this example, we substitute \( u = 10 - x^{2} \).
• Differentiate the chosen substitution: Differentiate \( u \) with respect to \( x \) to help replace \( dx \) in your integral. This gives \( du = -2x \, dx \), which can be rearranged as \( \frac{-1}{2} du = x \, dx \).
• Replace in the integral: Change all terms from the original variable to \( u \), converting your integral from a function of \( x \) to a function of \( u \). Our integral transforms into \(-\frac{3}{2} \int u^{-1/3} \, du\).

Substitution essentially rewrites the integral in a simpler form, making it possible to perform operations that may not have been straightforward before. Once the integral is solved, substitute back to the original variable to find the solution in suitable terms.

The power rule for integration is a straightforward yet integral tool in computing antiderivatives, which are the primary goal in integral calculus. The power rule dictates that for any function of the form \( u^n \), where \( n \) is not equal to \(-1\), the integral is expressed as:

• \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \)

Here, 'C' represents the constant of integration, acknowledging that multiple antiderivatives exist due to functions that differ by a constant having the same derivative. Let's see how it applies to our exercise:

• We simplified the integrand using substitution to get \(-\frac{3}{2} \int u^{-1/3} \, du \).
• Now, applying the power rule: Since \( n = -\frac{1}{3} \), we compute \(-\frac{1}{3} + 1 = \frac{2}{3} \).
• The antiderivative thus becomes \( \frac{u^{2/3}}{2/3} \), which when simplified with the constants results in \(-\frac{9}{4} u^{2/3} + C \).

By following the power rule and combining it with substitution, we've transformed a seemingly difficult integral into a more manageable expression. This illustrates the power and utility of integrating techniques in calculus.