91Ó°ÊÓ

The substitution method is a powerful technique used in calculus to simplify integration. It involves choosing a new variable to replace a complicated expression in the integrand. In our example, we tackled the integral \( \int_{-1}^{1} \frac{x^{2} dx}{\sqrt{x^{3}+9}} \) using the substitution \( u = x^3 + 9 \). This change makes the integral easier to evaluate by simplifying its denominator.

By calculating the derivative \( \frac{du}{dx} = 3x^2 \), we can express \( dx \) in terms of \( du \), giving \( dx = \frac{du}{3x^2} \). Substitution not only alters the integrand but also requires us to adjust the limits of integration. Calculating these new limits ensures that the range of \( u \) corresponds accurately to the original range of \( x \).

This substitution simplifies the integral to \( \frac{1}{3} \int_{8}^{10} u^{-1/2} \, du \). By converting the integral into a basic form, the substitution method allows for straightforward integration.

A definite integral evaluates the area under a curve within specified limits. In this case, from \( x = -1 \) to \( x = 1 \) in the original expression. After applying the substitution \( u = x^3 + 9 \), we calculate the corresponding new limits, making them \( u = 8 \) to \( u = 10 \).

The integration range is crucial in definite integrals, as it defines the endpoints of the interval. The definite integral effectively computes a sum of infinitely small quantities (areas) under the curve within these bounds.

Once the integral is expressed in terms of \( u \) and evaluated with its antiderivative, you compute the integral's value over the specified interval. By using the fundamental theorem of calculus, you find the difference \( F(b) - F(a) \). Here it translated to \( \frac{2}{3} (\sqrt{10} - \sqrt{8}) \), which gives the exact area under the curve.

An antiderivative is the reverse process of differentiation. In integration, finding an antiderivative means determining a function whose derivative yields the integrand. For the simplified integral \( \frac{1}{3} \int u^{-1/2} \, du \), the antiderivative is essential to solving it.

The antiderivative of \( u^{-1/2} \) was determined as \( 2u^{1/2} \). You integrate the function by multiplying the antiderivative by constants outside the integral, in this case, \( \frac{1}{3} \).

Antiderivatives help in calculating definite integrals by using them to evaluate the function at the upper and lower limits of integration. By evaluating \( 2u^{1/2} \) from \( u = 8 \) to \( u = 10 \), you achieve the result \( \frac{2}{3} (\sqrt{10} - \sqrt{8}) \). Understanding the concept and application of antiderivatives is essential for mastering integration.