U-Substitution with Definite Integrals

This paragraph introduces the concept of handling definite integrals using U-substitution. The example provided involves integrating from 1 to 2 of the function \( x \cdot (x^2 + 1)^3 \). The process begins by setting \( U = x^2 + 1 \) and solving for \( \frac{du}{dx} \), which leads to \( \frac{du}{dx} = 2x \). The integral is then transformed into \( \int_2^5 U^3 \cdot \frac{1}{2} du \) by substituting the bounds accordingly. The anti-derivative is found to be \( \frac{1}{4}U^4 \), and the final evaluation results in \( \frac{1}{4} \cdot 5^4 - \frac{1}{4} \cdot 2^4 \). Another example is discussed, where the integral involves a trigonometric function, and the process is similar, with \( U = 3x \) and \( \frac{du}{dx} = 3 \), leading to the integral \( \int_{\pi/4}^{\pi/3} \sin(U) \cdot \frac{1}{3} du \). The anti-derivative here is \( -\frac{1}{3} \cos(U) \), and the evaluation is performed from \( \pi/4 \) to \( \pi/3 \).

The second paragraph delves into the application of Riemann sums to approximate the area under a curve. The example involves a function \( x \cdot (x^2 + 1)^{1/2} \), which is integrated from 0 to 2. The process starts by setting \( U = x^2 + 1 \) and solving for \( \frac{du}{dx} \), resulting in \( \frac{du}{dx} = 2x \). The integral is then rewritten as \( \int_1^5 U^{1/2} \cdot \frac{2}{3} du \) after substituting the bounds. The anti-derivative is \( \frac{2}{3}U^{3/2} \), and the evaluation yields \( \frac{2}{3} \cdot 5^{3/2} - \frac{2}{3} \cdot 1^{3/2} \). Additionally, the paragraph discusses the calculation of left and right Riemann sums using a table of function values and intervals, illustrating how to approximate the area under the curve by summing the products of function values and interval widths.